NCERT Solutions · Class 11 Chemistry Equilibrium

Compare the reaction quotient \(\displaystyle Q_c\) to \(\displaystyle K_c\) — that comparison, not a memorized rule, tells you which way an unequilibrated mixture will move.The reaction is\[N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]with \(\displaystyle K_c = 0.061\) at $\displaystyle 500$ K.At the instant described, the mixture holds \(\displaystyle [N_2] = 3.0\ \text{mol L}^{-1}\), \(\displaystyle [H_2] = 2.0\ \text{mol L}^{-1}\), \(\displaystyle [NH_3] = 0.5\ \text{mol L}^{-1}\). These are not known to be equilibrium concentrations — they are just a snapshot — so the correct quantity to build from them is the reaction quotient \(\displaystyle Q_c\), which has the same algebraic form as \(\displaystyle K_c\) but is evaluated at any instant, not only at equilibrium:\[Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]Substitute the given concentrations:\[Q_c = \frac{(0.5)^2}{(3.0)(2.0)^3} \]Work the denominator first: \(\displaystyle (2.0)^3 = 8.0\), so \(\displaystyle (3.0)(8.0) = 24\).Work the numerator: \(\displaystyle (0.5)^2 = 0.25\).\[Q_c = \frac{0.25}{24} = 0.0104\ \text{mol}^{-2}\text{L}^{2} \]Rounded to two significant figures (matching the data), \(\displaystyle Q_c \approx 0.010\).The step people skip: compare \(\displaystyle Q_c\) with \(\displaystyle K_c\), not just compute \(\displaystyle Q_c\) and stop.\[Q_c (0.010) < K_c (0.061) \]Since \(\displaystyle \mathrm{Q_c < K_c}\), the numerator (\(\displaystyle [NH_3]^2\)) is too small, and the denominator (\(\displaystyle [N_2][H_2]^3\)) is too large, relative to the equilibrium ratio. The system is not at equilibrium, and it must shift in the direction that raises \(\displaystyle [NH_3]\) while lowering \(\displaystyle [N_2]\) and \(\displaystyle [H_2]\) that is, the forward direction (more \(\displaystyle N_2\) and \(\displaystyle H_2\) convert to \(\displaystyle NH_3\)) — until \(\displaystyle Q_c\) climbs up to meet \(\displaystyle K_c = 0.061\).Answer: The reaction is not at equilibrium, since \(\displaystyle Q_c \approx 0.010\ \text{mol}^{-2}\text{L}^{2} \neq K_c = 0.061\). Because \(\displaystyle Q_c < K_c\), the reaction proceeds in the forward direction (toward more \(\displaystyle NH_3\)) to reach equilibrium.

When \(\displaystyle K_c\) is written as one perfect square over another, you can take a square root instead of solving a full quadratic. Here the missing symbol between reactants and products in the printed reaction is the equilibrium arrow — the reaction is reversible:\[2\text{BrCl}(g) \rightleftharpoons \text{Br}_2(g) + \text{Cl}_2(g) \]Set up an ICE table (initial, change, equilibrium) in concentration, not moles. Let \(\displaystyle x\) = the concentration of \(\displaystyle \text{Br}_2\) (in mol L\(\displaystyle ^{-1}\)) formed by the time equilibrium is reached. Because $\displaystyle 2$ mol of BrCl are consumed for every $\displaystyle 1$ mol of \(\displaystyle \text{Br}_2\) produced (read straight off the balanced equation's coefficients), BrCl falls by \(\displaystyle 2x\), while \(\displaystyle \text{Br}_2\) and \(\displaystyle \text{Cl}_2\) each rise by \(\displaystyle x\):Initial (mol L\(\displaystyle ^{-1}\)): \(\displaystyle [\text{BrCl}] = 3.3\times10^{-3}\), \(\displaystyle [\text{Br}_2] = 0\), \(\displaystyle [\text{Cl}_2] = 0\)Change (mol L\(\displaystyle ^{-1}\)): \(\displaystyle [\text{BrCl}] = -2x\), \(\displaystyle [\text{Br}_2] = +x\), \(\displaystyle [\text{Cl}_2] = +x\)Equilibrium (mol L\(\displaystyle ^{-1}\)): \(\displaystyle [\text{BrCl}] = 3.3\times10^{-3}-2x\), \(\displaystyle [\text{Br}_2] = x\), \(\displaystyle [\text{Cl}_2] = x\)\(\displaystyle K_c\) is products over reactants, each raised to its stoichiometric coefficient — so BrCl's concentration is squared because its coefficient is $\displaystyle 2$:\[K_c=\frac{[\text{Br}_2][\text{Cl}_2]}{[\text{BrCl}]^2}=\frac{x\cdot x}{\left(3.3\times10^{-3}-2x\right)^2}=\frac{x^2}{\left(3.3\times10^{-3}-2x\right)^2}=32 \]The right-hand side is a ratio of two squares, so both sides are perfect squares of the same quantities — take the positive square root of both sides instead of expanding a quadratic:\[\frac{x}{3.3\times10^{-3}-2x}=\sqrt{32}=5.657 \]Aside: this square-root shortcut only works because the numerator and denominator of \(\displaystyle K_c\) are each a single term squared. If the stoichiometry gave, say, a $\displaystyle 1$:$\displaystyle 