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Chemical Equilibrium — Diagnostic Tests

Unit Tests​

UT-1: Kp Calculations with Unit Conversion​

Question:

Nitrogen and hydrogen react to form ammonia:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

1.00 mol1.00\,\text{mol} of N2\text{N}_2 and 3.00 mol3.00\,\text{mol} of H2\text{H}_2 are mixed in a sealed container of volume 5.00 dm35.00\,\text{dm}^3 and allowed to reach equilibrium at 500 ∘C500\,^\circ\text{C}. At equilibrium, the total pressure is 200 atm200\,\text{atm} and there is 0.400 mol0.400\,\text{mol} of NH3\text{NH}_3 present.

(a) Calculate the mole fractions and partial pressures of all three gases at equilibrium.

(b) Calculate KpK_p at this temperature, stating its units.

(c) Explain why changing the total pressure does not change the value of KpK_p (provided temperature remains constant).

Solution:

(a) From the stoichiometry, 0.400 mol0.400\,\text{mol} of NH3\text{NH}_3 requires 0.200 mol0.200\,\text{mol} of N2\text{N}_2 and 0.600 mol0.600\,\text{mol} of H2\text{H}_2 to react.

Equilibrium moles:

  • N2\text{N}_2: 1.00−0.200=0.800 mol1.00 - 0.200 = 0.800\,\text{mol}
  • H2\text{H}_2: 3.00−0.600=2.400 mol3.00 - 0.600 = 2.400\,\text{mol}
  • NH3\text{NH}_3: 0.400 mol0.400\,\text{mol}

Total moles at equilibrium: 0.800+2.400+0.400=3.600 mol0.800 + 2.400 + 0.400 = 3.600\,\text{mol}

Mole fractions:

  • x(N2)=0.800/3.600=0.2222x(\text{N}_2) = 0.800/3.600 = 0.2222
  • x(H2)=2.400/3.600=0.6667x(\text{H}_2) = 2.400/3.600 = 0.6667
  • x(NH3)=0.400/3.600=0.1111x(\text{NH}_3) = 0.400/3.600 = 0.1111

Partial pressures (pi=xi×ptotalp_i = x_i \times p_{\text{total}}):

  • p(N2)=0.2222×200=44.4 atmp(\text{N}_2) = 0.2222 \times 200 = 44.4\,\text{atm}
  • p(H2)=0.6667×200=133.3 atmp(\text{H}_2) = 0.6667 \times 200 = 133.3\,\text{atm}
  • p(NH3)=0.1111×200=22.2 atmp(\text{NH}_3) = 0.1111 \times 200 = 22.2\,\text{atm}

(b)

Kp=◆LB◆p(NH3)2◆RB◆◆LB◆p(N2)×p(H2)3◆RB◆=◆LB◆22.22◆RB◆◆LB◆44.4×133.33◆RB◆=◆LB◆492.8◆RB◆◆LB◆44.4×2368593◆RB◆=◆LB◆492.8◆RB◆◆LB◆1.0517×108◆RB◆=4.68×10−6K_p = \frac◆LB◆p(\text{NH}_3)^2◆RB◆◆LB◆p(\text{N}_2) \times p(\text{H}_2)^3◆RB◆ = \frac◆LB◆22.2^2◆RB◆◆LB◆44.4 \times 133.3^3◆RB◆ = \frac◆LB◆492.8◆RB◆◆LB◆44.4 \times 2368593◆RB◆ = \frac◆LB◆492.8◆RB◆◆LB◆1.0517 \times 10^8◆RB◆ = 4.68 \times 10^{-6}

Units: atm−2\text{atm}^{-2}

Kp=4.68×10−6 atm−2K_p = 4.68 \times 10^{-6}\,\text{atm}^{-2}

(c) KpK_p is a constant at a given temperature. When the total pressure changes, the system adjusts the equilibrium position (Le Chatelier's principle) to restore the original KpK_p value. Increasing pressure shifts the equilibrium towards the side with fewer moles of gas (the NH3\text{NH}_3 side in this case), increasing p(NH3)p(\text{NH}_3) and changing the individual partial pressures. However, the ratio defining KpK_p returns to its equilibrium value because KpK_p depends only on temperature, not pressure.

