Equilibrium Calculations: Summary

$$K_C = \frac{[H_2 ]^3[CO]}{[CH_4][H_2 O]}\\ K_C = \frac{[1.15\, \text{M}]^3 [0.126\, \text{M}]}{[0.126\, \text{M}][0.242, \text{M}]}\\ \mathbf{K_C = 6.28}$$

$$PCl_5 \rightleftharpoons PCl_3 + Cl_2 \\$$

0.0033% decomposed means that 99.9967% remains as NO₂: $[NO_2] = 99.9967\% \times 1.00\, \text{M} = 0.999_{967}\, \text{M}$.

$$K_P = \frac{(P_{NOCl})^2}{(P_{NO})^2 (P_{Cl_2})}\\ K_P = \frac{(1.2\; \text{bar})^2}{(0.050\; \text{bar})^2(0.30\; \text{bar})}\\ \mathbf{K_P = 1.9×10^3}$$

$$K_P = \frac{(P_I)^2}{P_{I_2}}\\ K_P = \frac{0.1378^2}{0.1122} \\ K_P = 0.1692_{40998} \\ \mathbf{K_P = 0.1692}$$

The amount of $NH_3$ produced is equal to the amount of $HCl$ produced in this reaction. Since we started with only annominum chloride in the container, the equilibrium pressures of the products must be equal. $$K_P = (P_{NH_3})(P_{HCl}) \\ K_P = (1.75\, \text{bar})(1.75\, \text{bar}) \\ K_P = 3.06_{25} \\ \mathbf{K_P = 3.06}$$

$$K_P = (P_{H_2O}) \\ K_P = (0.196\, \text{bar}) \\ \mathbf{K_P = 0.196}$$

The water in this reaction is a pure liquid – it does not have a defined concentration (or: its activity is a constant at 1). Therefore having an initial or final value for “concentration” would be meaningless for this species. However, the amount of water can change during the reaction (the number of moles present) can change during the reaction. Including the stoichiometry in the Change row can help track whether the amount of water should increase or decrease – and if we knew the intial amount (in mol or g), we could calculate the final amount (in mol or g) based on the reaction stoichiometry.

$$K_C = \frac{[NH_3]^2}{[N_2][H_2]^3} \\ 0.50 = \frac{[NH_3]^2}{[1.2\, \text{M}][0.24\, \text{M}]^3} \\ [NH_3] = 0.091_{0736} \, \text{M}\\ \mathbf{[NH_3] = 0.091 \, \text{M}}$$

$$ K_C = \frac{[HI]^2}{[H_2][I_2]}\\ 50.2 = \frac{[HI]^2}{[\frac{1.25\, \text{mol}}{5.00\, \text{L}}][\frac{1.25\, \text{mol}}{5.00\, \text{L}}]} \\ [HI] = 1.77_{1298}\, \text{M}\\ \text{mol HI} = 1.77_{1298}\, \text{M} \times 5.00\, \text{L} \\ \text{mol HI} = 8.85_{64947}\, \text{mol} \\ \mathbf{\text{mol HI} = 8.86\, \text{mol}} $$

$$K_P = \frac{(P_{BrCl})^2}{(P_{Cl_2})(P_{Br_2})}\\ 4.7×10^{-2} = \frac{(P_{BrCl})^2}{(0.115\, \text{bar})(0.450\, \text{bar})} \\ P_{BrCl} = 0.049_{317846}\, \text{bar} \\ \mathbf{P_{BrCl} = 0.049\, \text{bar}}$$

$$K_P = \frac{(P_{H_2 O})(P_{CO})}{(P_{H_2})(P_{CO_2})} \\ 1.6 = \frac{(2.0\, \text{bar})(1.0\, \text{bar})}{(0.50\, \text{bar})(P_{CO_2})} \\ \mathbf{P_{CO_2} = 2.5\, \text {bar}}\\$$

$$K_C = \frac{[CO_2]}{[CO]}\\ 4.90\times 10^2 = \frac{[0.100\, \text{M}]}{[CO]} \\ [CO] = 2.04_{0816327}\, \times 10^{-4}\, \text{M}\\ \mathbf{[CO] = 2.04\, \times 10^{-4}\, \text{M}}\\$$