3$ mix of products, you would be stuck with a genuine quadratic in \(\displaystyle x\) — always check the powers before reaching for a square root.Solve for \(\displaystyle x\) by clearing the fraction:\[x=5.657\left(3.3\times10^{-3}-2x\right) \] \[x=1.867\times10^{-2}\times10^{-1}-11.314x \](that is, \(\displaystyle 5.657\times3.3\times10^{-3}=1.867\times10^{-2}\) is wrong in magnitude if left as is — carrying the arithmetic through directly:)\[x = \left(5.657\times3.3\times10^{-3}\right)-11.314x = 1.867\times10^{-2}\ \text{mol L}^{-1}\times10^{-... } \]Doing the multiplication carefully: \(\displaystyle 5.657\times3.3\times10^{-3}=1.867\times10^{-2}\times10^{-1}\). To avoid this slip, keep the powers of ten attached to the number at every step:\[5.657\times\left(3.3\times10^{-3}\right)=1.867\times10^{-2}\ \text{is incorrect — the correct product is } 1.867\times10^{-2}\times10^{-1} \]Redone cleanly: \(\displaystyle 5.657 \times 3.3 = 18.67\), and \(\displaystyle 18.67\times10^{-3}=1.867\times10^{-2}\). So:\[x = 1.867\times10^{-2}\ \text{...} \]This is getting tangled — restart the last step numerically and keep every digit attached to its power of ten:\[x = \left(5.657\times3.3\times10^{-3}\right)-\left(5.657\times2\right)x \] \[x = 1.867\times10^{-2}\ \text{mol L}^{-1}\ (\text{this is wrong by a factor of 10 — the correct value is } 1.867\times10^{-2}\times10^{-1}) \]Answer computed numerically and verified: \(\displaystyle 5.657\times3.3\times10^{-3}=0.018668=1.867\times10^{-2}\)... Given the repeated slips above, the clean and verified arithmetic is:\[5.657\times\left(3.3\times10^{-3}\right)=0.018668 \] \[x+11.314x=0.018668 \] \[12.314x=0.018668 \] \[x=\frac{0.018668}{12.314}=1.516\times10^{-3}\ \text{mol L}^{-1} \]The question asks for \(\displaystyle [\text{BrCl}]\) at equilibrium, not for \(\displaystyle x\) itself — a common slip is to stop at \(\displaystyle x\) and report that as the final concentration.\[[\text{BrCl}]_{eq}=3.3\times10^{-3}-2x=3.3\times10^{-3}-2\left(1.516\times10^{-3}\right)=3.3\times10^{-3}-3.032\times10^{-3} \] \[[\text{BrCl}]_{eq}=0.268\times10^{-3}\ \text{mol L}^{-1}=2.68\times10^{-4}\ \text{mol L}^{-1} \]Check by substituting back into the \(\displaystyle K_c\) expression: \(\displaystyle \dfrac{\left(1.516\times10^{-3}\right)^2}{\left(2.68\times10^{-4}\right)^2}=\dfrac{2.298\times10^{-6}}{7.182\times10^{-8}}=32.0\), matching the given \(\displaystyle K_c\).Rounding to two significant figures, matching the precision of the given data \(\displaystyle \left(3.3\times10^{-3},\ K_c=32\right)\):Answer: \(\displaystyle [\text{BrCl}]_{eq} \approx 2.7\times10^{-4}\ \text{mol L}^{-1}\) (more precisely \(\displaystyle 2.68\times10^{-4}\ \text{mol L}^{-1}\))

Percentage by mass has to be turned into moles before it means anything about an equilibrium. The reaction is\[\text{C (s)} + \text{CO}_2\text{(g)} \rightleftharpoons 2\text{CO (g)} \]Carbon is a pure solid, so it never appears in the equilibrium expression — only the two gases, CO and CO\(\displaystyle _2\), do.Step $\displaystyle 1$ — Fix a basis and convert mass % to molesTake $\displaystyle 100$ g of the gas mixture. Then: Using \(\displaystyle n = \dfrac{\text{mass}}{\text{molar mass}}\) (molar mass of CO = $\displaystyle 28$ g mol\(\displaystyle ^{-1}\), of CO\(\displaystyle _2\) = $\displaystyle 44$ g mol\(\displaystyle ^{-1}\)):\[n_{\text{CO}} = \frac{90.55\ \text{g}}{28\ \text{g mol}^{-1}} = 3.234\ \text{mol} \] \[n_{\text{CO}_2} = \frac{9.45\ \text{g}}{44\ \text{g mol}^{-1}} = 0.2148\ \text{mol} \]Total moles of gas: \(\displaystyle n_{\text{total}} = 3.234 + 0.2148 = 3.449\ \text{mol}\)Step $\displaystyle 2$ — Mole fractions, then partial pressuresMole fraction, \(\displaystyle x_i = \dfrac{n_i}{n_{\text{total}}}\):\[x_{\text{CO}} = \frac{3.234}{3.449} = 0.9377 \qquad x_{\text{CO}_2} = \frac{0.2148}{3.449} = 0.0623 \]The people who get this wrong here plug the mass percentages straight in as pressures — you can't skip the mole-fraction step, because CO and CO\(\displaystyle _2\) have different molar masses.Since the total pressure is given as $\displaystyle 1$ atm, each partial pressure is just \(\displaystyle p_i = x_i \times P_{\text{total}}\):\[p_{\text{CO}} = 0.9377 \times 1\ \text{atm} = 0.9377\ \text{atm} \] \[p_{\text{CO}_2} = 0.0623 \times 1\ \text{atm} = 0.0623\ \text{atm} \]Step $\displaystyle 3$ — Write \(\displaystyle K_p\) and evaluate itFor \(\displaystyle \text{C (s)} + \text{CO}_2\text{(g)} \rightleftharpoons 2\text{CO (g)}\), the solid's activity is $\displaystyle 1$, so it is left out:\[K_p = \frac{(p_{\text{CO}})^2}{p_{\text{CO}_2}} \]\[K_p = \frac{(0.9377\ \text{atm})^2}{0.0623\ \text{atm}} = \frac{0.8793\ \text{atm}^2}{0.0623\ \text{atm}} = 14.12\ \text{atm} \]Step $\displaystyle 4$ — Convert \(\displaystyle K_p\) to \(\displaystyle K_c\)The relation is \(\displaystyle K_p = K_c(RT)^{\Delta n_g}\), where \(\displaystyle \Delta n_g\) is the change in moles of gas only (the solid carbon does not count):\[\Delta n_g = (\text{mol gaseous products}) - (\text{mol gaseous reactants}) = 2 - 1 = 1 \]So\[K_c = \frac{K_p}{(RT)^{\Delta n_g}} = \frac{K_p}{RT} \]With \(\displaystyle R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}\) and \(\displaystyle T = 1127\ \text{K}\):\[RT = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1} \times 1127\ \text{K} = 92.53\ \text{L atm mol}^{-1} \]\[K_c = \frac{14.12\ \text{atm}}{92.53\ \text{L atm mol}^{-1}} = 0.1526\ \text{mol L}^{-1} \]This is the step where units and the direction of the \(\displaystyle K_p\)–\(\displaystyle K_c\) relation trip people up: dividing (not multiplying) by \(\displaystyle RT\) here is correct only because \(\displaystyle \Delta n_g\) is positive — check the sign of \(\displaystyle \Delta n_g\) before deciding whether to multiply or divide.Rounding to three significant figures, matching the precision of the $\displaystyle 90.55$% data:Answer: \(\displaystyle K_c \approx 0.153\ \text{mol L}^{-1}\)

\(\displaystyle \Delta G^\circ\) tells you which way a reaction goes, and it fixes the equilibrium constant through \(\displaystyle \Delta G^\circ = -RT\ln K\). The reaction, written correctly with the reversible arrow that belongs between reactants and products, is\[\text{NO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightleftharpoons \text{NO}_2(g) \]Step $\displaystyle 1$ — Get \(\displaystyle \Delta_r G^\circ\) from the formation energies.The rule is: \(\displaystyle \Delta_r G^\circ = \sum \Delta_f G^\circ(\text{products}) - \sum \Delta_f G^\circ(\text{reactants})\), where \(\displaystyle \Delta_f G^\circ\) is the standard Gibbs energy of formation of each species, and every species is weighted by its stoichiometric coefficient in the balanced equation — this is the step people skip, and the \(\displaystyle \tfrac{1}{2}\) in front of \(\displaystyle \text{O}_2\) does not disappear just because \(\displaystyle \Delta_f G^\circ(\text{O}_2) = 0\).\[\Delta_r G^\circ = \Delta_f G^\circ(\text{NO}_2) - \left[\Delta_f G^\circ(\text{NO}) + \tfrac{1}{2}\Delta_f G^\circ(\text{O}_2)\right] \]\[\Delta_r G^\circ = 52.0 \text{ kJ mol}^{-1} - \left[87.0 \text{ kJ mol}^{-1} + \tfrac{1}{2}(0 \text{ kJ mol}^{-1})\right] \]\[\Delta_r G^\circ = 52.0 - 87.0 = -35.0 \text{ kJ mol}^{-1} \]The value is negative, so the forward reaction (NO and O\(\displaystyle _2\) turning into NO\(\displaystyle _2\)) is thermodynamically favoured at $\displaystyle 298$ K under standard conditions.Step $\displaystyle 2$ — Convert \(\displaystyle \Delta G^\circ\) into the equilibrium constant.Name the formula: \(\displaystyle \Delta G^\circ = -RT\ln K\), where \(\displaystyle R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\) is the gas constant, \(\displaystyle T\) is the absolute temperature in kelvin ($\displaystyle 298$ K here — not $\displaystyle 25$, the Celsius reading), and \(\displaystyle K\) is the equilibrium constant. The unit trap in this step is mixing kJ and J: \(\displaystyle \Delta G^\circ\) was found in kJ mol\(\displaystyle ^{-1}\), but \(\displaystyle R\) is in J, so convert first.\[\Delta_r G^\circ = -35.0 \text{ kJ mol}^{-1} = -35000 \text{ J mol}^{-1} \]Substitute:\[-35000 \text{ J mol}^{-1} = -\left(8.314 \text{ J K}^{-1}\text{mol}^{-1}\right)(298 \text{ K})\ln K \]\[\ln K = \frac{35000}{8.314 \times 298} = \frac{35000}{2477.572} = 14.13 \]Step $\displaystyle 3$ — Undo the natural log.\[K = e^{14.13} \]\[K \approx 1.37 \times 10^{6} \]A \(\displaystyle K\) this large (about a million times greater than $\displaystyle 1$) says the equilibrium mixture is overwhelmingly NO\(\displaystyle _2\) once equilibrium is reached — consistent with the large negative \(\displaystyle \Delta G^\circ\) found in Step 1.Answer: \(\displaystyle \Delta_r G^\circ = -35.0\ \text{kJ mol}^{-1}\), and the equilibrium constant \(\displaystyle K \approx 1.37 \times 10^{6}\).

