What is the approximate value of log Kp for the reaction: (g) + 3(g) ⇌ 2(g) at 25°C. The standard en — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate the Standard Enthalpy of the Reaction ($\Delta_r H^\circ$) The given reaction for the synthesis of ammonia is: $$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$ The standard enthalpy of reaction ($\Delta_r H^\circ$) is calculated using the standard enthalpies of formation ($\Delta_f H^\circ$) of the products and reactants: $$\Delta_r H^\circ = \sum \left( n \cdot \Delta_f H^\circ \right)_{\text{products}} - \sum \left( m \cdot \Delta_f H^\circ \right)_{\text{reactants}}$$ Since the standard enthalpy of formation of pure elemental gases in their standard reference states ($\ce{N2(g)}$ and $\ce{H2(g)}$) is zero, we substitute the values as follows: $$\Delta_r H^\circ = 2 \cdot \Delta_f H^\circ(\ce{NH3(g)}) - \left[ \Delta_f H^\circ(\ce{N2(g)}) + 3 \cdot \Delta_f H^\circ(\ce{H2(g)}) \right]$$ $$\Delta_r H^\circ = 2 \times \left( -40.0\text{ kJ mol}^{-1} \right) - \left[ 0\text{ kJ mol}^{-1} + 3 \times 0\text{ kJ mol}^{-1} \right]$$ $$\Delta_r H^\circ = -80.0\text{ kJ mol}^{-1} = -80,000\text{ J mol}^{-1}$$ Step 2 - Calculate the Standard Entropy of the Reaction ($\Delta_r S^\circ$) The standard entropy of the reaction ($\Delta_r S^\circ$) is calculated using the absolute standard entropies ($S^\circ$) of the products and reactants: $$\Delta_r S^\circ = \sum \left( n \cdot S^\circ \right)_{\text{products}} - \sum \left( m \cdot S^\circ \right)_{\text{reactants}}$$ $$\Delta_r S^\circ = 2 \cdot S^\circ(\ce{NH3(g)}) - \left[ S^\circ(\ce{N2(g)}) + 3 \cdot S^\circ(\ce{H2(g)}) \right]$$ Substituting the given values: $$\Delta_r S^\circ = 2 \times 192\text{ J K}^{-1}\text{ mol}^{-1} - \left[ 191\text{ J K}^{-1}\text{ mol}^{-1} + 3 \times 130\text{ J K}^{-1}\text{ mol}^{-1} \right]$$ $$\Delta_r S^\circ = 384\text{ J K}^{-1}\text{ mol}^{-1} - \left[ 191 + 390 \right]\text{ J K}^{-1}\text{ mol}^{-1}$$ $$\Delta_r S^\circ = 384\text{ J K}^{-1}\text{ mol}^{-1} - 581\text{ J K}^{-1}\text{ mol}^{-1}$$ $$\Delta_r S^\circ = -197\text{ J K}^{-1}\text{ mol}^{-1}$$ Step 3 - Calculate the Standard Gibbs Free Energy Change ($\Delta_r G^\circ$) We use the Gibbs-Helmholtz equation to find the standard free energy change of the reaction at $T = 25^\circ\text{C} = 298\text{ K}$: $$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ$$ Substituting the calculated values of enthalpy and entropy: $$\Delta_r G^\circ = -80,000\text{ J mol}^{-1} - 298\text{ K} \times \left( -197\text{ J K}^{-1}\text{ mol}^{-1} \right)$$ $$\Delta_r G^\circ = -80,000\text{ J mol}^{-1} + 58,706\text{ J mol}^{-1}$$ $$\Delta_r G^\circ = -21,294\text{ J mol}^{-1}$$ Step 4 - Relate standard free energy change to the equilibrium constant ($\log_{10} K_p$) The standard Gibbs free energy change of a reaction is related to its equilibrium constant $K_p$ by the relation: $$\Delta_r G^\circ = -2.303 RT \log_{10} K_p$$ Rearranging the formula to solve for $\log_{10} K_p$: $$\log_{10} K_p = \frac{-\Delta_r G^\circ}{2.303 RT}$$ Substituting $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$, $T = 298\text{ K}$, and $\Delta_r G^\circ = -21,294\text{ J mol}^{-1}$: $$\log_{10} K_p = \frac{-(-21,294\text{ J mol}^{-1})}{2.303 \times 8.314\text{ J K}^{-1}\text{ mol}^{-1} \times 298\text{ K}}$$ $$\log_{10} K_p = \frac{21,294}{5,705.85}$$ $$\log_{10} K_p \approx \boxed{3.73}$$ Step 5 - Analysis of Options * **(A) 0.04**: Incorrect. This extremely small value of log Kp would correspond to a standard free energy change very close to zero, which is not supported by our calculations. * **(B) 7.05**: Incorrect. This could arise from calculation errors such as not doubling the enthalpy of formation of ammonia or adding the term $T\Delta_r S^\circ$ instead of subtracting it. * **(C) 8.6**: Incorrect. This value does not match any proper substitution path. * **(D) 3.73**: Correct. This is the precise value computed using the integrated relationship between standard enthalpy, standard entropy, and the equilibrium constant.