UT-2: Effect of Temperature on Equilibrium Constant​

Question:

For the equilibrium:

PCl5(g)⇌PCl3(g)+Cl2(g)ΔH=+124 kJ mol−1\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \quad \Delta H = +124\,\text{kJ mol}^{-1}

Kc=0.0211 mol dm−3K_c = 0.0211\,\text{mol dm}^{-3} at 500 K500\,\text{K}.

(a) State and explain the effect of increasing temperature on the equilibrium yield of PCl3\text{PCl}_3.

(b) Kc=0.0420 mol dm−3K_c = 0.0420\,\text{mol dm}^{-3} at 600 K600\,\text{K}. Calculate the value of KcK_c at 550 K550\,\text{K}, using the van't Hoff equation approximation: ln⁡(K2/K1)≈◆LB◆ΔH∘◆RB◆◆LB◆R◆RB◆(1T1−1T2)\ln(K_2/K_1) \approx \frac◆LB◆\Delta H^\circ◆RB◆◆LB◆R◆RB◆\left(\frac{1}{T_1} - \frac{1}{T_2}\right).

(c) Explain why adding a catalyst does not change the equilibrium position or the value of KcK_c.

Solution:

(a) The forward reaction is endothermic (ΔH=+124 kJ mol−1\Delta H = +124\,\text{kJ mol}^{-1}). By Le Chatelier's principle, increasing temperature favours the endothermic (forward) direction. The equilibrium shifts to the right, producing more PCl3\text{PCl}_3 (and Cl2\text{Cl}_2). This is confirmed by KcK_c increasing with temperature: KcK_c increases when temperature increases for endothermic reactions.

(b) Using the van't Hoff equation:

ln⁡(K550K500)=1240008.31(1500−1550)\ln\left(\frac{K_{550}}{K_{500}}\right) = \frac{124000}{8.31}\left(\frac{1}{500} - \frac{1}{550}\right)

ln⁡(K5500.0211)=14922×(0.002000−0.001818)=14922×1.818×10−4=2.713\ln\left(\frac{K_{550}}{0.0211}\right) = 14922 \times \left(0.002000 - 0.001818\right) = 14922 \times 1.818 \times 10^{-4} = 2.713

K5500.0211=e2.713=15.08\frac{K_{550}}{0.0211} = e^{2.713} = 15.08

K550=15.08×0.0211=0.318 mol dm−3K_{550} = 15.08 \times 0.0211 = 0.318\,\text{mol dm}^{-3}

Verification: K600=0.0420K_{600} = 0.0420 should also be consistent. Using the equation between 500 K and 600 K:

ln⁡(0.0420/0.0211)=1240008.31(1/500−1/600)=14922×3.333×10−4=4.974\ln(0.0420/0.0211) = \frac{124000}{8.31}(1/500 - 1/600) = 14922 \times 3.333 \times 10^{-4} = 4.974

ln⁡(1.990)=0.688\ln(1.990) = 0.688

This gives 0.6880.688 vs the expected 4.9744.974, which indicates the van't Hoff equation is being applied over too wide a temperature range. The values given at 500 K and 600 K may not be perfectly consistent. The calculation at 550 K using 500 K data gives K550=0.318 mol dm−3K_{550} = 0.318\,\text{mol dm}^{-3}.

(c) A catalyst provides an alternative pathway with lower activation energy for both the forward and reverse reactions equally. It increases the rate at which equilibrium is achieved but does not change the thermodynamics of the reaction. The position of equilibrium (and hence KcK_c) depends on the relative energy levels of reactants and products, which are unchanged by the catalyst. The catalyst accelerates both forward and reverse rates by the same factor, so the equilibrium constant remains the same.

UT-3: Heterogeneous Equilibrium and Solids​

Question:

For the thermal decomposition of calcium carbonate:

CaCO3(s)⇌CaO(s)+CO2(g)ΔH=+178 kJ mol−1\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad \Delta H = +178\,\text{kJ mol}^{-1}

(a) Write the expression for KpK_p for this equilibrium and explain why the concentrations of the solids are not included.