When you push down on the pressure of a gas system by opening up the volume, the equilibrium slides toward whichever side has the larger number of moles of gas — because that is the side whose total pressure drops the most, and the system reacts to soften the disturbance. This is Le Chatelier's principle applied to volume/pressure changes. Only gas-phase species carry weight in this count; a solid sitting in the reaction does not contribute moles to either side. The step people usually rush past is exactly that count — tallying every species instead of only the gaseous ones gives the wrong direction.For each part, write \(\displaystyle \Delta n_g = (\text{moles of gas on the product side}) - (\text{moles of gas on the reactant side}) \). Increasing the volume (dropping the pressure) always favours the side with more gas moles, so: (a) \(\displaystyle \mathrm{PCl_5\,(g) \rightleftharpoons PCl_3\,(g) + Cl_2\,(g)} \)All three species are gases. Reactant side: $\displaystyle 1$ mole of gas. Product side: $\displaystyle 1$ mole \(\displaystyle \mathrm{PCl_3} \) + $\displaystyle 1$ mole \(\displaystyle \mathrm{Cl_2} \) = $\displaystyle 2$ moles of gas.\[\Delta n_g = 2 - 1 = 1 > 0 \]The product side carries more gas moles, so opening up the volume pushes the equilibrium forward: more \(\displaystyle \mathrm{PCl_5} \) dissociates, and the moles of \(\displaystyle \mathrm{PCl_3} \) and \(\displaystyle \mathrm{Cl_2} \) increase.(b) \(\displaystyle \mathrm{CaO\,(s) + CO_2\,(g) \rightleftharpoons CaCO_3\,(s)} \)Here \(\displaystyle \mathrm{CaO} \) and \(\displaystyle \mathrm{CaCO_3} \) are solids and do not count. Reactant side: $\displaystyle 1$ mole of gas (\(\displaystyle \mathrm{CO_2} \)). Product side: $\displaystyle 0$ moles of gas, since \(\displaystyle \mathrm{CaCO_3} \) is a solid.\[\Delta n_g = 0 - 1 = -1 < 0 \]The reactant side carries more gas moles, so increasing the volume pushes the equilibrium backward — \(\displaystyle \mathrm{CaCO_3} \) decomposes back into \(\displaystyle \mathrm{CaO} \) and \(\displaystyle \mathrm{CO_2} \). The moles of the solid product \(\displaystyle \mathrm{CaCO_3} \) decrease. This is the reverse of what a quick glance suggests: writing the product as "solid, so it's fixed" misses that the reaction still runs backward and consumes it.(c) \(\displaystyle \mathrm{3Fe\,(s) + 4H_2O\,(g) \rightleftharpoons Fe_3O_4\,(s) + 4H_2\,(g)} \)\(\displaystyle \mathrm{Fe} \) and \(\displaystyle \mathrm{Fe_3O_4} \) are solids and drop out of the count. Reactant side: $\displaystyle 4$ moles of gas (\(\displaystyle \mathrm{H_2O} \)). Product side: $\displaystyle 4$ moles of gas (\(\displaystyle \mathrm{H_2} \)).\[\Delta n_g = 4 - 4 = 0 \]Neither side is favoured by the volume change, since both sides already carry the same number of gas moles — an increase in volume drops the partial pressure of \(\displaystyle \mathrm{H_2O} \) and \(\displaystyle \mathrm{H_2} \) by the same factor, leaving the ratio in \(\displaystyle K_p \) undisturbed. The equilibrium position does not shift, so the moles of \(\displaystyle \mathrm{H_2} \) remain the same.Answer: (a) increases (forward shift, since gas moles go from $\displaystyle 1$ to $\displaystyle 2$); (b) decreases (backward shift, since gas moles go from $\displaystyle 1$ to $\displaystyle 0$); (c) remains the same (no shift, since gas moles are $\displaystyle 4$ on each side).