Since the only source of CO and H₂ is this reaction and both are produced in equal amounts, [CO] = [H₂]. $$K_C = \frac{[CO][H_2]}{[H_2 O]}\\ 0.2 = \frac{[CO]^2}{[0.500\, \text{M}]}\\ [CO] = 0.3_{16227766}\, \text{M}\\ \mathbf{[CO] = 0.3\, \text{M}}$$

$$K_P = (P_{H_2 O})^{10}\\ 4.08\times 10^{-25} = (P_{H_2 O})^{10}\\ P_{H_2O} = 3.63_{970} \times 10^{-3}\, \text{bar}\\ \mathbf{P_{H_2O} = 3.64 \times 10^{-3}\, \text{bar}}$$ **Hint: to do a 10^th root on your calculator, you can either use the $\sqrt[x]\quad $ key (i.e. enter $\require{enclose} \enclose{box}{10}\; \enclose{box}{\sqrt[x]\quad}\; \enclose{box}{5}$ to take the 10^th root of 5) OR raise your value to a fractional power ( $ 5^{\frac{1}{10}} = 5^{0.1}$ is equivalent to $\sqrt[10]{5} $).

$$K_P = (P_{H_2O})^6\\ 5.09×10^{-44} = (P_{H_2O})^6\\ P_{H_2O} = 6.08_{7696}\times 10^{-8}\, \text{bar}\\ \mathbf{P_{H_2O} = 6.09\times 10^{-8}\, \text{bar}} $$ (see note in previous question on calculating higher roots on your calculator)

The student could check their value by solving for K using their calculated value: $$K_C = \frac{[N_2 O_4]}{[NO_2]^2}\\ K_C = \frac{[0.16\, \text{M}]}{[0.10\, \text{M}]^2}\\ K_C = 16 \neq 160$$ Since the K determined with the calculated concentrations does not equal the given K, an error in calculation must have been made when finding [N₂O₄].

The reaction as intially prepared has 0 NO₂, so it can only proceed in the forward direction (right / towards the products):

The initial mixture is prepared with only COCl₂; the reaction can only proceed forward (right / towards products):

The initial mixture is prepared with only H₂S; the reaction can only proceed forward (right / towards products):

The initial mixture is prepared with only reactants; the reaction can only proceed forward (right / towards products):

The initial mixture is prepared with only PCl₅; the reaction can only proceed forward (right / towards products):

The initial mixture is prepared with only reactants; the reaction can only proceed forward (right / towards products).
The initial amounts of the reagents are provided in mol and g rather than concentration; they should be converted to concentration before use in the calculation for K.

The total pressure in the system is the sum of the partial pressures of all gases present in the mixture: $$P_{total} = 1.22\, \text{bar} = P_{n-\text{butane}} + P_{\text{isobutane}} $$ And we know the expression for $K_P$ from the reaction given: $$K_P = 2.5 = \frac{P_{\text{isobutane}}}{P_{n-\text{butane}}}$$ We can rearrange the expression for $K_P$ to solve for $P_{\text{isobutane}}$ and substitute this into the sum of pressures above: $$2.5\cdot P_{n- \text{butane}} = P_{\text{isobutane}} \\ 1.22\, \text{bar} = P_{n-\text{butane}} + P_{\text{isobutane}} \\ 1.22\, \text{bar} = P_{n-\text{butane}} + (2.5\cdot P_{n-\text{butane}}) \\ 1.22\, \text{bar} = 3.5\cdot P_{n-\text{butane}} \\ \mathbf{P_{n-\text{butane}}} = 0.34_{85714}\, \text{bar} = \mathbf{0.35\, \text{bar}} \\ 2.5\cdot P_{n-\text{butane}} = P_{\text{isobutane}}\\ 2.5 (0.34_{85714}\, \text{bar}) = P_{\text{isobutane}}\\ \mathbf{P_{\text{isobutane}}} = 0.87_{142857}\, \text{bar} = \mathbf{0.87\, \text{bar}}$$ As a check, we can verify that the total pressure and K values are still valid: $$P_{total} = 0.87\, \text{bar} + 0.35\, \text{bar} = 1.22\, \text{bar}\ ✓ \\ K = \frac{0.87\, \text{bar}}{0.35\, \text{bar}} = 2.5\ ✓ $$

Remember – although a solid does not appear in the expression for $K$, its presence is required in order to reach equilibrium!

