NCERT Solutions · Class 11 Chemistry Equilibrium

When a substance decomposes on its own in a sealed flask, the pressure it loses is exactly what pairs up (in mole ratio) to form the products — track that loss with an ICE table in partial pressures.The equilibrium is\[2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \]Only HI(g) is placed in the flask, so at the start there is no \(\displaystyle \text{H}_2 \) or \(\displaystyle \text{I}_2 \) present — they appear only as HI breaks down.Step $\displaystyle 1$: Set up the ICE table in partial pressures.Let \(\displaystyle p \) be the fall in the partial pressure of HI(g) as it reacts. Since $\displaystyle 2$ mol HI produce $\displaystyle 1$ mol \(\displaystyle \text{H}_2 \) and $\displaystyle 1$ mol \(\displaystyle \text{I}_2 \), the pressure of each product rises by \(\displaystyle p/2 \).

	\(\displaystyle \text{HI(g)} \)	\(\displaystyle \text{H}_2\text{(g)} \)	\(\displaystyle \text{I}_2\text{(g)} \)
Initial	\(\displaystyle 0.2\ \text{atm} \)	\(\displaystyle 0 \)	\(\displaystyle 0 \)
Change	\(\displaystyle -p \)	\(\displaystyle +p/2 \)	\(\displaystyle +p/2 \)
Equilibrium	\(\displaystyle 0.2-p \)	\(\displaystyle p/2 \)	\(\displaystyle p/2 \)

Step $\displaystyle 2$: Use the given equilibrium pressure of HI(g) to find \(\displaystyle p \).The problem states the equilibrium partial pressure of HI(g) is \(\displaystyle 0.04\ \text{atm} \), so\[0.2 - p = 0.04 \ \text{atm} \quad\Rightarrow\quad p = 0.2 - 0.04 = 0.16\ \text{atm} \]This \(\displaystyle p \) is the pressure of HI consumed, not the pressure of either product — that is the substitution step that trips people up here, because the stoichiometric coefficient of HI is $\displaystyle 2$ while those of \(\displaystyle \text{H}_2 \) and \(\displaystyle \text{I}_2 \) are $\displaystyle 1$ each.Step $\displaystyle 3$: Find the equilibrium partial pressures of the products.\[p_{\text{H}_2} = p_{\text{I}_2} = \frac{p}{2} = \frac{0.16\ \text{atm}}{2} = 0.08\ \text{atm} \]So at equilibrium: \[p_{\text{HI}} = 0.04\ \text{atm}, \qquad p_{\text{H}_2} = 0.08\ \text{atm}, \qquad p_{\text{I}_2} = 0.08\ \text{atm} \]Step $\displaystyle 4$: Write \(\displaystyle K_p \) from the balanced equation and substitute.For \(\displaystyle 2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \), the equilibrium-constant expression (products over reactants, each partial pressure raised to its stoichiometric coefficient) is\[K_p = \frac{p_{\text{H}_2}\, p_{\text{I}_2}}{\left(p_{\text{HI}}\right)^2} \]Substituting the equilibrium pressures:\[K_p = \frac{(0.08\ \text{atm})(0.08\ \text{atm})}{(0.04\ \text{atm})^2} = \frac{6.4\times10^{-3}\ \text{atm}^2}{1.6\times10^{-3}\ \text{atm}^2} \]\[K_p = 4 \]The atm\(\displaystyle ^2\) in the numerator cancels exactly against the atm\(\displaystyle ^2\) in the denominator, so \(\displaystyle K_p \) here comes out as a pure number with no unit — this happens because the number of gas moles does not change across the reaction (\(\displaystyle \Delta n_g = (1+1) - 2 = 0 \)), so the pressure units in \(\displaystyle K_p = K_c(RT)^{\Delta n_g} \) never appear in the first place.Answer: \(\displaystyle K_p = 4 \) (no units, since \(\displaystyle \Delta n_g = 0 \))