(b) At 800 ∘C800\,^\circ\text{C}, Kp=0.220 atmK_p = 0.220\,\text{atm}. Calculate the minimum pressure of CO2\text{CO}_2 that must be applied to prevent decomposition of CaCO3\text{CaCO}_3 at this temperature.

(c) Explain how the decomposition temperature changes if the experiment is carried out at a lower total pressure.

Solution:

(a)

Kp=p(CO2)K_p = p(\text{CO}_2)

Solids (CaCO3\text{CaCO}_3 and CaO\text{CaO}) are excluded from the equilibrium expression because their concentrations (and hence activities) are effectively constant. The concentration of a pure solid depends only on its density, which is fixed at a given temperature. Since these constants are incorporated into the equilibrium constant, only the gaseous species appears.

(b) For decomposition to be prevented, the system must not reach equilibrium (i.e., Qp>KpQ_p \gt K_p, where QpQ_p is the reaction quotient). The minimum CO2\text{CO}_2 pressure to prevent decomposition is when Qp=KpQ_p = K_p:

p(CO2)min=Kp=0.220 atmp(\text{CO}_2)_{\text{min}} = K_p = 0.220\,\text{atm}

If the partial pressure of CO2\text{CO}_2 exceeds 0.220 atm0.220\,\text{atm}, the reverse reaction is favoured and decomposition is suppressed.

(c) Lower total pressure means the partial pressure of CO2\text{CO}_2 in the surroundings is lower. Since the equilibrium requires p(CO2)=Kpp(\text{CO}_2) = K_p for decomposition, and KpK_p increases with temperature (endothermic reaction), at lower total pressures the CO2\text{CO}_2 partial pressure needed to prevent decomposition (=Kp= K_p at that temperature) is reached at a lower temperature. Therefore, CaCO3\text{CaCO}_3 decomposes at a lower temperature when the total pressure is reduced. This is consistent with Le Chatelier's principle: reducing pressure favours the side with more moles of gas (the products side, which has 1 mol of gas vs 0 mol on the reactant side).

Integration Tests​

IT-1: Industrial Process Optimisation (with Kinetics and Thermodynamics)​

Question:

The Haber process for ammonia synthesis:

N2(g)+3H2(g)⇌2NH3(g)ΔH=−92 kJ mol−1\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92\,\text{kJ mol}^{-1}

(a) Explain why industrial conditions of approximately 450 ∘C450\,^\circ\text{C} and 200 atm200\,\text{atm} are used, given that low temperature and high pressure would give a higher equilibrium yield.

(b) The iron catalyst used in the Haber process increases the rate of reaction. Explain, with reference to activation energy, how the catalyst works, and explain why it does not change the equilibrium yield.

(c) Unreacted N2\text{N}_2 and H2\text{H}_2 are recycled. Explain the effect of this on the overall yield of the process.

Solution:

(a) Low temperature would give a higher equilibrium yield (exothermic forward reaction), but the rate would be impractically slow. At very low temperatures, the reaction would take far too long to reach equilibrium, making the process economically unviable. 450 ∘C450\,^\circ\text{C} is a compromise: high enough for a reasonable rate, low enough for a useful equilibrium yield.

High pressure favours the forward reaction (4 mol of gas on the left vs 2 mol on the right), but very high pressures require expensive, thick-walled vessels that are costly to build and maintain. 200 atm200\,\text{atm} is a compromise between equilibrium yield and economic/ safety considerations.

(b) The iron catalyst provides an alternative reaction pathway with a lower activation energy. N2\text{N}_2 and H2\text{H}_2 are adsorbed onto the catalyst surface, where the N≡\equivN triple bond is weakened, allowing easier reaction with hydrogen. The catalyst does not change the equilibrium yield because it lowers the activation energy equally for both forward and reverse reactions. The ratio of forward and reverse rate constants (which equals KcK_c) remains unchanged.

(c) Recycling unreacted N2\text{N}_2 and H2\text{H}_2 means that over time, essentially all reactants are converted to product. Although the equilibrium conversion per pass is only about 15--20% at 450 ∘C450\,^\circ\text{C} and 200 atm200\,\text{atm}, the recycled gases pass through the reactor repeatedly until they react. This gives a much higher overall yield than a single pass through the reactor, making the process economically efficient.

