NCERT Solutions · Class 11 Chemistry Equilibrium

Vapour pressure of a pure liquid depends only on temperature, not on the volume of the container — so once equilibrium is restored, the pressure comes back to exactly where it started. The trick in this question is not to confuse the momentary dip right after the container expands with the final state after equilibrium re-establishes itself — these are two different things and the question asks about both.(a) Initial effect on vapour pressureAt the instant the volume is increased, no new liquid has had time to evaporate yet — the same number of vapour molecules, \(\displaystyle n\), are now spread through a larger volume \(\displaystyle V\). Treating the vapour as an ideal gas, \[PV = nRT \quad\Rightarrow\quad P = \frac{nRT}{V} \] where \(\displaystyle P\) is the vapour pressure, \(\displaystyle n\) is the moles of vapour present, \(\displaystyle R\) is the gas constant, \(\displaystyle T\) is the (constant) temperature, and \(\displaystyle V\) is the volume of the vapour space.Since \(\displaystyle n\) and \(\displaystyle T\) haven't changed yet, but \(\displaystyle V\) has suddenly gone up, \(\displaystyle P\) must drop immediately — the same amount of gas is simply less crowded. This is the instantaneous effect, before evaporation has a chance to respond.The step people get wrong: it's tempting to say "vapour pressure only depends on temperature, so nothing changes." That's true only once equilibrium is re-established — right after the disturbance, before evaporation catches up, the pressure genuinely falls.(b) Rates of evaporation and condensation, immediately after So immediately after the expansion, rate of evaporation \(\displaystyle >\) rate of condensation — the two rates are no longer equal, and the system is out of equilibrium in the direction of net evaporation.(c) Restoring equilibrium and the final vapour pressureBecause evaporation now outpaces condensation, liquid keeps evaporating into the larger space. As it does, the vapour's concentration — and hence its pressure — rises back up, which in turn raises the rate of condensation. This continues until the rate of condensation again equals the rate of evaporation, at which point dynamic equilibrium is restored.Since temperature hasn't changed anywhere in this process, and the vapour pressure of a liquid in equilibrium with its vapour depends only on temperature (provided enough liquid remains to keep supplying vapour), the final equilibrium vapour pressure comes back to the same value it had before the volume was increased — the extra space just gets filled with more vapour until the original pressure is reached again.Answer: (a) Vapour pressure decreases immediately after the volume increase. (b) Rate of evaporation is unchanged; rate of condensation decreases, so evaporation outpaces condensation. (c) Net evaporation continues until the two rates become equal again, and the final vapour pressure equals the original (initial) vapour pressure, since it depends only on temperature.

\(\displaystyle K_c\) is built straight from the balanced equation — each concentration raised to its own stoichiometric coefficient, and the reaction must be read as reversible. The reactants and products here run together with no symbol between them because the arrow got lost in transcription; the equilibrium is\[2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \]For a general equilibrium \(\displaystyle aA + bB \rightleftharpoons cC\), the equilibrium constant is\[K_c = \frac{[\text{C}]^{c}}{[\text{A}]^{a}[\text{B}]^{b}} \]where each bracket is the equilibrium concentration (mol L\(\displaystyle ^{-1}\)) of that species, and the exponent is its coefficient in the balanced equation — not the number "$\displaystyle 2$" in front of SO\(\displaystyle _2\) meaning "multiply by $\displaystyle 2$," but "raise to the power 2."For this reaction, SO\(\displaystyle _3\) is the product (coefficient $\displaystyle 2$) and SO\(\displaystyle _2\), O\(\displaystyle _2\) are reactants (coefficients $\displaystyle 2$ and $\displaystyle 1$):\[K_c = \frac{[\text{SO}_3]^{2}}{[\text{SO}_2]^{2}[\text{O}_2]} \]Substituting the given equilibrium concentrations — \(\displaystyle [\text{SO}_2] = 0.60\ \text{M}\), \(\displaystyle [\text{O}_2] = 0.82\ \text{M}\), \(\displaystyle [\text{SO}_3] = 1.90\ \text{M}\):\[K_c = \frac{(1.90\ \text{M})^{2}}{(0.60\ \text{M})^{2}(0.82\ \text{M})} \]Work out the powers first, keeping units attached at every step:\[(1.90\ \text{M})^{2} = 3.61\ \text{M}^{2} \] \[(0.60\ \text{M})^{2} = 0.36\ \text{M}^{2} \]Then multiply the denominator terms:\[(0.36\ \text{M}^{2})(0.82\ \text{M}) = 0.2952\ \text{M}^{3} \]Now divide:\[K_c = \frac{3.61\ \text{M}^{2}}{0.2952\ \text{M}^{3}} = 12.229...\ \text{M}^{-1} \]Watch the units here — this is the step people skip. \(\displaystyle K_c\) is not automatically unitless. Its units come from the net power of concentration left over after the exponents cancel: on top there are $\displaystyle 2$ powers of M, on the bottom there are \(\displaystyle 2+1=3\) powers of M, leaving \(\displaystyle \text{M}^{2}/\text{M}^{3} = \text{M}^{-1} = \text{mol}^{-1}\,\text{L}\). Only when the total moles of gas are the same on both sides of the equation does \(\displaystyle K_c\) come out dimensionless — that is not the case here ($\displaystyle 3$ mol of gas → $\displaystyle 2$ mol of gas).Rounding once, at the end: the two least-precise inputs, \(\displaystyle 0.60\ \text{M}\) and \(\displaystyle 0.82\ \text{M}\), each carry only two significant figures, so the answer is honestly known to two significant figures:\[K_c \approx 12\ \text{mol}^{-1}\,\text{L} \]Answer: \(\displaystyle K_c \approx 12\ \text{mol}^{-1}\,\text{L}\) (unrounded, \(\displaystyle 12.229\ \text{mol}^{-1}\,\text{L}\)).

For gases mixed at the same total pressure, each gas's partial pressure is its volume fraction times the total pressure — volume fraction and mole fraction are the same thing for an ideal gas.The equilibrium is\[\text{I}_2(g) \rightleftharpoons 2\text{I}(g) \]Iodine vapour is reported as $\displaystyle 40$% I atoms by volume, so the rest — $\displaystyle 60$% — is undissociated \(\displaystyle \text{I}_2\).Step $\displaystyle 1$: Convert volume percentages to partial pressures.For a gas mixture, partial pressure of a component \(\displaystyle p_i\) is given by\[p_i = x_i \times P_{\text{total}} \]where \(\displaystyle x_i\) is the mole (= volume) fraction of that component and \(\displaystyle P_{\text{total}}\) is the total pressure of the mixture.Here \(\displaystyle P_{\text{total}} = 10^{5}\ \text{Pa}\), \(\displaystyle x_{\text{I}} = 0.40\), and \(\displaystyle x_{\text{I}_2} = 0.60\).\[p_{\text{I}} = 0.40 \times 10^{5}\ \text{Pa} = 4\times10^{4}\ \text{Pa} \]\[p_{\text{I}_2} = 0.60 \times 10^{5}\ \text{Pa} = 6\times10^{4}\ \text{Pa} \]The step people skip: a mixture's total pressure is the sum of the partial pressures of every species present — you cannot use the total pressure itself in place of either partial pressure in the equilibrium expression.Step $\displaystyle 2$: Write \(\displaystyle K_p\) for the reaction.For \(\displaystyle \text{I}_2(g) \rightleftharpoons 2\text{I}(g)\), the equilibrium constant in terms of pressure is\[K_p = \frac{(p_{\text{I}})^{2}}{p_{\text{I}_2}} \]with \(\displaystyle p_{\text{I}}\) and \(\displaystyle p_{\text{I}_2}\) the equilibrium partial pressures of I(g) and \(\displaystyle \text{I}_2\)(g) found above. The exponent $\displaystyle 2$ on \(\displaystyle p_{\text{I}}\) comes directly from the stoichiometric coefficient of I(g) in the balanced equation — this is the mistake to watch for: forgetting to square the product's pressure because its coefficient is $\displaystyle 2$, not 1.Step $\displaystyle 3$: Substitute and evaluate.\[K_p = \frac{\left(4\times10^{4}\ \text{Pa}\right)^{2}}{6\times10^{4}\ \text{Pa}} = \frac{16\times10^{8}\ \text{Pa}^{2}}{6\times10^{4}\ \text{Pa}} \]\[K_p = \frac{16}{6}\times10^{8-4}\ \text{Pa} = 2.667\times10^{4}\ \text{Pa} \]Since the given data ($\displaystyle 40$%) carries about three significant figures at best, the result is rounded to three significant figures.Answer: \(\displaystyle K_p \approx 2.67 \times 10^{4}\ \text{Pa}\)