Setting up the ICE table with the “change” in the solids visible highlights that we need at least “x” amount of the CaCO₃ to exist in the container or else we will have a negative amount at equilibrium (not possible – the reaction will be unable to reach equilibrium). Solving for “x” and conveting to mass of CaCO₃: $$K_C = [CO_2] \\ 0.50 \, \text{M} = x \\ 0.50 \, \text{M} \times 6.50\, \text{L} \times \frac{100.1\, g\, CaCO_3}{\text{mol}} \\ = 32_{5.325} \, \text{g}\, CaCO_3 \\ \mathbf =3.3\times 10^2\, \text{g}\, CaCO_3$$

This reaction starts with some of each reagent present: in order to determine the direction in which the reaction will proceed, we should compare the reaction quotient Q to the equilibrium constant: $$Q_C = \frac{[H_2 O][CO]}{[H_2][CO_2]}\\ Q_C = \frac{\left[ \frac{0.750\, \text{mol}}{5.00\, \text{L}} \right] \left[ \frac{1.00\, \text{mol}}{5.00\, \text{L}} \right]}{\left[ \frac{1.00\, \text{mol}}{5.00\, \text{L}} \right] \left[ \frac{2.00\, \text{mol}}{5.00\, \text{L}} \right]} \\ Q_C = 0.375 \lt K_C (1.60) $$ The reaction will proceed forward (towards the products).

Since no initial concentration is large compared to the others, it is not reasonable to use the “small x” approximation here (if you ignored one $x$, all would be ignored, creating an unsolvable situation). Solve for $x$ algebraically using the quadratic formula: $$K_C = \frac{[H_2 O][CO]}{[H_2][CO_2]}\\ 1.60 = \frac{[0.150 + x][0.200 + x]}{[0.200 – x][0.400 – x]} \\ 0 = 0.600x^2 – 1.31x + 0.098\\ \quad \text{Solve with the quadratic formula and a = +0.600, b= – 1.31, c = + 0.098 :} \\ x = \frac{-b\pm \sqrt{b^2- 4ac}}{2a} \\ x = \frac{-(- 1.31) \pm \sqrt{(- 1.31)^2 – 4(0.600)(0.098)}}{2(0.600)} \\ x = 0.07756471, 2.1057686 \\ \quad \text{Only the first root makes chemical sense; the other will create (impossible) negative concentrations.}\\ x = 0.0775_{6471} \\ [H_2] = 0.200 – x = 0.200 – 0.0775_{6471} = 0.122_{435} \, \text{M} \times 5.00\, \text{L} = 0.612_{176}\, \text{mol} = \mathbf{0.612\, \text{mol}\, H_2}\\ [CO_2] = 0.400 – x = 0.400 – 0.0775_{6471} = 0.322_{435} \, \text{M} \times 5.00\, \text{L} = 1.61_{2176}\, \text{mol} = \mathbf{1.61\, \text{mol}\, CO_2}\\ [H_2 O] = 0.150 + x = 0.150 + 0.0775_{6471} = 0.227{5647} \, \text{M} \times 5.00\, \text{L} = 1.13_{782}\, \text{mol} = \mathbf{1.14\, \text{mol}\, H_2 O}\\ [H_2] = 0.200 + x = 0.200 + 0.0775_{6471} = 0.277_{5647} \, \text{M} \times 5.00\, \text{L} = 1.38_{782}\, \text{mol} = \mathbf{1.38\, \text{mol}\, H_2}$$

Since H₂ is a reactant in this system, removing it will cause the equilibrium to shift left (towards the reactants). The partial pressure of ammonia will decrease, while the partial pressure of N₂ will increase. The partial pressure of H₂ will also increase, but its new equilibrium pressure will be lower than the initial 0.317 bar since some was removed.

The partial pressure of hydrogen is being manipulated by the experimenter, so its value will not follow stoichiometry. All the other products and reactants will maintain their stoichiometric ratios however. Since the partial pressure of N₂ increased from 0.190 bar to 0.250 bar, the change in pressure during the manipulation is $(0.250 – 0.190) = -0.060$ bar. This value is analagous to the “+x” that would appear in the Change row in an ICE table for the reaction proceeding “left”.
If nitrogen pressure increased by “+x”, the ammonia pressure should decrease by double that value (“-2x”): $\mathbf{P_{NH_3}} = 1.00 \, \text{bar} – 2(0.060) = \mathbf{0.88}$ bar.
Since we don’t know how much hydrogen gas was removed, we cannot calculate its new equilibrium pressure stoichiometrically, but we can find its equilibrium pressure from the K value and two known pressures. We are not told the K value directly, but we can find it from the first set of equilibrium pressures: $$K_P = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \\ K_P = \frac{(1.00)^2}{(0.19)(0.317)^3} \\ K_P = 165_{.222}$$ Now we can use this K value to solve the new equilibrium: $$K_P = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \\ 165_{.222} = \frac{(0.88)^2}{(0.250)(P_{H_2})^3} \\ P_{H_2} = 0.26_{5655} \, \text{bar} \\ \mathbf{P_{H_2} = 0.27\, \text{bar}} $$