IT-2: Equilibrium and Gas Volume Calculations (with Quantitative Chemistry)​

Question:

2.00 mol2.00\,\text{mol} of SO2\text{SO}_2 and 1.50 mol1.50\,\text{mol} of O2\text{O}_2 are mixed in a 10.0 dm310.0\,\text{dm}^3 container and allowed to reach equilibrium:

2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)

At equilibrium, 1.20 mol1.20\,\text{mol} of SO3\text{SO}_3 is present.

(a) Calculate KcK_c for this equilibrium.

(b) Calculate KpK_p at the same temperature, given that the total pressure at equilibrium is 5.00 atm5.00\,\text{atm}.

(c) If the volume of the container is halved at constant temperature, predict the effect on the equilibrium moles of SO3\text{SO}_3 and calculate the new equilibrium moles of all species.

Solution:

(a) Let xx be the moles of O2\text{O}_2 that react. From stoichiometry, 2x2x mol of SO2\text{SO}_2 reacts and 2x2x mol of SO3\text{SO}_3 forms.

2x=1.202x = 1.20, so x=0.600x = 0.600

Equilibrium moles:

  • SO2\text{SO}_2: 2.00−1.20=0.800 mol2.00 - 1.20 = 0.800\,\text{mol}
  • O2\text{O}_2: 1.50−0.600=0.900 mol1.50 - 0.600 = 0.900\,\text{mol}
  • SO3\text{SO}_3: 1.20 mol1.20\,\text{mol}

Equilibrium concentrations (dividing by 10.0 dm310.0\,\text{dm}^3):

  • [SO2]=0.0800 mol dm−3[\text{SO}_2] = 0.0800\,\text{mol dm}^{-3}
  • [O2]=0.0900 mol dm−3[\text{O}_2] = 0.0900\,\text{mol dm}^{-3}
  • [SO3]=0.120 mol dm−3[\text{SO}_3] = 0.120\,\text{mol dm}^{-3}

Kc=◆LB◆[SO3]2◆RB◆◆LB◆[SO2]2[O2]◆RB◆=◆LB◆0.1202◆RB◆◆LB◆0.08002×0.0900◆RB◆=◆LB◆0.01440◆RB◆◆LB◆5.760×10−4◆RB◆=25.0 mol−1 dm3K_c = \frac◆LB◆[\text{SO}_3]^2◆RB◆◆LB◆[\text{SO}_2]^2[\text{O}_2]◆RB◆ = \frac◆LB◆0.120^2◆RB◆◆LB◆0.0800^2 \times 0.0900◆RB◆ = \frac◆LB◆0.01440◆RB◆◆LB◆5.760 \times 10^{-4}◆RB◆ = 25.0\,\text{mol}^{-1}\text{ dm}^3

(b) Total equilibrium moles: 0.800+0.900+1.20=2.90 mol0.800 + 0.900 + 1.20 = 2.90\,\text{mol}

Partial pressures:

  • p(SO2)=(0.800/2.90)×5.00=1.379 atmp(\text{SO}_2) = (0.800/2.90) \times 5.00 = 1.379\,\text{atm}
  • p(O2)=(0.900/2.90)×5.00=1.552 atmp(\text{O}_2) = (0.900/2.90) \times 5.00 = 1.552\,\text{atm}
  • p(SO3)=(1.20/2.90)×5.00=2.069 atmp(\text{SO}_3) = (1.20/2.90) \times 5.00 = 2.069\,\text{atm}

Kp=◆LB◆p(SO3)2◆RB◆◆LB◆p(SO2)2×p(O2)◆RB◆=◆LB◆2.0692◆RB◆◆LB◆1.3792×1.552◆RB◆=◆LB◆4.281◆RB◆◆LB◆2.952×1.552◆RB◆=4.2814.583=0.934 atm−1K_p = \frac◆LB◆p(\text{SO}_3)^2◆RB◆◆LB◆p(\text{SO}_2)^2 \times p(\text{O}_2)◆RB◆ = \frac◆LB◆2.069^2◆RB◆◆LB◆1.379^2 \times 1.552◆RB◆ = \frac◆LB◆4.281◆RB◆◆LB◆2.952 \times 1.552◆RB◆ = \frac{4.281}{4.583} = 0.934\,\text{atm}^{-1}

(c) When volume is halved (5.00 dm35.00\,\text{dm}^3), all concentrations initially double, so Qc>KcQ_c \gt K_c (since the denominator increases more due to the squared terms). The equilibrium shifts to the right (fewer moles of gas) to restore KcK_c.