\(\displaystyle K_p \) and \(\displaystyle K_c \) are linked by \(\displaystyle K_p = K_c(RT)^{\Delta n} \), and only GASEOUS species count in \(\displaystyle \Delta n \) — solids drop out of the expression entirely.Here \(\displaystyle K_p \) is the equilibrium constant written in terms of partial pressures (in bar), \(\displaystyle K_c \) is the equilibrium constant written in terms of molar concentrations, \(\displaystyle R = 0.0831\ \text{bar L K}^{-1}\text{mol}^{-1} \) is the gas constant (matched to the bar unit \(\displaystyle K_p \) is given in), \(\displaystyle T \) is the absolute temperature in kelvin, and \(\displaystyle \Delta n \) is (moles of gaseous products) \(\displaystyle - \) (moles of gaseous reactants), read straight off the balanced equation — the arrow missing from the printed equations here is a reversible arrow, \(\displaystyle \rightleftharpoons \), since every species is present at equilibrium simultaneously.Solving the relation for \(\displaystyle K_c \): \[K_c = \dfrac{K_p}{(RT)^{\Delta n}} \]Part (i): \(\displaystyle 2\text{NOCl(g)} \rightleftharpoons 2\text{NO(g)} + \text{Cl}_2\text{(g)} \), \(\displaystyle K_p = 1.8\times10^{-2} \) at \(\displaystyle T = 500\ \text{K} \)Every species here is a gas, so \(\displaystyle \Delta n \) is just (moles of gas on the product side) minus (moles of gas on the reactant side): \[\Delta n = (2 + 1) - 2 = 1 \]Substituting into \(\displaystyle K_c = \dfrac{K_p}{(RT)^{\Delta n}} \): \[K_c = \dfrac{1.8\times10^{-2}}{(0.0831 \times 500)^{1}} = \dfrac{1.8\times10^{-2}}{41.55} \]\[K_c = 4.33\times10^{-4}\ \text{mol L}^{-1} \]Part (ii): \(\displaystyle \text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)} \), \(\displaystyle K_p = 167 \) at \(\displaystyle T = 1073\ \text{K} \)This is the step people get wrong: CaCO\(\displaystyle _3\)(s) and CaO(s) are pure solids, and a pure solid has fixed "concentration" (its density divided by its molar mass), so it never appears in an equilibrium-constant expression at all. Only CO\(\displaystyle _2\)(g) is counted — not zero (from "everything cancels"), and not two (from counting all three formulas written in the equation): \[\Delta n = 1 - 0 = 1 \]Substituting into \(\displaystyle K_c = \dfrac{K_p}{(RT)^{\Delta n}} \): \[K_c = \dfrac{167}{(0.0831 \times 1073)^{1}} = \dfrac{167}{89.17} \]\[K_c = 1.87\ \text{mol L}^{-1} \]Both results are rounded only at this last step, to three significant figures, matching the precision of the given \(\displaystyle K_p \) values and temperatures.Answer: \(\displaystyle K_c = 4.33\times10^{-4}\ \text{mol L}^{-1} \) for (i), and \(\displaystyle K_c = 1.87\ \text{mol L}^{-1} \) for (ii)

Reversing an equilibrium flips its equilibrium constant — the new \(\displaystyle K_c\) is the reciprocal of the old one.The equilibrium is\[\text{NO (g)} + \text{O}_3\text{ (g)} \rightleftharpoons \text{NO}_2\text{ (g)} + \text{O}_2\text{ (g)} \]with the forward reaction having\[K_c = 6.3 \times 10^{14} \text{ at } 1000\text{ K} \]For any equilibrium, the law of mass action gives the forward constant as\[K_c(\text{forward}) = \frac{[\text{NO}_2][\text{O}_2]}{[\text{NO}][\text{O}_3]} \]where the square brackets denote equilibrium molar concentrations of each species.The reverse reaction is the same equilibrium written the other way round,\[\text{NO}_2\text{ (g)} + \text{O}_2\text{ (g)} \rightleftharpoons \text{NO (g)} + \text{O}_3\text{ (g)} \]so its equilibrium constant expression is just the reciprocal of the ratio above:\[K_c(\text{reverse}) = \frac{[\text{NO}][\text{O}_3]}{[\text{NO}_2][\text{O}_2]} = \frac{1}{K_c(\text{forward})} \]This reciprocal relationship holds for any equilibrium — it does not depend on whether the reactions are elementary