\(\displaystyle K_c\) is defined in terms of concentrations (mol per litre), never in terms of raw mole counts — so before you can compare anything to \(\displaystyle K_c\), you have to divide every mole amount by the volume of the vessel.The reaction is\[N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]Step $\displaystyle 1$: convert moles to molar concentrationsUse \(\displaystyle [X] = \dfrac{n_X}{V} \), where \(\displaystyle n_X\) is the number of moles of species \(\displaystyle X\) present right now and \(\displaystyle V\) is the volume of the container in litres. Here \(\displaystyle V = 20\ \text{L}\):\[[N_2] = \frac{1.57\ \text{mol}}{20\ \text{L}} = 0.0785\ \text{mol L}^{-1} \] \[[H_2] = \frac{1.92\ \text{mol}}{20\ \text{L}} = 0.0960\ \text{mol L}^{-1} \] \[[NH_3] = \frac{8.13\ \text{mol}}{20\ \text{L}} = 0.4065\ \text{mol L}^{-1} \]Step $\displaystyle 2$: write the reaction quotient \(\displaystyle Q_c\), not \(\displaystyle K_c\)Since you don't yet know whether the system is at equilibrium, you plug the current concentrations into the equilibrium-constant expression and call the result \(\displaystyle Q_c\):\[Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]The exponents $\displaystyle 2$, $\displaystyle 1$, $\displaystyle 3$ come straight from the balanced equation's coefficients — this is the step people skip and then get a wrong power of \(\displaystyle [H_2]\).Step $\displaystyle 3$: substitute and evaluate\[[H_2]^3 = (0.0960\ \text{mol L}^{-1})^3 = 8.847\times10^{-4}\ \text{mol}^3\text{L}^{-3} \]\[[N_2][H_2]^3 = (0.0785\ \text{mol L}^{-1})(8.847\times10^{-4}\ \text{mol}^3\text{L}^{-3}) = 6.945\times10^{-5}\ \text{mol}^4\text{L}^{-4} \]\[[NH_3]^2 = (0.4065\ \text{mol L}^{-1})^2 = 0.16524\ \text{mol}^2\text{L}^{-2} \]\[Q_c = \frac{0.16524\ \text{mol}^2\text{L}^{-2}}{6.945\times10^{-5}\ \text{mol}^4\text{L}^{-4}} = 2379\ \text{mol}^{-2}\text{L}^{2} \]Rounding to two significant figures — matching the precision of the given \(\displaystyle K_c\) — gives\[Q_c \approx 2.4\times10^{3} \](The units on \(\displaystyle Q_c\) are the same as the units on \(\displaystyle K_c\) for this reaction, since both come from the identical expression, so they cancel out of the comparison in Step $\displaystyle 4$ either way.)Step $\displaystyle 4$: compare \(\displaystyle Q_c\) to \(\displaystyle K_c\)The given equilibrium constant is \(\displaystyle K_c = 1.7\times10^{2}\). Since\[Q_c\ (2.4\times10^{3}) \;>\; K_c\ (1.7\times10^{2}) \]the mixture is not at equilibrium.A larger-than-equilibrium \(\displaystyle Q_c\) means the numerator — the product, \(\displaystyle NH_3\) — is present in excess relative to what equilibrium allows, given how much \(\displaystyle N_2\) and \(\displaystyle H_2\) are on hand. This is the point where it's easy to get the direction backwards: \(\displaystyle Q_c > K_c\) does not push the reaction forward to make more product — it pushes it in the direction that consumes the excess product. So the net reaction runs in reverse, from right to left: some \(\displaystyle NH_3\) decomposes back into \(\displaystyle N_2\) and \(\displaystyle H_2\) until the concentrations rearrange themselves so that \(\displaystyle Q_c\) falls to \(\displaystyle 1.7\times10^{2}\).**Answer: \(\displaystyle Q_c \approx 2.4\times10^{3}\), which is greater than \(\displaystyle K_c = 1.7\times10^{2}\), so the mixture is not at equilibrium; the net reaction proceeds in the reverse direction (\(\displaystyle NH_3\) decomposing to \(\displaystyle N_2\) and \(\displaystyle H_2\)) until equilibrium is reached.

In an equilibrium-constant expression, every species written in the numerator is a product and every species in the denominator is a reactant — the exponent it carries is its coefficient in the balanced equation.The general form of an equilibrium-constant expression (the law of mass action) for a reaction\[aA + bB \rightleftharpoons cC + dD \]is\[K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \]Here the square brackets mean equilibrium molar concentration, and the exponents \(\displaystyle a, b, c, d\) are exactly the stoichiometric coefficients from the balanced equation: the products \(\displaystyle C\) and \(\displaystyle D\) sit on top, the reactants \(\displaystyle A\) and \(\displaystyle B\) sit on the bottom.The given expression is\[K_c = \dfrac{[\mathrm{NH_3}]^{4}[\mathrm{O_2}]^{5}}{[\mathrm{NO}]^{4}[\mathrm{H_2O}]^{6}} \]Matching this against the general form: This is the step that trips people up: it feels natural to put \(\displaystyle \mathrm{NH_3}\) and \(\displaystyle \mathrm{O_2}\) on the reactant side, because the familiar combustion of ammonia is written \(\displaystyle 4\mathrm{NH_3} + 5\mathrm{O_2} \rightarrow 4\mathrm{NO} + 6\mathrm{H_2O}\). But the \(\displaystyle K_c\) expression you are given is what decides the direction — whatever sits above the line is a product of this equilibrium, regardless of which arrangement looks more familiar chemically. Reading the exponents straight off the fraction, the equation runs the other way:\[4\,\mathrm{NO(g)} + 6\,\mathrm{H_2O(g)} \rightleftharpoons 4\,\mathrm{NH_3(g)} + 5\,\mathrm{O_2(g)} \]Check this is actually balanced, atom for atom, since a set of exponents only means something if it comes from a real balanced equation: All three atoms balance, confirming the equation.Answer: \(\displaystyle 4\,\mathrm{NO(g)} + 6\,\mathrm{H_2O(g)} \rightleftharpoons 4\,\mathrm{NH_3(g)} + 5\,\mathrm{O_2(g)}\)

