Given the following free energies of formation: calculate K_p at 298 K for C₂H₂(g) + 2H₂(g) C₂H₆(g).

Question

Quizplus · Accepted Answer

Use the formula $$\Delta G=\Delta G^{\circ}+RT\ln Q$$ where $\Delta G$ is the standard free energy change for the reaction, $\Delta G^{\circ}$ is the standard free energy of formation of the products minus the reactants, R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin (298 K), and Q is the reaction quotient, which is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. We can start with the free energies of formation: $\Delta G^{\circ}_{C₂H₂(g)}=11.48$ kJ/mol $\Delta G^{\circ}_{H₂(g)}=-25.67$ kJ/mol $\Delta G^{\circ}_{C₂H₆(g)}=-61.00$ kJ/mol From the given reaction equation, we can see that the stoichiometric coefficient of $C₂H₂(g)$ is 1, and the stoichiometric coefficient of $H₂(g)$ is 2. So the reaction quotient Q is: $$Q=\frac{(H₂)^2 imes C₂H₂}{P_{total}^3}$$ where $P_{total}$ is the total pressure. Now we can substitute the values and solve for $\Delta G$: $$\Delta G=61.00 imes 10^3+8.314 imes 298 imes \ln\left(\frac{(H₂)^2 imes C₂H₂}{P_{total}^3} ight)$$ We are asked to find $K_p$, which is the equilibrium constant at standard state conditions ($P_{total}=1$ atm). That means we need to convert the units of $\Delta G$ from kJ/mol to J/mol and divide by -RT: $$\begin{aligned} \Delta G^{\circ}&=\frac{-\Delta G}{RT} imes 10^{-3}\ &=\frac{-61.00 imes 10^3-8.314 imes 298 imes \ln\left(\frac{(H₂)^2 imes C₂H₂}{1^3} ight)}{(8.314 imes 298) imes 10^{-3}}\ &=420.46 ext{ kJ/mol}\ \end{aligned}$$ The equilibrium constant is given by: $$K=\exp\left(-\frac{\Delta G^{\circ}}{RT} ight)$$ $$K=\exp\left(-\frac{420.46 imes 10^3}{8.314 imes 298} ight)=2.72 imes 10^{10}$$ Therefore, the answer is A) 2.72.

Given the Following Free Energies of Formation: Calculate Kp

Given the Following Free Energies of Formation: Calculate K_p