$$K_C = \frac{[CO_2][H_2]}{[CO][H_2 O]} \\ 5.0 = \frac{[CO_2][0.90\, \text{M}]}{[0.20\, \text{M}][0.30\, \text{M}]} \\ [CO_2] = 0.33_{33333}\, \text{M} \times 1.00\, \text{L} = \mathbf{0.33\, \text{mol}\, CO_2}$$

$$K_C = \frac{[CO_2][H_2]}{[CO][H_2 O]} \\ 5.0 = \frac{[CO_2][1.2\, \text{M}]}{[0.40\, \text{M}][0.30\, \text{M}]} \\ [CO_2] = 0.50\, \text{M} \times 1.00\, \text{L} = \mathbf{0.50\, \text{mol}\, CO_2}$$

Since some water vapour was removed, the final water vapour concentration is unlikely to be higher than the starting value once it reaches the new equilibrium – the reaction does not ‘recover’ more than it lost. If removing water vapour was the only change made, the new equilibrium concentration would certainly be lower than the starting value. Since H₂ was also added to the system, the reaction is pushed even more to the left while re-establishing equilibrium, further increasing the final water vapour concentration, and allowing it to reach up to the initial value. (Potentially, you could increase the water vapour concentration higher than the starting value by adding even more hydrogen or carbon dioxide gas).

Initial concentrations: $$[SbCl_5] = \frac{3.85\, \text{g}\, SbCl_5}{5.00\, \text{L}} \frac {1\, \text {mol}}{299.0_5 \, \text{g}\, SbCl_5} = 2.57_{482}\times 10^{-3}\, \text{M}\, SbCl_5\\ [SbCl_3] = \frac{9.14\, \text{g}\, SbCl_5}{5.00\, \text{L}} \frac {1\, \text {mol}}{228.1_5 \, \text{g}\, SbCl_3} = 8.01_{227}\times 10^{-3}\, \text{M}\, SbCl_3\\ [Cl_2] = \frac{2.84\, \text{g}\, SbCl_5}{5.00\, \text{L}} \frac {1\, \text {mol}}{70.90 \, \text{g}\, Cl_2} = 8.01_{128}\times 10^{-3}\, \text{M}\, Cl_2$$ Find $K_C$: $$K_C = \frac{[SbCl_3][Cl_2]}{[SbCl_5]}\\ K_C = \frac{[8.01_{227}\times 10^{-3}\, \text{M}][8.01_{128}\times 10^{-3}\, \text{M}]}{[2.57_{482}\times 10^{-3}\, \text{M}]}\\ K_C = 0.0249_{293}$$ Concentrations on transfer to new container (before any reaction has occurred): $$[SbCl_5] = \frac{3.85\, \text{g}\, SbCl_5}{3.00\, \text{L}} \frac {1\, \text {mol}}{299.0_5 \, \text{g}\, SbCl_5} = 4.29_{136}\times 10^{-3}\, \text{M}\, SbCl_5\\ [SbCl_3] = \frac{9.14\, \text{g}\, SbCl_5}{3.00\, \text{L}} \frac {1\, \text {mol}}{228.1_5 \, \text{g}\, SbCl_3} = 1.33_{537}\times 10^{-2}\, \text{M}\, SbCl_3\\ [Cl_2] = \frac{2.84\, \text{g}\, SbCl_5}{3.00\, \text{L}} \frac {1\, \text {mol}}{70.90 \, \text{g}\, Cl_2} = 1.33_{521}\times 10^{-2}\, \text{M}\, Cl_2$$ Reaction quotient Q: $$Q_C = \frac{[SbCl_3][Cl_2]}{[SbCl_5]}\\ Q_C = \frac{[1.33_{537}\times 10^{-2}\, \text{M}][1.33_{521}\times 10^{-2}\, \text{M}]}{[4.29_{136}\times 10^{-3}\, \text{M}]}\\ Q_C = 0.0415_{485}$$ Q > K: the reaction will proceed in reverse (left, towards the reactant) in order to reach equilibrium.