A gas-phase equilibrium constant is built from equilibrium concentrations, not from the moles you started with, so the first job is turning "$\displaystyle 40$ percent reacted" into an ICE table.The reaction is\[\text{H}_2\text{O}(g) + \text{CO}(g) \rightleftharpoons \text{H}_2(g) + \text{CO}_2(g) \]Step $\displaystyle 1$: Convert "$\displaystyle 40$ percent of water reacts" into moles.You start with $\displaystyle 1$ mole of \(\displaystyle \mathrm{H_{2}O}\) and $\displaystyle 1$ mole of CO in a $\displaystyle 10$ L vessel. Since water is a single substance here, $\displaystyle 40$ percent by mass reacting is the same as $\displaystyle 40$ percent of its moles reacting (mass and moles are proportional through one fixed molar mass):\[\text{moles of H}_2\text{O reacted} = 0.40 \times 1\ \text{mol} = 0.4\ \text{mol} \]From the $\displaystyle 1$:$\displaystyle 1$ stoichiometry, exactly $\displaystyle 0.4$ mol of CO is also consumed, and $\displaystyle 0.4$ mol each of \(\displaystyle \mathrm{H_{2}}\) and \(\displaystyle \mathrm{CO_{2}}\) is produced.Step $\displaystyle 2$: Build the ICE table (in moles).

	\(\displaystyle \mathrm{H_{2}O}\)	CO	\(\displaystyle \mathrm{H_{2}}\)	\(\displaystyle \mathrm{CO_{2}}\)
Initial	$\displaystyle 1$	$\displaystyle 1$	$\displaystyle 0$	$\displaystyle 0$
Change	-$\displaystyle 0.4$	-$\displaystyle 0.4$	+$\displaystyle 0.4$	+$\displaystyle 0.4$
Equilibrium	$\displaystyle 0.6$	$\displaystyle 0.6$	$\displaystyle 0.4$	$\displaystyle 0.4$

A step people get wrong here: forgetting that CO is consumed by the same amount as H2O. The equation shows $\displaystyle 1$ mol \(\displaystyle \mathrm{H_{2}O}\) reacting with $\displaystyle 1$ mol CO, so whatever moles of water react, exactly that many moles of CO react too. It is not "$\displaystyle 40$ percent of CO" independently; it is the CO that reacted alongside that water.Step $\displaystyle 3$: Convert moles to concentrations.Concentration is moles divided by the volume of the vessel, V = $\displaystyle 10$ L:\[[\text{H}_2\text{O}] = \frac{0.6\ \text{mol}}{10\ \text{L}} = 0.06\ \text{mol L}^{-1}, \qquad [\text{CO}] = \frac{0.6\ \text{mol}}{10\ \text{L}} = 0.06\ \text{mol L}^{-1} \]\[[\text{H}_2] = \frac{0.4\ \text{mol}}{10\ \text{L}} = 0.04\ \text{mol L}^{-1}, \qquad [\text{CO}_2] = \frac{0.4\ \text{mol}}{10\ \text{L}} = 0.04\ \text{mol L}^{-1} \]Step $\displaystyle 4$: Write the equilibrium constant expression and substitute.Kc is the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient:\[K_c = \frac{[\text{H}_2][\text{CO}_2]}{[\text{H}_2\text{O}][\text{CO}]} \]Substituting the equilibrium concentrations:\[K_c = \frac{(0.04\ \text{mol L}^{-1})(0.04\ \text{mol L}^{-1})}{(0.06\ \text{mol L}^{-1})(0.06\ \text{mol L}^{-1})} = \frac{0.0016}{0.0036} \]The unit aside people skip: here every concentration term carries mol per litre, and there are two of them on top and two on the bottom, so the units cancel completely. This happens because the reaction has equal moles of gas on both sides ($\displaystyle 2$ mol reactants to $\displaystyle 2$ mol products, so the change in gas moles is zero), which is also why you could have used moles directly instead of concentrations. The factor of $\displaystyle 1$/V (here $\displaystyle 1$/$\displaystyle 10$) appears twice in the numerator and twice in the denominator and cancels exactly:\[K_c = \frac{(0.4)(0.4)}{(0.6)(0.6)} = \frac{0.16}{0.36} = 0.444\ldots \]Rounding to three significant figures (matching the precision of the given $\displaystyle 40$ percent):\[K_c = 0.444 \]Answer: Kc = $\displaystyle 0.444$ (dimensionless, since the moles of gas are equal on both sides of the equation).