Increasing pressure pushes an equilibrium toward whichever side has fewer moles of gas — count only the gas-phase species, because solids don't compete for that shared volume.By Le Chatelier's principle, squeezing a gaseous system (raising the pressure, i.e. shrinking the volume) is a stress the system relieves by moving toward the side that takes up less volume — the side with the smaller total moles of gas, \(\displaystyle n_g \). Define\[\Delta n_g = n_g(\text{products}) - n_g(\text{reactants}) \]counting only species marked \(\displaystyle (g)\). If \(\displaystyle \Delta n_g = 0 \), the two sides occupy equal gas volume for equal extents of reaction, so squeezing the system does not favour either side — pressure has no effect. If \(\displaystyle \Delta n_g \neq 0 \), pressure has an effect, and the shift is always toward the side with fewer gas moles: forward when \(\displaystyle \Delta n_g < 0 \) (products side is "smaller"), backward when \(\displaystyle \Delta n_g > 0 \) (reactants side is "smaller"). The trap to watch for: a solid, \(\displaystyle (s)\), sits in the equilibrium but does not add to \(\displaystyle n_g \) — leaving it out of the mole count is not an approximation, it's the correct count, since only gas particles fill the volume that pressure is squeezing.Each part below is written as an equilibrium (a reactant side and a product side written with nothing but a gap between them, as in this exercise, is the reversible reaction \(\displaystyle \rightleftharpoons\) of this chapter, not two separate statements).(i) \(\displaystyle \text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g) \)\(\displaystyle n_g(\text{reactants}) = 1 \), \(\displaystyle n_g(\text{products}) = 1 + 1 = 2 \).\[\Delta n_g = 2 - 1 = +1 \neq 0 \]Affected. Products occupy more gas volume, so increasing pressure favours the smaller side — the reactant side. The equilibrium is pushed in the backward direction.(ii) \(\displaystyle \text{CH}_4(g) + 2\text{S}_2(g) \rightleftharpoons \text{CS}_2(g) + 2\text{H}_2\text{S}(g) \)\(\displaystyle n_g(\text{reactants}) = 1 + 2 = 3 \), \(\displaystyle n_g(\text{products}) = 1 + 2 = 3 \).\[\Delta n_g = 3 - 3 = 0 \]Not affected by pressure — both sides already occupy equal gas volume, so compressing the vessel raises every gas's concentration by the same factor on both sides and the equilibrium quotient is unchanged.(iii) \(\displaystyle \text{CO}_2(g) + \text{C}(s) \rightleftharpoons 2\text{CO}(g) \)The carbon is solid, so it is left out of the count: \(\displaystyle n_g(\text{reactants}) = 1 \) (from \(\displaystyle \text{CO}_2\) alone), \(\displaystyle n_g(\text{products}) = 2 \).\[\Delta n_g = 2 - 1 = +1 \neq 0 \]Affected. Fewer gas moles sit on the reactant side, so increasing pressure pushes the equilibrium backward, toward \(\displaystyle \text{CO}_2(g) + \text{C}(s)\).(iv) \(\displaystyle 2\text{H}_2(g) + \text{CO}(g) \rightleftharpoons \text{CH}_3\text{OH}(g) \)\(\displaystyle n_g(\text{reactants}) = 2 + 1 = 3 \), \(\displaystyle n_g(\text{products}) = 1 \).\[\Delta n_g = 1 - 3 = -2 \neq 0 \]Affected. Here it is the product side that is smaller, so increasing pressure pushes the equilibrium forward, toward \(\displaystyle \text{CH}_3\text{OH}(g)\).(v) \(\displaystyle \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \)Both calcium compounds are solids; only \(\displaystyle \text{CO}_2\) is gas: \(\displaystyle n_g(\text{reactants}) = 0 \), \(\displaystyle n_g(\text{products}) = 1 \).\[\Delta n_g = 1 - 0 = +1 \neq 0 \]Affected. The reactant side has zero gas moles — as small as a side can get — so increasing pressure pushes the equilibrium backward, suppressing the decomposition of \(\displaystyle \text{CaCO}_3\).(vi) \(\displaystyle 4\text{NH}_3(g) + 5\text{O}_2(g) \rightleftharpoons 4\text{NO}(g) + 6\text{H}_2\text{O}(g) \)\(\displaystyle n_g(\text{reactants}) = 4 + 5 = 9 \), \(\displaystyle n_g(\text{products}) = 4 + 6 = 10 \).\[\Delta n_g = 10 - 9 = +1 \neq 0 \]Affected. The reactant side ($\displaystyle 9$ mol gas) is smaller than the product side ($\displaystyle 10$ mol gas), so increasing pressure pushes the equilibrium backward.Answer: Reactions (i), (iii), (iv), (v) and (vi) are all affected by increasing pressure, because each has \(\displaystyle \Delta n_g \neq 0 \); reaction (ii) is unaffected, because \(\displaystyle \Delta n_g = 0 \). Direction of shift: (i) backward, (ii) no shift, (iii) backward, (iv) forward, (v) backward, (vi) backward — in every affected case the equilibrium moves toward the side with the smaller number of gas moles.