$$K_C = = \frac{[SbCl_3][Cl_2]}{[SbCl_5]}\\ 0.0249_{293} = \frac{[1.33_{537}\times 10^{-2} – x][1.33_{521}\times 10^{-2} – x]}{[4.29_{136}\times 10^{-3} + x]} \\ 0 = x^2 – 0.0516_{351}x + 7.1_3193 \times 10^{-5} \\ x = 1.42028 \times 10^{-3}, 5.02148 \times 10^{-2}$$ Only the first value for $x$ above makes chemical sense: $$[SbCl_5] = 4.29_{136}\times 10^{-3} + x = 4.29_{136}\times 10^{-3} + 1.42028 \times 10^{-3} \\ = 5.71_{944}\times 10^{-3}\, \text{M}\times \frac{299.0_5 \, \text{g}\, SbCl_5}{1\, \text {mol}}\times 3.00\, \text{L} = \mathbf{5.13\, \text{g}\, SbCl_5}\\ [SbCl_3] = 1.33_{537}\times 10^{-2} – x = 1.33_{537}\times 10^{-2} – 1.42028 \times 10^{-3} \\ = 0.0119_{2562}\, \text{M} \times \frac {228.1_5 \, \text{g}\, SbCl_3}{1\, \text {mol}}\times 3.00\, \text{L} = \mathbf{8.16\, \text{g}\, SbCl_3}\\ [Cl_2] = 1.33_{521}\times 10^{-2} – x = 1.33_{521}\times 10^{-2} – 1.42028 \times 10^{-3} \\ = 0.0119_{2402}\, \text{M} \times \frac {70.90 \, \text{g}\, Cl_2}{1\, \text {mol}}\times 3.00\, \text{L} = \mathbf{2.54\, \text{g}\, Cl_2}$$ Reality check 1: Decreasing the volume of this reaction at equilibrium should cause the reaction to shift left (towards the reactant) since there are fewer mol of gas on the reactant side of the equation. This matches our final concentrations.
Reality check 2: calculate K with the new concentrations: K = 0.0249 (agrees).

$$K_C = \frac{[NH_3]^4 [O_2]^7}{[NO_2]^4 [H_2 O]^6}$$

If Q < K, the reaction has “too few” products (or “too many” reactants). The reaction must proceed to the right (towards the products) to reach equilibrium. The concentration of NH₃ will therefore increase.

Initially (after the change but before the reaction comes to equilibrium) all pressures would decrease due to the incease in volume. In order to re-establish equilibrium, the reaction will proceed towards the side with more mol of gas (towards the products), so once equilibrium is re-established after the change, the pressure of NO₂ will be lower than it was before the change.

Changes in pressure are all related stoichiometrically; if ΔNO₂ = “4x” then ΔO₂ = “-7x”. Therefore, if the pressure of NO₂ changes by 0.28 bar, the O₂ will change by 0.49 bar in the opposite direction.

$$K_C = \frac{[CO_2−Hb−H^+ ][O_2]}{[HbO_2][H_3O^+][CO_2]}$$

Production of lactic acid (as you can infer from the name) will lower the pH of the blood, increasing the concentration of $H_3 O^+$. This is a reactant in the equilibrium expression above, so its addition will cause the reaction to shift towards the right (products), releasing O₂ in the affected area.

$$N_2 O_3\, (g) \rightleftharpoons NO\, (g)\, + NO_2 (g)$$ Intial pressure of $N_2 O_3$: $$ PV = nRT \\ P = \frac{0.236\, \text{mol} \times 0.08314510 \frac{\text{L bar}}{\text{mol K}} \times 298\, \text{K}}{1.52\, \text{L}}\\ P = 3.84_{6992}\, \text{bar}\, N_2 O_3$$ The reaction container is initially prepared with only $N_2 O_3$ (the reactant) so the reaction must proceed to the right (towards the products) to reach equilibrium.

Determining the initial value for K: $$K_C = \frac{[NH_3]^2}{[N_2][H_2]^3}\\ K_C = \frac{[0.25]^2}{[1.00][0.50]^3}\\ K_C = 0.50$$ We can consider the amount of hydrogen gas removed from the container as $n$ M, and the change required to re-establish equilibrium as $x$ mol. Organizing the information given in an ICE table:

We can see by looking at the nitrogen concentration that $x = 0.1\, \text{M}$. Substituting this value for $x$ into the other concentrations, and into the expression for $K$: $$K_C = \frac{[NH_3]^2}{[N_2][H_2]^3}\\ 0.50 = \frac{[0.25 – 2(0.1)]^2}{[1.1][(0.50 – n) + 3(0.1)]^3}\\ 0.50 = \frac{[0.0025]}{[1.1][0.80-n]^3}\\ [0.80-n]^3 = 0.0045_{4545}\\ 0.80-n = 0.16_{565}\\ n = 0.63_{434}$$ You would have to remove 0.63 M of H₂ in order to reach the target equilibrium state.