steps or not. The extra fact given here, that both the forward and reverse steps are elementary bimolecular reactions, is what lets you also write each rate law directly from its stoichiometry (rate \(\displaystyle = k_f[\text{NO}][\text{O}_3]\) forward, rate \(\displaystyle = k_b[\text{NO}_2][\text{O}_2]\) reverse, with \(\displaystyle K_c = k_f/k_b\)) — but you don't need the rate constants at all to answer this question, since \(\displaystyle K_c(\text{reverse})\) is fixed the moment \(\displaystyle K_c(\text{forward})\) is known.A common slip here is to think a "different" reaction needs a fresh calculation from scratch. It doesn't — reversing a reaction only inverts the constant.Substituting the given value:\[K_c(\text{reverse}) = \frac{1}{6.3 \times 10^{14}} \]\[K_c(\text{reverse}) = 0.1587 \times 10^{-14} = 1.587 \times 10^{-15} \]The given \(\displaystyle K_c\) has $\displaystyle 2$ significant figures, so the result is rounded to $\displaystyle 2$ significant figures:\[K_c(\text{reverse}) \approx 1.6 \times 10^{-15} \]Answer: \(\displaystyle K_c\text{(reverse)} = 1.6 \times 10^{-15}\)

The concentration of a pure solid or a pure liquid is a fixed number, not a variable — so it never actually appears as a "moving part" in the equilibrium expression.Concentration is defined as\[c = \dfrac{n}{V} = \dfrac{\text{mass}}{\text{molar mass} \times V} \]where \(\displaystyle n\) is moles, \(\displaystyle V\) is volume, and mass/\(\displaystyle V\) is density. For a pure solid or a pure liquid, density and molar mass are properties of the substance itself at a given temperature — they do not depend on how much of the solid or liquid is sitting in the flask. Whether there is $\displaystyle 1$ g or $\displaystyle 100$ g of a pure solid present, its concentration (moles per unit volume of that solid) works out to the same value, because both the mass and the volume it occupies scale together. So\[c_{\text{pure solid/liquid}} = \dfrac{\text{density}}{\text{molar mass}} = \text{constant at a given }T \]This is the step people trip on: concentration is only a meaningful, changing quantity for something whose amount can vary independently of its volume — true for gases and for solutes in a solution, but not true for a pure solid or liquid, which simply occupies whatever volume its fixed density dictates.Take the decomposition of calcium carbonate: \[\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]Written naively, the equilibrium constant would be\[K_c' = \dfrac{[\text{CaO}][\text{CO}_2]}{[\text{CaCO}_3]} \]But \(\displaystyle [\text{CaO}]\) and \(\displaystyle [\text{CaCO}_3]\) are both constants (fixed density/molar mass values, unaffected by how much of each solid is present). Since \(\displaystyle K_c'\) is itself a constant at a given temperature, the two solid-concentration constants can be multiplied through and absorbed into it, defining a new constant:\[K_c = K_c' \times \dfrac{[\text{CaCO}_3]}{[\text{CaO}]} = [\text{CO}_2] \]So the equilibrium constant expression is simply \(\displaystyle \mathrm{K_c = [\text{CO}_2]^{-}}\) the pure solids drop out not because their concentrations are zero or unimportant to the chemistry, but because their concentrations are constants that get folded into the value of \(\displaystyle K\) itself rather than appearing as variables in it. The same reasoning applies to any pure liquid taking part in a reaction (for example, \(\displaystyle \text{H}_2\text{O}(l)\) in a reaction where water is a reactant/product but not the solvent).Answer: Pure solids and pure liquids have a fixed concentration (density ÷ molar mass) at a given temperature, independent of how much of them is present. Since this constant value would only multiply into the already-constant \(\displaystyle K\), it is absorbed into \(\displaystyle K\) itself rather than written explicitly — so only the concentrations of gases and dissolved species, whose concentrations actually vary, appear in the equilibrium constant expression.