Let yy mol of additional SO3\text{SO}_3 form:

  • SO2\text{SO}_2: 0.800−y0.800 - y
  • O2\text{O}_2: 0.900−y/20.900 - y/2
  • SO3\text{SO}_3: 1.20+y1.20 + y

Concentrations (dividing by 5.00 dm35.00\,\text{dm}^3):

Kc=25.0=◆LB◆((1.20+y)/5.00)2◆RB◆◆LB◆((0.800−y)/5.00)2×((0.900−y/2)/5.00)◆RB◆K_c = 25.0 = \frac◆LB◆((1.20+y)/5.00)^2◆RB◆◆LB◆((0.800-y)/5.00)^2 \times ((0.900-y/2)/5.00)◆RB◆

25.0=◆LB◆(1.20+y)2×5.00◆RB◆◆LB◆(0.800−y)2×(0.900−y/2)◆RB◆25.0 = \frac◆LB◆(1.20+y)^2 \times 5.00◆RB◆◆LB◆(0.800-y)^2 \times (0.900 - y/2)◆RB◆

5.0(1.20+y)2=25.0(0.800−y)2(0.900−y/2)5.0(1.20+y)^2 = 25.0(0.800-y)^2(0.900-y/2)

This is a cubic equation. Solving iteratively or by approximation: since the volume halved (significant change), the equilibrium shifts substantially. An approximate solution gives y≈0.08y \approx 0.08:

New equilibrium moles: SO2≈0.72\text{SO}_2 \approx 0.72, O2≈0.86\text{O}_2 \approx 0.86, SO3≈1.28\text{SO}_3 \approx 1.28.

The moles of SO3\text{SO}_3 have increased from 1.201.20 to approximately 1.28 mol1.28\,\text{mol}, consistent with Le Chatelier's principle (increasing pressure favours the side with fewer moles of gas: 3 mol →\to 2 mol).

IT-3: Acid Dissociation Equilibrium (with Acids and Bases)​

Question:

Ethanoic acid dissociates in water:

CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq)

Ka=1.74×10−5 mol dm−3K_a = 1.74 \times 10^{-5}\,\text{mol dm}^{-3} at 298 K298\,\text{K}.

(a) Calculate the pH of a 0.100 mol dm−30.100\,\text{mol dm}^{-3} solution of ethanoic acid, stating any approximation you make.

(b) Calculate the percentage dissociation of ethanoic acid at this concentration.

(c) A student dilutes the solution to 0.00100 mol dm−30.00100\,\text{mol dm}^{-3} and claims the pH should decrease by exactly 2 units (since concentration decreased by a factor of 100). Show that this is incorrect and calculate the actual pH.

Solution:

(a) Let xx be the concentration of H+\text{H}^+ at equilibrium:

Ka=x20.100−x=1.74×10−5K_a = \frac{x^2}{0.100 - x} = 1.74 \times 10^{-5}

Approximation: Since KaK_a is very small, x≪0.100x \ll 0.100, so 0.100−x≈0.1000.100 - x \approx 0.100:

x2=1.74×10−5×0.100=1.74×10−6x^2 = 1.74 \times 10^{-5} \times 0.100 = 1.74 \times 10^{-6}

x=◆LB◆1.74×10−6◆RB◆=1.319×10−3 mol dm−3x = \sqrt◆LB◆1.74 \times 10^{-6}◆RB◆ = 1.319 \times 10^{-3}\,\text{mol dm}^{-3}

pH=−log⁡(1.319×10−3)=2.88\text{pH} = -\log(1.319 \times 10^{-3}) = 2.88

Verification: x/0.100=0.01319=1.32%x/0.100 = 0.01319 = 1.32\%, so the approximation is valid (less than 5%).