When a reaction is run in reverse, the same \(\displaystyle K_c\) still governs the equilibrium mixture — you just have to write the ICE table for the reaction as it is actually proceeding.Here you start with pure \(\displaystyle \text{HI}(g)\) and let it dissociate, so the reaction that is actually happening is the reverse of the one you were given the constant for:\[2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]The forward reaction \(\displaystyle \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)\) has\[K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = 54.8 \text{ at } 700\text{ K} \]This is the same equilibrium state whichever direction you approach it from, so this same expression (with this same numerical \(\displaystyle K_c\)) still connects the equilibrium concentrations — you do not need a separate "reverse" constant, you just solve for \(\displaystyle [\text{H}_2]\) and \(\displaystyle [\text{I}_2]\) from it.Setting up the ICE table. Because the vessel started with only \(\displaystyle \text{HI}(g)\) (no \(\displaystyle \text{H}_2\) or \(\displaystyle \text{I}_2\) present initially), the stoichiometry \(\displaystyle 2\text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2\) tells you that for every \(\displaystyle 2x\) mol L\(\displaystyle ^{-1}\) of HI that decomposes, exactly \(\displaystyle x\) mol L\(\displaystyle ^{-1}\) of \(\displaystyle \text{H}_2\) and \(\displaystyle x\) mol L\(\displaystyle ^{-1}\) of \(\displaystyle \text{I}_2\) are produced — in a $\displaystyle 1$:$\displaystyle 1$ ratio with each other, because both come from the same starting substance:

	\(\displaystyle \text{H}_2\)	\(\displaystyle \text{I}_2\)
At equilibrium	\(\displaystyle x\)	\(\displaystyle x\)

This is the step people skip: \(\displaystyle [\text{H}_2]=[\text{I}_2]\) here is not an assumption, it is forced by the fact that H\(\displaystyle _2\) and I\(\displaystyle _2\) only entered the mixture by decomposing the same HI, in equal amounts.Substituting into \(\displaystyle K_c\). With \(\displaystyle [\text{HI}] = 0.5\ \text{mol L}^{-1}\) at equilibrium and \(\displaystyle [\text{H}_2] = [\text{I}_2] = x\):\[54.8 = \frac{(0.5)^2}{x \cdot x} = \frac{0.25}{x^2} \]Solve for \(\displaystyle x^2\):\[x^2 = \frac{0.25}{54.8} = 4.562 \times 10^{-3}\ \text{mol}^2\text{L}^{-2} \]Take the square root (carrying the full value, not yet rounded):\[x = \sqrt{4.562 \times 10^{-3}} = 6.754 \times 10^{-2}\ \text{mol L}^{-1} \]The data (\(\displaystyle K_c = 54.8\), three significant figures) justifies rounding only at this last step, to three significant figures:\[x = 6.75 \times 10^{-2}\ \text{mol L}^{-1} \]Since \(\displaystyle [\text{H}_2] = [\text{I}_2] = x\):\[[\text{H}_2] = [\text{I}_2] = 6.75 \times 10^{-2}\ \text{mol L}^{-1} \]Answer: \(\displaystyle [\text{H}_2] = [\text{I}_2] = 6.75 \times 10^{-2}\ \text{mol L}^{-1}\)

When both product coefficients are $\displaystyle 1$ and the reactant term is squared on both sides, you can take a square root instead of grinding through the quadratic formula. That is the shortcut this equilibrium is built for.The reaction is\[2\text{ICl}(g) \rightleftharpoons \text{I}_2(g) + \text{Cl}_2(g), \qquad K_c = 0.14 \]Only ICl is present at the start, at $\displaystyle 0.78$ M, so the reaction must go forward (left to right) to reach equilibrium — there is no I\(\displaystyle _2\) or Cl\(\displaystyle _2\) yet for it to go backward.Set up the ICE table. Let \(\displaystyle x\) be the concentration (in mol/L) of I\(\displaystyle _2\) formed at equilibrium. Because $\displaystyle 2$ mol ICl is consumed for every $\displaystyle 1$ mol I\(\displaystyle _2\) (and $\displaystyle 1$ mol Cl\(\displaystyle _2\)) produced, ICl drops by \(\displaystyle 2x\):