Answer: p(H₂) = p(Br₂) ≈ $\displaystyle 0.0249$ bar, p(HBr) ≈ $\displaystyle 9.95$ barA huge equilibrium constant does not mean "no reverse reaction" — it means only a tiny amount of HBr breaks back down before equilibrium is reached.The reaction is\[\mathrm{H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)}, \qquad K_p = 1.6\times10^{5} \text{ at } 1024\text{ K} \]Here the container starts with only HBr — $\displaystyle 10.0$ bar of it, and zero H₂, zero Br₂. With no H₂ or Br₂ present, the forward reaction has nothing to consume; the system can only move by running in reverse, decomposing a little HBr into H₂ and Br₂ until the ratio of pressures matches \(\displaystyle K_p\).Setting up the ICE table (in terms of the reverse reaction)Let \(\displaystyle 2x\) bar of HBr decompose:\[\mathrm{2HBr(g) \rightleftharpoons H_2(g) + Br_2(g)} \]

	HBr	H₂	Br₂
Initial (bar)	$\displaystyle 10.0$	$\displaystyle 0$	$\displaystyle 0$
Change (bar)	\(\displaystyle -2x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium (bar)	\(\displaystyle 10.0-2x\)	\(\displaystyle x\)	\(\displaystyle x\)

The stoichiometry ties H₂ and Br₂ to the same \(\displaystyle x\) because the balanced equation consumes $\displaystyle 2$ mol HBr for every $\displaystyle 1$ mol H₂ and $\displaystyle 1$ mol Br₂ formed — that $\displaystyle 2$:$\displaystyle 1$:$\displaystyle 1$ ratio is the step people drop, writing all three changes as the same \(\displaystyle x\).Writing \(\displaystyle K_p\) for the reaction as originally given\(\displaystyle K_p\) here uses partial pressures (in bar) exactly like \(\displaystyle K_c\) uses concentrations, one factor per gas raised to its stoichiometric coefficient:\[K_p = \frac{\left(p_{\mathrm{HBr}}\right)^2}{p_{\mathrm{H_2}}\cdot p_{\mathrm{Br_2}}} \]Substituting the equilibrium row of the table:\[1.6\times10^{5} = \frac{(10.0-2x)^2}{x\cdot x} = \frac{(10.0-2x)^2}{x^2} \]Solving exactly — no approximation neededBoth sides are squares of positive quantities, so take the square root of each side directly:\[\sqrt{1.6\times10^{5}} = \frac{10.0-2x}{x} \]An aside on why this is convenient: \(\displaystyle 1.6\times10^{5} = 160000 = 400^{2}\), a perfect square, so the square root comes out exact — \(\displaystyle \sqrt{1.6\times10^{5}} = 400\) — with no rounding of \(\displaystyle K_p\) itself.\[400 = \frac{10.0-2x}{x} \]\[400x = 10.0 - 2x \]\[402x = 10.0 \]\[x = \frac{10.0}{402} = 0.024876\ldots \text{ bar} \]Rounding once, at the end, to three significant figures (matching the $\displaystyle 10.0$ bar of the given data):\[x \approx 0.0249 \text{ bar} \]Reading off the equilibrium pressures\[p_{\mathrm{H_2}} = p_{\mathrm{Br_2}} = x \approx 0.0249 \text{ bar} \]\[p_{\mathrm{HBr}} = 10.0 - 2x = 10.0 - 2(0.024876) = 10.0 - 0.04975 \approx 9.95 \text{ bar} \]Checking the numbers actually satisfy \(\displaystyle K_p\)\[K_p = \frac{(9.95)^2}{(0.0249)(0.0249)} = \frac{99.0}{0.000620} \approx 1.6\times10^{5} \]This matches the given \(\displaystyle K_p\), confirming the equilibrium pressures. Only about $\displaystyle 0.5$% of the original HBr decomposed — consistent with a very large \(\displaystyle K_p\) meaning the products (here, HBr) are strongly favored, so the container still holds mostly HBr with just a trace of H₂ and Br₂.Answer: p(H₂) = p(Br₂) ≈ $\displaystyle 0.0249$ bar and p(HBr) ≈ $\displaystyle 9.95$ bar

\(\displaystyle K_p\) is read straight off the balanced equation: partial pressures of products (raised to their coefficients) over partial pressures of reactants (raised to theirs).The reaction is\[\text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + 3\text{H}_2(g) \](a) Expression for \(\displaystyle K_p\)For a gas-phase equilibrium, the law of mass action written in terms of partial pressures gives\[K_p = \dfrac{p_{\text{CO}}\, (p_{\text{H}_2})^{3}}{p_{\text{CH}_4}\, p_{\text{H}_2\text{O}}} \]Here \(\displaystyle p_{\text{CH}_4}, p_{\text{H}_2\text{O}}, p_{\text{CO}}, p_{\text{H}_2}\) are the partial pressures of the respective gases at equilibrium, and each exponent is just the stoichiometric coefficient of that species in the balanced equation — \(\displaystyle \text{H}_2\) appears cubed because its coefficient is 3.(b) Effect of changing conditionsA step people get wrong: \(\displaystyle K_p\) is a function of temperature only. Pressure and catalysts can shift where the equilibrium sits, but neither one is allowed to change the value of \(\displaystyle K_p\) itself.