A vanishingly small \(\displaystyle K_c\) is the answer before you even solve for \(\displaystyle x\) — it tells you the reaction barely moves forward, so almost all of the \(\displaystyle \mathrm{N_2}\) and \(\displaystyle \mathrm{O_2}\) you started with is still there at equilibrium. The missing symbol in the printed reaction is a reversible arrow: this is an equilibrium, not a one-way reaction.\[2\mathrm{N_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{N_2O(g)}, \qquad K_c = 2.0\times10^{-37} \]Step $\displaystyle 1$ — turn moles into concentrations. Concentration is moles per litre of the reaction mixture's volume, \(\displaystyle c = n/V\), not per mole of anything else:\[[\mathrm{N_2}]_0=\dfrac{0.482\ \text{mol}}{10\ \text{L}}=0.0482\ \text{mol L}^{-1},\qquad [\mathrm{O_2}]_0=\dfrac{0.933\ \text{mol}}{10\ \text{L}}=0.0933\ \text{mol L}^{-1},\qquad [\mathrm{N_2O}]_0=0 \]Step $\displaystyle 2$ — set up the ICE table. Let \(\displaystyle x\) be the mol L\(\displaystyle ^{-1}\) of \(\displaystyle \mathrm{O_2}\) consumed (its coefficient is $\displaystyle 1$, so the changes in the other species are just multiples of \(\displaystyle x\)):

	\(\displaystyle \mathrm{N_2}\)	\(\displaystyle \mathrm{O_2}\)	\(\displaystyle \mathrm{N_2O}\)
Initial	\(\displaystyle 0.0482\)	\(\displaystyle 0.0933\)	\(\displaystyle 0\)
Change	\(\displaystyle -2x\)	\(\displaystyle -x\)	\(\displaystyle +2x\)
Equilibrium	\(\displaystyle 0.0482-2x\)	\(\displaystyle 0.0933-x\)	\(\displaystyle 2x\)

Step $\displaystyle 3$ — write \(\displaystyle K_c\), matching every exponent to its stoichiometric coefficient. This is the step people get wrong: \(\displaystyle \mathrm{N_2}\) and \(\displaystyle \mathrm{N_2O}\) both carry coefficient $\displaystyle 2$, so both get squared, while \(\displaystyle \mathrm{O_2}\) has coefficient $\displaystyle 1$ and stays unsquared.\[K_c=\frac{[\mathrm{N_2O}]^2}{[\mathrm{N_2}]^2[\mathrm{O_2}]}=\frac{(2x)^2}{(0.0482-2x)^2(0.0933-x)}=2.0\times10^{-37} \]Step $\displaystyle 4$ — exploit how small \(\displaystyle K_c\) is. A \(\displaystyle K_c\) of \(\displaystyle 10^{-37}\) means the forward reaction essentially does not happen; \(\displaystyle x\) will turn out to be many orders of magnitude smaller than \(\displaystyle 0.0482\) or \(\displaystyle 0.0933\). So approximate \(\displaystyle 0.0482-2x\approx0.0482\) and \(\displaystyle 0.0933-x\approx0.0933\) — an approximation that is only safe because \(\displaystyle K_c\) is this tiny; you would never drop \(\displaystyle x\) like this if \(\displaystyle K_c\) were close to 1.\[\frac{(2x)^2}{(0.0482)^2(0.0933)}=2.0\times10^{-37} \]Step $\displaystyle 5$ — solve for \(\displaystyle x\).\[(0.0482)^2(0.0933)=(2.323\times10^{-3})(0.0933)=2.168\times10^{-4} \]\[(2x)^2=2.0\times10^{-37}\times2.168\times10^{-4}=4.335\times10^{-41} \]\[2x=\sqrt{4.335\times10^{-41}}=6.58\times10^{-21}\ \text{mol L}^{-1} \qquad\Rightarrow\qquad x=3.29\times10^{-21}\ \text{mol L}^{-1} \]Checking the approximation: \(\displaystyle x\sim10^{-21}\) is indeed utterly negligible next to \(\displaystyle 0.0482\) and \(\displaystyle 0.0933\), so dropping it was justified.Step $\displaystyle 6$ — read off the equilibrium composition.\[[\mathrm{N_2O}] = 2x = 6.6\times10^{-21}\ \text{mol L}^{-1} \quad\Rightarrow\quad n(\mathrm{N_2O}) = 6.6\times10^{-21}\ \text{mol L}^{-1}\times10\ \text{L} = 6.6\times10^{-20}\ \text{mol} \]\[[\mathrm{N_2}] \approx 0.0482\ \text{mol L}^{-1} \Rightarrow n(\mathrm{N_2})\approx0.482\ \text{mol (unchanged to 3 s.f.)