(b)

Percentage dissociation=◆LB◆1.319×10−3◆RB◆◆LB◆0.100◆RB◆×100=1.32%\text{Percentage dissociation} = \frac◆LB◆1.319 \times 10^{-3}◆RB◆◆LB◆0.100◆RB◆ \times 100 = 1.32\%

(c) At 0.00100 mol dm−30.00100\,\text{mol dm}^{-3}:

Ka=x20.00100−x=1.74×10−5K_a = \frac{x^2}{0.00100 - x} = 1.74 \times 10^{-5}

Here, xx may not be negligible compared to 0.001000.00100. Solving the quadratic:

x2+1.74×10−5x−1.74×10−8=0x^2 + 1.74 \times 10^{-5}x - 1.74 \times 10^{-8} = 0

x=◆LB◆−1.74×10−5+◆LB◆(1.74×10−5)2+4×1.74×10−8◆RB◆◆RB◆◆LB◆2◆RB◆x = \frac◆LB◆-1.74 \times 10^{-5} + \sqrt◆LB◆(1.74 \times 10^{-5})^2 + 4 \times 1.74 \times 10^{-8}◆RB◆◆RB◆◆LB◆2◆RB◆

x=◆LB◆−1.74×10−5+◆LB◆3.028×10−10+6.960×10−8◆RB◆◆RB◆◆LB◆2◆RB◆x = \frac◆LB◆-1.74 \times 10^{-5} + \sqrt◆LB◆3.028 \times 10^{-10} + 6.960 \times 10^{-8}◆RB◆◆RB◆◆LB◆2◆RB◆

x=◆LB◆−1.74×10−5+◆LB◆6.990×10−8◆RB◆◆RB◆◆LB◆2◆RB◆=◆LB◆−1.74×10−5+2.644×10−4◆RB◆◆LB◆2◆RB◆=1.235×10−4 mol dm−3x = \frac◆LB◆-1.74 \times 10^{-5} + \sqrt◆LB◆6.990 \times 10^{-8}◆RB◆◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆-1.74 \times 10^{-5} + 2.644 \times 10^{-4}◆RB◆◆LB◆2◆RB◆ = 1.235 \times 10^{-4}\,\text{mol dm}^{-3}

pH=−log⁡(1.235×10−4)=3.91\text{pH} = -\log(1.235 \times 10^{-4}) = 3.91

The pH change is from 2.882.88 to 3.913.91, a change of 1.031.03 units (not 2 units). This is because diluting a weak acid increases the percentage dissociation (from 1.32%1.32\% to 12.4%12.4\%), partially compensating for the lower concentration. The student's reasoning would only apply to a strong acid.

Additional Practice Problems​

UT-4: Heterogeneous Equilibrium​

Question: For the equilibrium NH4HS(s)⇌NH3(g)+H2S(g)\mathrm{NH}_4\mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_3(g) + \mathrm{H}_2\mathrm{S}(g), the total pressure at equilibrium is 0.660 atm0.660\,\mathrm{atm} at 298 K298\,\mathrm{K}. Calculate KpK_p.

Solution:

Since the solid does not appear in the expression:

Kp=p(NH3)×p(H2S)K_p = p(\mathrm{NH}_3) \times p(\mathrm{H}_2\mathrm{S})

From stoichiometry, p(NH3)=p(H2S)=0.6602=0.330 atmp(\mathrm{NH}_3) = p(\mathrm{H}_2\mathrm{S}) = \frac{0.660}{2} = 0.330\,\mathrm{atm} (1 mark).

Kp=0.330×0.330=0.109 atm2K_p = 0.330 \times 0.330 = 0.109\,\mathrm{atm}^2 (1 mark).

UT-5: Le Chatelier Applied​

Question: For the equilibrium N2(g)+3H2(g)⇌2NH3(g)ΔH=−92 kJ mol−1\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g) \quad \Delta H = -92\,\mathrm{kJ\,mol^{-1}}, predict and explain the effect of each change on the equilibrium yield of ammonia:

(a) Increasing pressure (b) Increasing temperature (c) Adding a catalyst (d) Removing NH3\mathrm{NH}_3 as it is formed

Solution:

(a) Increasing pressure favours the side with fewer moles of gas. There are 4 mol of gas on the left and 2 mol on the right. The equilibrium shifts to the right, increasing the ammonia yield (1 mark).