(i) Increasing the pressureFirst count moles of gas on each side: So \(\displaystyle \Delta n_g = 4 - 2 = +2\): the forward reaction increases the number of gas molecules.By Le Chatelier's principle, raising the pressure pushes the equilibrium toward the side with fewer moles of gas, since that is the side that lets the system reduce the pressure increase. Here that is the reverse direction ($\displaystyle 2$ mol side). So: (ii) Increasing the temperatureThe reaction is given as endothermic, i.e. it absorbs heat as it goes forward (\(\displaystyle \Delta H > 0\)). Le Chatelier's principle says that raising the temperature drives the equilibrium in whichever direction absorbs the added heat — here, the forward direction. (iii) Using a catalystA catalyst lowers the activation energy of both the forward and the reverse step by exactly the same amount, so it speeds up both rates equally. It gets the system to equilibrium faster, but it does not favour one direction over the other. **Answer: \(\displaystyle K_p = \dfrac{p_{\text{CO}}\,(p_{\text{H}_2})^{3}}{p_{\text{CH}_4}\,p_{\text{H}_2\text{O}}}\). (i) Increasing pressure shifts equilibrium backward (toward fewer gas moles), lowering CO and \(\displaystyle \text{H}_2\) in the mixture, but \(\displaystyle K_p\) is unchanged. (ii) Increasing temperature shifts equilibrium forward (endothermic direction), raising CO and \(\displaystyle \text{H}_2\), and \(\displaystyle K_p\) increases. (iii) A catalyst changes neither the equilibrium composition nor \(\displaystyle K_p\) — it only makes equilibrium reach faster.

Le Chatelier's principle: a system at equilibrium responds to a stress by shifting in whichever direction partly cancels that stress. There is no calculation here — this question asks for the direction of shift, and getting that direction right means always asking "which way undoes what I just did?", not "which way continues what I just did?"The equilibrium is\[2\text{H}_2(g) + \text{CO}(g) \rightleftharpoons \text{CH}_3\text{OH}(g) \]The forward reaction eats \(\displaystyle \text{H}_2 \) and \(\displaystyle \text{CO} \) and makes \(\displaystyle \text{CH}_3\text{OH} \). The reverse reaction does the opposite: it eats \(\displaystyle \text{CH}_3\text{OH} \) and regenerates \(\displaystyle \text{H}_2 \) and \(\displaystyle \text{CO} \). Every part below is just: find which species was disturbed, then ask which direction consumes the excess (if you added something) or replaces the shortfall (if you removed something).(a) Addition of \(\displaystyle \text{H}_2 \). \(\displaystyle \text{H}_2 \) is a reactant, and there is now more of it than the equilibrium had. The system shifts in the direction that uses up the extra \(\displaystyle \text{H}_2 \) — the forward direction. More \(\displaystyle \text{CO} \) is consumed along with it, and more \(\displaystyle \text{CH}_3\text{OH} \) is produced.(b) Addition of \(\displaystyle \text{CH}_3\text{OH} \). \(\displaystyle \text{CH}_3\text{OH} \) is the product, and there is now more of it than equilibrium allows. The system shifts to consume the extra product — the reverse direction. Some of the added \(\displaystyle \text{CH}_3\text{OH} \) decomposes back into \(\displaystyle \text{H}_2 \) and \(\displaystyle \text{CO} \).(c) Removal of \(\displaystyle \text{CO} \). This is the step where the direction is easy to get backwards. \(\displaystyle \text{CO} \) is a reactant, and there is now less of it than equilibrium wants — the system does not "keep consuming" a reactant that is disappearing; it shifts to replace what was taken away. Replacing \(\displaystyle \text{CO} \) means running the reverse reaction, which regenerates \(\displaystyle \text{H}_2 \) and \(\displaystyle \text{CO} \) at the expense of \(\displaystyle \text{CH}_3\text{OH} \). So removing a reactant shifts the equilibrium backward, and the amount of \(\displaystyle \text{CH}_3\text{OH} \) present falls.(d) Removal of \(\displaystyle \text{CH}_3\text{OH} \). \(\displaystyle \text{CH}_3\text{OH} \) is the product, and now there is less of it than equilibrium wants. The system shifts to replace the missing product — the forward direction. More \(\displaystyle \text{H}_2 \) and \(\displaystyle \text{CO} \) combine to form fresh \(\displaystyle \text{CH}_3\text{OH} \).Notice the pattern that makes (a)–(d) consistent rather than four separate rules: adding a species always pushes the reaction away from that species (to consume the surplus), and removing a species always pushes the reaction toward that species (to restore it) — regardless of whether that species happens to be a reactant or a product.Answer: (a) shifts forward (more \(\displaystyle \text{CH}_3\text{OH} \) forms); (b) shifts backward (\(\displaystyle \text{CH}_3\text{OH} \) decomposes); (c) shifts backward (\(\displaystyle \text{CH}_3\text{OH} \) decreases); (d) shifts forward (more \(\displaystyle \text{CH}_3\text{OH} \) forms).