} \]\[[\mathrm{O_2}] \approx 0.0933\ \text{mol L}^{-1} \Rightarrow n(\mathrm{O_2})\approx0.933\ \text{mol (unchanged to 3 s.f.)} \]The two sig figs in the final \(\displaystyle x\) come from the two sig figs given in \(\displaystyle K_c\); the physical picture is that with \(\displaystyle K_c\) this small, the "equilibrium mixture" is, for all practical purposes, still just the starting \(\displaystyle 0.482\ \text{mol}\ \mathrm{N_2}\) and \(\displaystyle 0.933\ \text{mol}\ \mathrm{O_2}\), with only a trace — about \(\displaystyle 6.6\times10^{-20}\ \text{mol}\) — of \(\displaystyle \mathrm{N_2O}\) formed.Answer: at equilibrium, \(\displaystyle n(\mathrm{N_2})\approx0.482\ \text{mol}\), \(\displaystyle n(\mathrm{O_2})\approx0.933\ \text{mol}\) (both essentially unchanged), and \(\displaystyle n(\mathrm{N_2O})\approx6.6\times10^{-20}\ \text{mol}\) — a negligible amount, consistent with \(\displaystyle K_c=2.0\times10^{-37}\).

Use the reaction stoichiometry to convert "how much product formed" into "how much reactant was used up" — the mole ratios in the balanced equation are the conversion factors.The reaction is\[2\text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2\text{NOBr}(g) \]Because the coefficients are \(\displaystyle 2 : 1 : 2\), every $\displaystyle 2$ mol of NOBr formed corresponds to $\displaystyle 2$ mol of NO consumed and $\displaystyle 1$ mol of \(\displaystyle \text{Br}_2\) consumed. In other words, moles of NO used up equal moles of NOBr formed (both carry coefficient $\displaystyle 2$), while moles of \(\displaystyle \text{Br}_2\) used up are half the moles of NOBr formed.Set up an ICE table (Initial, Change, Equilibrium) in moles.Let \(\displaystyle x\) = mol of \(\displaystyle \text{Br}_2\) consumed. Then, by the \(\displaystyle 2:1:2\) ratio, \(\displaystyle 2x\) mol of NO is consumed and \(\displaystyle 2x\) mol of NOBr is formed.\[\begin{array}{c|ccc} & 2\text{NO} & \text{Br}_2 & 2\text{NOBr} \\\hline \text{Initial (mol)} & 0.087 & 0.0437 & 0 \\ \text{Change (mol)} & -2x & -x & +2x \\ \text{Equilibrium (mol)} & 0.087-2x & 0.0437-x & 2x \end{array} \]Use the given equilibrium amount of NOBr to find \(\displaystyle x\).At equilibrium, NOBr = $\displaystyle 0.0518$ mol, so\[2x = 0.0518 \text{ mol} \quad\Rightarrow\quad x = \frac{0.0518}{2} \text{ mol} = 0.0259 \text{ mol} \]This \(\displaystyle x\) is the moles of \(\displaystyle \text{Br}_2\) consumed; \(\displaystyle 2x = 0.0518\) mol is the moles of NO consumed — note that the NO consumed is numerically equal to the NOBr formed only because both have the same stoichiometric coefficient ($\displaystyle 2$). This is the step people get wrong: they sometimes halve the NOBr value again for NO, but NO and NOBr share the same coefficient, so no halving is needed for NO — the halving applies only to \(\displaystyle \text{Br}_2\).Subtract the amount consumed from the initial amount to get the equilibrium amount of each reactant.For NO: \[n(\text{NO})_{eq} = 0.087 \text{ mol} - 2x = 0.087 \text{ mol} - 0.0518 \text{ mol} = 0.0352 \text{ mol} \]For \(\displaystyle \text{Br}_2\): \[n(\text{Br}_2)_{eq} = 0.0437 \text{ mol} - x = 0.0437 \text{ mol} - 0.0259 \text{ mol} = 0.0178 \text{ mol} \]Both results are reported to three significant figures, matching the precision of the given data ($\displaystyle 0.087$, $\displaystyle 0.0437$, $\displaystyle 0.0518$ mol).Answer: equilibrium amount of NO \(\displaystyle = 0.0352\) mol, equilibrium amount of \(\displaystyle \text{Br}_2\) \(\displaystyle = 0.0178\) mol.