(b) Increasing temperature favours the endothermic direction. Since the forward reaction is exothermic (ΔH<0\Delta H < 0), the equilibrium shifts to the left, decreasing the ammonia yield (1 mark).

(c) A catalyst increases the rate of both forward and reverse reactions equally. It has no effect on the equilibrium position or yield. The system reaches equilibrium faster but the yield is unchanged (1 mark).

(d) Removing NH3\mathrm{NH}_3 decreases its partial pressure (concentration). The equilibrium shifts to the right to replace the NH3\mathrm{NH}_3, increasing the yield of NH3\mathrm{NH}_3 per pass through the reactor (1 mark). This is the principle behind the industrial Haber process, where NH3\mathrm{NH}_3 is continually condensed out.

IT-4: Equilibrium and Thermodynamics Combined​

Question: For the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g)\mathrm{CO}(g) + \mathrm{H}_2\mathrm{O}(g) \rightleftharpoons \mathrm{CO}_2(g) + \mathrm{H}_2(g):

ΔH∘=−41 kJ mol−1\Delta H^\circ = -41\,\mathrm{kJ\,mol^{-1}}, ΔS∘=−42 J mol−1 K−1\Delta S^\circ = -42\,\mathrm{J\,mol^{-1}\,K^{-1}}.

(a) Calculate KpK_p at 298 K298\,\mathrm{K}.

(b) At what temperature does Kp=1K_p = 1?

(c) If the reaction starts with 1.00 atm1.00\,\mathrm{atm} of CO\mathrm{CO} and 1.00 atm1.00\,\mathrm{atm} of H2O\mathrm{H}_2\mathrm{O} and no products, calculate the equilibrium partial pressures at 298 K298\,\mathrm{K}.

Solution:

(a) ΔG∘=−41000−298×(−42)=−41000+12516=−28484 J mol−1\Delta G^\circ = -41000 - 298 \times (-42) = -41000 + 12516 = -28484\,\mathrm{J\,mol^{-1}}

Kp=exp⁡(◆LB◆28484◆RB◆◆LB◆8.314×298◆RB◆)=exp⁡(11.50)=9.89×104K_p = \exp\left(\frac◆LB◆28484◆RB◆◆LB◆8.314 \times 298◆RB◆\right) = \exp(11.50) = 9.89 \times 10^4

Since Δn=0\Delta n = 0, KpK_p is dimensionless (1 mark).

(b) Kp=1K_p = 1 when ΔG∘=0\Delta G^\circ = 0:

T=◆LB◆ΔH∘◆RB◆◆LB◆ΔS∘◆RB◆=−41000−0.042=976 KT = \frac◆LB◆\Delta H^\circ◆RB◆◆LB◆\Delta S^\circ◆RB◆ = \frac{-41000}{-0.042} = 976\,\mathrm{K} (1 mark).

(c) ICE table (pressures in atm):

CO\mathrm{CO}H2O\mathrm{H}_2\mathrm{O}CO2\mathrm{CO}_2H2\mathrm{H}_2
Initial1.001.0000
Change−x-x−x-x+x+x+x+x
Equilibrium1−x1-x1−x1-xxxxx

Kp=x2(1−x)2=9.89×104K_p = \frac{x^2}{(1-x)^2} = 9.89 \times 10^4

x1−x=◆LB◆9.89×104◆RB◆=314.5\frac{x}{1-x} = \sqrt◆LB◆9.89 \times 10^4◆RB◆ = 314.5

x=314.5(1−x)=314.5−314.5xx = 314.5(1-x) = 314.5 - 314.5x

315.5x=314.5315.5x = 314.5

x=0.997 atmx = 0.997\,\mathrm{atm}

Equilibrium: p(CO)=p(H2O)=0.003 atmp(\mathrm{CO}) = p(\mathrm{H}_2\mathrm{O}) = 0.003\,\mathrm{atm}, p(CO2)=p(H2)=0.997 atmp(\mathrm{CO}_2) = p(\mathrm{H}_2) = 0.997\,\mathrm{atm} (1 mark).