Let's break down the solution step by step--- # **a) Calculation Equilibrium Constant at K** ##Given Data** Standard heat of formation of NH₃ at 298 K, ΔHf°(NH₃, 298) = **–46100 J/mol** - Standard free energy of formation of NH₃ at 298 K, ΔGf°(NH₃, 298) = **–16500 J/mol** - Temperature for equilibrium constant calculation, **T = 500 K** - Reaction: \(\mathrm{N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g)}\) ## **Step 1: Calculate Standard Free Energy Change (ΔG°) for the Reaction at 298 K** The standard free energy change: \[ \Delta G^\circ_{298} = \sum \nu_i \Delta G_f^\circ (\text{products}) - \sum \nu_j \Delta G_f^\circ (\text{reactants}) \] For the reaction, - ΔGf°(N₂) = (element in standard state) - ΔGf°(H₂) = (element in standard state) - ΔGf°(NH₃) = –16500 J/mol \[ \Delta G^\circ_{298} = [2 \times (-16500)] - [ + 3 \times ] = -33000\, \text{J/mol} \] --- ## **Step 2: Calculate ΔH° for the Reaction at 298 K** Similarly, - ΔHf°(N₂) = - ΔHf°(H₂) = - ΔHf°(NH₃) = –46100 J/mol \[ \Delta H^\circ_{298} = [2 \times (-46100)] - [ + 3 \times ] = -92200\, \text{J/mol} \] --- ## **Step 3: Calculate ΔG° at 500 K Using the van't Hoff Equation** Assume ΔH° is **constant** from 298 K to 500 K. The temperature dependence of ΔG° is given by: \[ \Delta G^\circ_T = \Delta H^\circ - T \Delta S^\circ \] First, calculate ΔS° at 298 K: \[ \Delta G^\circ_{298} = \Delta H^\circ_{298} - 298 \Delta S^\circ \] \[ -33000 = -92200 - 298 \Delta S^\circ \] \[ 298 \Delta S^\circ = -92200 + 33000 = -59200 \] \[ \Delta S^\circ = \frac{-59200}{298} = -198.66\, \text{J/mol·K} \] Now, calculate ΔG° at 500 K: \[ \Delta G^\circ_{500} = \Delta H^\circ_{298} - 500 \cdot \Delta S^\circ \] \[ = -92200 - 500 \times (-198.66) \] \[ = -92200 + 99330 \] \[ = 713\, \text{J/mol} \] --- ## **Step 4: Calculate the Equilibrium Constant \( K \) at 500 K** \[ \Delta G^\circ = -RT \ln K \] \[ K = \exp\left(-\frac{\Delta G^\circ}{RT}\right) \] Where \( R = 8.314 \) J/mol·K, \( T = 500 \) K, \( \Delta G^\circ = 713 \) J/mol, \[ K = \exp\left(-\frac{713}{8.314 \times 500}\right) \] \[ = \exp\left(-\frac{713}{4157}\right) \] \[ = \exp(-1.715) \] \[ = .180 \] --- ### **Final Answer for (a):** \[ \boxed{K = .18} \] **at 500 K (to 2 significant figures)** --- # **(b) Derivation of Relationship for Equilibrium Constant** The general relationship for the equilibrium constant in terms of activities or fugacities is: \[ K = \prod_i \left(\frac{f_i}{f_i^}\right)^{\nu_i} \] Where: - \( f_i \) = fugacity of component \( i \) - \( f_i^ \) = standard-state fugacity of component \( i \) - \( \nu_i \) = stoichiometric coefficient (positive for products, negative for reactants) This is the standard thermodynamic relationship for the equilibrium constant in terms of fugacities. --- ### **Final Answers (Summary):** **(a) The equilibrium constant at 500 K is** \(\boxed{.18}\). **(b) The relationship for the equilibrium constant is** \[ K = \prod_i \left(\frac{f_i}{f_i^}\right)^{\nu_i} \]