	\(\displaystyle 2\text{ICl}(g)\)	\(\displaystyle \text{I}_2(g)\)	\(\displaystyle \text{Cl}_2(g)\)
Initial (M)	$\displaystyle 0.78$	$\displaystyle 0$	$\displaystyle 0$
Change (M)	\(\displaystyle -2x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium (M)	\(\displaystyle 0.78-2x\)	\(\displaystyle x\)	\(\displaystyle x\)

Write the equilibrium-constant expression. For this reaction,\[K_c=\dfrac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^{2}} \]where each bracket is the equilibrium molar concentration of that species. Substituting the ICE-table entries:\[0.14=\dfrac{x\cdot x}{(0.78-2x)^{2}}=\dfrac{x^{2}}{(0.78-2x)^{2}} \]Take the square root of both sides. Both numerator and denominator are already perfect squares, so instead of expanding \(\displaystyle (0.78-2x)^2\) into a quadratic in \(\displaystyle x\), take the square root of each side directly — it's the same equation with far less algebra:\[\sqrt{0.14}=\dfrac{x}{0.78-2x} \]\[0.3742=\dfrac{x}{0.78-2x} \](A negative root is rejected here because a square root of a concentration ratio must be positive.)Solve for \(\displaystyle x\). Cross-multiplying:\[x=0.3742\,(0.78-2x)=0.2918-0.7483x \]\[x+0.7483x=0.2918 \]\[1.7483x=0.2918 \]\[x=\dfrac{0.2918}{1.7483}=0.1669\ \text{M} \]Read off the equilibrium concentrations from the ICE table using this \(\displaystyle x\):\[[\text{I}_2]=[\text{Cl}_2]=x=0.1669\ \text{M}\approx 0.167\ \text{M} \]\[[\text{ICl}]=0.78-2x=0.78-2(0.1669)=0.78-0.3339=0.4461\ \text{M}\approx 0.446\ \text{M} \]Check the arithmetic by plugging back into \(\displaystyle K_c\): the concentration of ICl left over ($\displaystyle 0.446$ M) must be far bigger than what converted (about $\displaystyle 0.334$ M out of $\displaystyle 0.78$ M), which matches a \(\displaystyle K_c\) below $\displaystyle 1$ — the equilibrium favours the reactant side, so most of the ICl should remain unreacted, and it does.\[K_c=\dfrac{(0.1669)^2}{(0.4461)^2}=\dfrac{0.02786}{0.1990}=0.140\ \checkmark \]The data ($\displaystyle 0.78$ M, \(\displaystyle K_c=0.14\)) justify three significant figures in the final concentrations.Answer: \(\displaystyle [\text{I}_2] = [\text{Cl}_2] \approx 0.167\ \text{M}\), and \(\displaystyle [\text{ICl}] \approx 0.446\ \text{M}\) at equilibrium.

\(\displaystyle K_p\) connects equilibrium PARTIAL PRESSURES — you only switch to concentration at the very last step, using the ideal-gas relation, never earlier.The missing symbol in the printed reaction is the equilibrium arrow — the equation is\[\mathrm{C_2H_6(g) \rightleftharpoons C_2H_4(g) + H_2(g)} \]Setting up the pressures. Let the total initial pressure of pure \(\displaystyle \mathrm{C_2H_6}\) be \(\displaystyle P_0 = 4.0\ \text{atm}\), and let \(\displaystyle x\) be the drop in its partial pressure as it dissociates. Since one mole of \(\displaystyle \mathrm{C_2H_6}\) makes one mole each of \(\displaystyle \mathrm{C_2H_4}\) and \(\displaystyle \mathrm{H_2}\), both products rise by the same \(\displaystyle x\):

	\(\displaystyle \mathrm{C_2H_6}\)	\(\displaystyle \mathrm{C_2H_4}\)	\(\displaystyle \mathrm{H_2}\)
Initial (atm)	\(\displaystyle 4.0\)	\(\displaystyle 0\)	\(\displaystyle 0\)
Change (atm)	\(\displaystyle -x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium (atm)	\(\displaystyle 4.0-x\)	\(\displaystyle x\)	\(\displaystyle x\)