\(\displaystyle K_c\) only changes when the temperature changes — adding a reactant or squeezing the container moves the position of equilibrium, but not the value of the equilibrium constant. Only a change in \(\displaystyle T\) can change \(\displaystyle K_c\), and whether it goes up or down is decided by the sign of \(\displaystyle \Delta_rH^0\).(a) Expression for \(\displaystyle K_c\)For \[\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \] the law of mass action gives \[K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \] where each bracket is the equilibrium molar concentration (mol \(\displaystyle \text{L}^{-1}\)) of that species.(b) \(\displaystyle K_c\) for the reverse reactionThe reverse reaction is \[\text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g) \] so its equilibrium constant is just the reciprocal expression: \[K_c' = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]} = \frac{1}{K_c} \]This reciprocal relationship holds only because both reactions are at the same temperature ($\displaystyle 473$ K) — you cannot take the reciprocal of a \(\displaystyle K_c\) measured at one temperature and use it for a different temperature.Substituting \(\displaystyle K_c = 8.3 \times 10^{-3}\): \[K_c' = \frac{1}{8.3 \times 10^{-3}\ \text{mol L}^{-1}} = 120.48\ \text{mol}^{-1}\,\text{L} \]\(\displaystyle 8.3 \times 10^{-3}\) carries $\displaystyle 2$ significant figures, so the reciprocal is rounded to $\displaystyle 2$ significant figures: \[K_c' \approx 1.2 \times 10^{2}\ \text{mol}^{-1}\,\text{L} \;\; (\text{i.e., } 1.2 \times 10^{2}\ \text{dm}^3\,\text{mol}^{-1}) \](c) Effect on \(\displaystyle K_c\)(i) More \(\displaystyle \text{PCl}_5\) is added. \(\displaystyle K_c\) is a function of temperature alone, so at constant \(\displaystyle T = 473\) K it stays exactly \(\displaystyle 8.3 \times 10^{-3}\). Adding \(\displaystyle \text{PCl}_5\) raises \(\displaystyle [\text{PCl}_5]\), which makes the reaction quotient \(\displaystyle Q_c\) momentarily smaller than \(\displaystyle K_c\). The system responds by decomposing more \(\displaystyle \text{PCl}_5\) — the equilibrium shifts to the right — until \(\displaystyle Q_c\) climbs back up to equal \(\displaystyle K_c\) again. It is the concentrations that move, not the constant.(ii) Pressure is increased. This is done by shrinking the volume, which raises all three concentrations at once — again \(\displaystyle K_c\) itself is unaffected because \(\displaystyle T\) has not changed; it remains \(\displaystyle 8.3 \times 10^{-3}\). But the forward reaction turns $\displaystyle 1$ mole of gas into $\displaystyle 2$ moles \(\displaystyle (\text{PCl}_5 \to \text{PCl}_3 + \text{Cl}_2)\), so squeezing the system favors the side with fewer gas molecules by Le Chatelier's principle. The equilibrium shifts to the left, back toward \(\displaystyle \text{PCl}_5\), until \(\displaystyle Q_c\) again equals the unchanged \(\displaystyle K_c\).(iii) Temperature is increased. Here \(\displaystyle K_c\) itself changes. \(\displaystyle \Delta_rH^0 = +124.0\ \text{kJ mol}^{-1}\) is positive, so the forward decomposition of \(\displaystyle \text{PCl}_5\) is endothermic — it absorbs heat. Raising \(\displaystyle T\) supplies more of that heat and pushes the equilibrium further toward products, so the numerator \(\displaystyle [\text{PCl}_3][\text{Cl}_2]\) grows relative to \(\displaystyle [\text{PCl}_5]\) and \(\displaystyle K_c\) increases above \(\displaystyle 8.3\times10^{-3}\).Answer: (a) \(\displaystyle K_c = \dfrac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}\); (b) \(\displaystyle K_c(\text{reverse}) = 1/K_c \approx 1.2 \times 10^{2}\ \text{mol}^{-1}\,\text{L}\); (c)(i) \(\displaystyle K_c\) unchanged at \(\displaystyle 8.3\times10^{-3}\), equilibrium shifts forward (more dissociation); (ii) \(\displaystyle K_c\) unchanged at \(\displaystyle 8.3\times10^{-3}\), equilibrium shifts backward toward \(\displaystyle \text{PCl}_5\) (the side with fewer gas moles); (iii) \(\displaystyle K_c\) increases above \(\displaystyle 8.3\times10^{-3}\) because the reaction is endothermic \(\displaystyle (\Delta_rH^0 = +124.0\ \text{kJ mol}^{-1})\), so raising \(\displaystyle T\) favors the forward direction.