The relationship between \(\displaystyle K_p\) and \(\displaystyle K_c\) hinges on \(\displaystyle \Delta n_g\), the change in moles of gas as reactants become products — get that number wrong (or its sign) and every later step flips.The reaction, written with the reversible arrow that belongs at equilibrium, is\[2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]Step $\displaystyle 1$ — Relate \(\displaystyle K_p\) and \(\displaystyle K_c\).For a gas-phase equilibrium, \[K_p = K_c (RT)^{\Delta n_g} \] where \(\displaystyle K_p\) is the equilibrium constant in terms of partial pressures, \(\displaystyle K_c\) is the equilibrium constant in terms of molar concentrations, \(\displaystyle R\) is the gas constant, \(\displaystyle T\) is the absolute temperature, and \(\displaystyle \Delta n_g\) is (moles of gaseous products) − (moles of gaseous reactants) in the balanced equation.Step $\displaystyle 2$ — Find \(\displaystyle \Delta n_g\).From the balanced equation, gaseous products = $\displaystyle 2$ mol (\(\displaystyle SO_3\)), and gaseous reactants = $\displaystyle 2$ + $\displaystyle 1$ = $\displaystyle 3$ mol (\(\displaystyle SO_2\) and \(\displaystyle O_2\)): \[\Delta n_g = 2 - (2+1) = -1 \]A negative \(\displaystyle \Delta n_g\) just says the reaction consumes more moles of gas than it makes — nothing to be alarmed about, but it is exactly the number that must carry its minus sign into the next step.Step $\displaystyle 3$ — Solve for \(\displaystyle K_c\).Rearranging \(\displaystyle K_p = K_c(RT)^{\Delta n_g}\) for \(\displaystyle K_c\): \[K_c = \dfrac{K_p}{(RT)^{\Delta n_g}} = K_p (RT)^{-\Delta n_g} \] Since \(\displaystyle \Delta n_g = -1\), the exponent \(\displaystyle -\Delta n_g = 1\), so this simplifies to \[K_c = K_p \times RT \]This is the step people get backwards: with \(\displaystyle \Delta n_g\) negative, dividing by \(\displaystyle (RT)^{\Delta n_g}\) is the same as multiplying by \(\displaystyle RT\) — not dividing by it.Step $\displaystyle 4$ — Substitute the numbers.Because \(\displaystyle K_p\) here is built from pressures in bar, \(\displaystyle R\) must be used in the matching units: \[R = 0.0831 \ \text{bar dm}^3\text{K}^{-1}\text{mol}^{-1}, \qquad T = 450\ \text{K}, \qquad K_p = 2.0 \times 10^{10}\ \text{bar}^{-1} \]First find \(\displaystyle RT\): \[RT = 0.0831 \ \text{bar dm}^3\text{K}^{-1}\text{mol}^{-1} \times 450\ \text{K} = 37.395\ \text{bar dm}^3\text{mol}^{-1} \]Now substitute into \(\displaystyle K_c = K_p \times RT\): \[K_c = \left(2.0 \times 10^{10}\ \text{bar}^{-1}\right) \times \left(37.395\ \text{bar dm}^3\text{mol}^{-1}\right) \]The bar and \(\displaystyle \text{bar}^{-1}\) cancel, leaving units of \(\displaystyle \text{dm}^3\text{mol}^{-1}\) — exactly what is expected for \(\displaystyle K_c\) when \(\displaystyle \Delta n_g = -1\), since \(\displaystyle K_c\) then carries units of \(\displaystyle (\text{mol dm}^{-3})^{-1}\).\[K_c = 2.0 \times 10^{10} \times 37.395\ \text{dm}^3\text{mol}^{-1} = 7.479 \times 10^{11}\ \text{dm}^3\text{mol}^{-1} \]\(\displaystyle K_p\) was given to $\displaystyle 2$ significant figures, so the final value is rounded to $\displaystyle 2$ significant figures:\[K_c \approx 7.5 \times 10^{11}\ \text{dm}^3\,\text{mol}^{-1} \]Answer: \(\displaystyle K_c \approx 7.5 \times 10^{11}\ \text{dm}^3\,\text{mol}^{-1}\) (equivalently \(\displaystyle \text{L mol}^{-1}\)).