Writing \(\displaystyle K_p\). For this reaction, \(\displaystyle K_p = \dfrac{p_{\mathrm{C_2H_4}}\cdot p_{\mathrm{H_2}}}{p_{\mathrm{C_2H_6}}}\), where each \(\displaystyle p\) is the equilibrium partial pressure in atm. Substituting the table values and the given \(\displaystyle K_p = 0.04\ \text{atm}\) at \(\displaystyle 899\ \text{K}\):\[0.04 = \frac{x\cdot x}{4.0 - x} = \frac{x^2}{4.0-x} \]Solving the quadratic. Clearing the denominator:\[x^2 = 0.04(4.0 - x) = 0.16 - 0.04x \] \[x^2 + 0.04x - 0.16 = 0 \]This is \(\displaystyle ax^2+bx+c=0\) with \(\displaystyle a=1,\ b=0.04,\ c=-0.16\), so\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-0.04 \pm \sqrt{(0.04)^2 - 4(1)(-0.16)}}{2} = \frac{-0.04 \pm \sqrt{0.0016 + 0.64}}{2} = \frac{-0.04 \pm \sqrt{0.6416}}{2} \]\[x = \frac{-0.04 \pm 0.8010}{2} \]A partial pressure change can't be negative, so only the \(\displaystyle +\) root is physical:\[x = \frac{-0.04 + 0.8010}{2} = \frac{0.7610}{2} = 0.3805\ \text{atm} \]Equilibrium partial pressure of \(\displaystyle \mathrm{C_2H_6}\).\[p_{\mathrm{C_2H_6}} = 4.0 - x = 4.0 - 0.3805 = 3.6195\ \text{atm} \]Converting pressure to concentration. \(\displaystyle K_p\) and the ICE table only ever gave a pressure — the question asks for a concentration, so this is the step people skip. Use the ideal gas law \(\displaystyle pV = nRT\), rearranged to give molar concentration directly:\[\frac{n}{V} = \frac{p}{RT} \]where \(\displaystyle p\) is the partial pressure (atm), \(\displaystyle R = 0.0821\ \text{L\,atm\,K}^{-1}\text{mol}^{-1}\) is the gas constant, \(\displaystyle T = 899\ \text{K}\) is the given temperature, and \(\displaystyle n/V\) is the molar concentration (mol L\(\displaystyle ^{-1}\)).\[[\mathrm{C_2H_6}] = \frac{p_{\mathrm{C_2H_6}}}{RT} = \frac{3.6195\ \text{atm}}{(0.0821\ \text{L\,atm\,K}^{-1}\text{mol}^{-1})(899\ \text{K})} \]\[RT = 0.0821 \times 899 = 73.81\ \text{L\,atm\,mol}^{-1} \]\[[\mathrm{C_2H_6}] = \frac{3.6195}{73.81}\ \text{mol L}^{-1} = 0.04904\ \text{mol L}^{-1} \]The data (\(\displaystyle K_p = 0.04\), \(\displaystyle P_0 = 4.0\ \text{atm}\)) carries two significant figures, so the concentration rounds to that precision.Answer: \(\displaystyle [\mathrm{C_2H_6}]_{\text{eq}} \approx 4.9 \times 10^{-2}\ \text{mol L}^{-1}\)

\(\displaystyle Q_c\) is the same product-over-reactant ratio as \(\displaystyle K_c\), but taken at any instant — it only tells you whether a system has reached equilibrium once you compare its value to the true \(\displaystyle K_c\).(i) Writing \(\displaystyle Q_c\)The equilibrium is \[\text{CH}_3\text{COOH}(l) + \text{C}_2\text{H}_5\text{OH}(l) \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5(l) + \text{H}_2\text{O}(l) \] The reaction quotient is built exactly like \(\displaystyle K_c\) — products over reactants, each raised to its stoichiometric coefficient (all coefficients here are $\displaystyle 1$): \[Q_c = \dfrac{[\text{CH}_3\text{COOC}_2\text{H}_5]\,[\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}]\,[\text{C}_2\text{H}_5\text{OH}]} \] where each bracket is a molar concentration (mol L\(\displaystyle ^{-1}\)) at the instant you evaluate it — not necessarily at equilibrium. The question's note that water is "not in excess and not a solvent" is exactly why \(\displaystyle [\text{H}_2\text{O}]\) is written out here — if water were the solvent (as in an aqueous acid-base equilibrium), its concentration would stay essentially constant and get absorbed into \(\displaystyle K_c\) instead of appearing on its own.(ii) Finding \(\displaystyle K_c\) at $\displaystyle 293$ KAll four substances share one flask, so every concentration equals (moles)/\(\displaystyle V\) for the same volume \(\displaystyle V\). In the \(\displaystyle Q_c\) ratio, \(\displaystyle V\) appears twice on top and twice on the bottom, so it cancels completely — moles can be used directly in place of concentrations without changing the numerical value of \(\displaystyle K_c\).Let \(\displaystyle x\) be the moles of ester formed: You are told the equilibrium ester amount is $\displaystyle 0.171$ mol, so \(\displaystyle x = 0.171\): The ethanol is the step people slip on — you start with only $\displaystyle 0.18$ mol of it, so after $\displaystyle 0.171$ mol reacts, just $\displaystyle 0.009$ mol survives. That is a subtraction of two nearly equal numbers, so keep all three decimal places; rounding $\displaystyle 0.18$ or $\displaystyle 0.171$ even slightly changes this small leftover a lot, and it changes \(\displaystyle K_c\) by that same large fraction.Substitute into \(\displaystyle K_c = Q_c\) at equilibrium: \[K_c = \frac{(0.171)(0.171)}{(0.829)(0.009)} = \frac{0.029241}{0.007461} \] \[K_c \approx 3.92 \] Both sides of the reaction have two moles of species ($\displaystyle 1$ + $\displaystyle 1$ reactants, $\displaystyle 1$ + $\displaystyle 1$ products), so the concentration units cancel exactly in the ratio — \(\displaystyle K_c\) here is a pure number, with no units to carry.(iii) Has the second mixture reached equilibrium?This run starts from different amounts — $\displaystyle 1.0$ mol acetic acid and $\displaystyle 0.5$ mol ethanol — and reports $\displaystyle 0.214$ mol of ester "after sometime," language that should prompt you to check rather than assume equilibrium has been reached.At this instant: Compute \(\displaystyle Q_c\) for this snapshot (moles again stand in for concentrations, by the same volume-cancellation as in part (ii)): \[Q_c = \frac{(0.214)(0.214)}{(0.786)(0.286)} = \frac{0.045796}{0.224796} \approx 0.204 \]A system sits at equilibrium only when \(\displaystyle Q_c = K_c\). Here \(\displaystyle Q_c \approx 0.204\) is far below \(\displaystyle K_c \approx 3.92\) — nowhere close to equal — so this mixture has not reached equilibrium. Because \(\displaystyle Q_c < K_c\), the numerator (ester and water) is still too small relative to the denominator (acid and alcohol); the forward, ester-forming reaction keeps running, raising \(\displaystyle Q_c\) until it climbs up to 3.92.Answer: (i) \(\displaystyle Q_c = \dfrac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}\); (ii) \(\displaystyle K_c \approx 3.92\) (unitless); (iii) No — \(\displaystyle Q_c \approx 0.204 \ne K_c \approx 3.92\), so equilibrium has not been reached, and the forward reaction continues.

At equilibrium, PCl₃ and Cl₂ are produced in a strict $\displaystyle 1$ : $\displaystyle 1$ ratio, so both concentrations can be written as the same unknown \(\displaystyle x\).The reaction is a genuine equilibrium (both directions occur at once), so the missing symbol between reactant and products is the equilibrium arrow, not a plus sign or an omission: \[\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]Set up an ICE table. Pure PCl₅ was placed in the vessel, so at the start there is no PCl₃ or Cl₂ — every mole of each that appears has come from the decomposition of PCl₅, one mole of PCl₅ giving one mole of PCl₃ and one mole of Cl₂.

	\(\displaystyle \text{PCl}_5\)	\(\displaystyle \text{PCl}_3\)	\(\displaystyle \text{Cl}_2\)
Initial	\(\displaystyle c_0\)	$\displaystyle 0$	$\displaystyle 0$
Change	\(\displaystyle -x\)	\(\displaystyle +x\)	\(\displaystyle +x\)
Equilibrium	\(\displaystyle c_0-x\)	\(\displaystyle x\)	\(\displaystyle x\)

The problem already gives the equilibrium value of PCl₅ directly, so \(\displaystyle c_0-x\) does not need to be found separately: \[[\text{PCl}_5]_{eq} = 0.5\times10^{-1}\ \text{mol L}^{-1} = 5.0\times10^{-2}\ \text{mol L}^{-1} \] and, by the stoichiometry above, \[[\text{PCl}_3]_{eq} = [\text{Cl}_2]_{eq} = x \]Write the equilibrium constant expression. \(\displaystyle K_c\) is the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient (here all coefficients are $\displaystyle 1$, and PCl₅ being a gas still appears because this is \(\displaystyle K_c\), not \(\displaystyle K_p\)): \[K_c=\dfrac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]This is the step it's easy to get wrong: because \(\displaystyle [\text{PCl}_3]=[\text{Cl}_2]=x\), the numerator is \(\displaystyle x\times x = x^{2}\), not \(\displaystyle 2x\) — the two concentrations are multiplied together, not added.Substitute the known numbers. \[8.3\times10^{-3}=\dfrac{x\times x}{5.0\times10^{-2}}=\dfrac{x^{2}}{5.0\times10^{-2}} \]Solve for \(\displaystyle x\). Multiply both sides by \(\displaystyle 5.0\times10^{-2}\ \text{mol L}^{-1}\): \[x^{2}=8.3\times10^{-3}\times5.0\times10^{-2}\ \text{mol}^2\text{L}^{-2}=4.15\times10^{-4}\ \text{mol}^2\text{L}^{-2} \]Take the square root of both sides (units come along under the root, so \(\displaystyle \text{mol}^2\text{L}^{-2}\) becomes \(\displaystyle \text{mol L}^{-1}\)): \[x=\sqrt{4.15\times10^{-4}\ \text{mol}^2\text{L}^{-2}} = 2.04\times10^{-2}\ \text{mol L}^{-1} \](only the positive root is kept, since \(\displaystyle x\) is a concentration and cannot be negative)The data (\(\displaystyle K_c\) to two significant figures) justifies quoting the result to three significant figures at most, so it is left as \(\displaystyle 2.04\times10^{-2}\ \text{mol L}^{-1}\) without further rounding.Both PCl₃ and Cl₂ carry this same value, since the reaction produces them in equal amounts: \[[\text{PCl}_3]_{eq}=[\text{Cl}_2]_{eq}=2.04\times10^{-2}\ \text{mol L}^{-1} \]Answer: \(\displaystyle [\text{PCl}_3] = [\text{Cl}_2] = 2.04\times10^{-2}\ \text{mol L}^{-1}\)

\(\displaystyle K_p\) only involves the gases — solids like FeO(s) and Fe(s) never appear in the equilibrium expression.For \[\text{FeO (s)} + \text{CO (g)} \rightleftharpoons \text{Fe (s)} + \text{CO}_2\text{ (g)} \] the equilibrium constant in terms of pressure is \[K_p = \frac{p_{\text{CO}_2}}{p_{\text{CO}}} \] where \(\displaystyle p_{\text{CO}_2}\) and \(\displaystyle p_{\text{CO}}\) are the partial pressures of the two gases at equilibrium (the pure solids FeO and Fe are left out because their activity is taken as $\displaystyle 1$).Step $\displaystyle 1$: Check which way the reaction has to run.Before assuming the system is already at equilibrium, compare the given pressures with \(\displaystyle K_p\) by forming the reaction quotient \(\displaystyle Q_p\): \[Q_p = \frac{p_{\text{CO}_2}}{p_{\text{CO}}} = \frac{0.80\ \text{atm}}{1.4\ \text{atm}} = 0.571 \] Since \(\displaystyle Q_p (0.571) > K_p (0.265)\), the system has too much \(\displaystyle \text{CO}_2\) relative to CO for equilibrium. The reaction must run in the reverse direction, converting some \(\displaystyle \text{CO}_2\) back into CO, until \(\displaystyle Q_p\) drops to 0.265.This is the step people skip — jumping straight to an ICE table without checking \(\displaystyle Q_p\) vs \(\displaystyle K_p\) can leave you assuming the wrong direction of change, which flips the sign of \(\displaystyle x\).Step $\displaystyle 2$: Set up the ICE table.Let \(\displaystyle x\) (atm) be the drop in \(\displaystyle p_{\text{CO}_2}\) as the reverse reaction proceeds; the same amount of CO is regenerated ($\displaystyle 1$:$\displaystyle 1$ stoichiometry).

	\(\displaystyle p_{\text{CO}}\)	\(\displaystyle p_{\text{CO}_2}\)
Initial	\(\displaystyle 1.4\)	\(\displaystyle 0.80\)
Change	\(\displaystyle +x\)	\(\displaystyle -x\)
Equilibrium	\(\displaystyle 1.4+x\)	\(\displaystyle 0.80-x\)

Step $\displaystyle 3$: Substitute into \(\displaystyle K_p\) and solve for \(\displaystyle x\).\[K_p = \frac{0.80 - x}{1.4 + x} = 0.265 \]Cross-multiplying: \[0.80 - x = 0.265(1.4+x) \] \[0.80 - x = 0.371 + 0.265x \] \[0.80 - 0.371 = x + 0.265x \] \[0.429 = 1.265x \] \[x = \frac{0.429}{1.265\ } = 0.339\ \text{atm} \]Step $\displaystyle 4$: Get the equilibrium partial pressures.\[p_{\text{CO}} = 1.4 + 0.339 = 1.739\ \text{atm} \approx 1.74\ \text{atm} \] \[p_{\text{CO}_2} = 0.80 - 0.339 = 0.461\ \text{atm} \approx 0.46\ \text{atm} \]Check: \(\displaystyle \dfrac{0.461}{1.739} = 0.265\), which matches the given \(\displaystyle K_p\), confirming the values are consistent.The data (two significant figures in the initial pressures) justifies rounding the final pressures to three significant figures.Answer: \(\displaystyle p_{\text{CO}} \approx 1.74\ \text{atm}\) and \(\displaystyle p_{\text{CO}_2} \approx 0.46\ \text{atm}\) at equilibrium.