Question

consider the following reaction at 25 °c:

5 so₃(g) + 2 nh₃(g) → 2 no(g) + 5 so₂(g) + 3 h₂o(g)

if δh° = 42.4 kj/mol and δs° = 562.3 j/mol•k, estimate the temperature at which this reaction would be at equilibrium assuming that enthalpy and entropy are independent of temperature.

Explanation:

Step1: Recall the equilibrium condition for Gibbs free energy

At equilibrium, \(\Delta G^{\circ} = 0\). The formula relating Gibbs free energy (\(\Delta G^{\circ}\)), enthalpy (\(\Delta H^{\circ}\)), entropy (\(\Delta S^{\circ}\)), and temperature (\(T\)) is \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\). Setting \(\Delta G^{\circ} = 0\) gives \(0=\Delta H^{\circ}-T\Delta S^{\circ}\), which can be rearranged to \(T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}\).

Step2: Convert units of \(\Delta H^{\circ}\) to be consistent with \(\Delta S^{\circ}\)

Given \(\Delta H^{\circ}=42.4\space kJ/mol = 42400\space J/mol\) (since \(1\space kJ = 1000\space J\)) and \(\Delta S^{\circ}=562.3\space J/mol\cdot K\).

Step3: Calculate the temperature \(T\)

Substitute the values into the formula \(T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}\):
\(T=\frac{42400\space J/mol}{562.3\space J/mol\cdot K}\approx75.4\space K\) (Wait, this seems low. Wait, no, wait. Wait, the reaction is at 25°C initially, but we are finding the temperature where \(\Delta G = 0\). Wait, let's check the formula again. Wait, \(\Delta G=\Delta H - T\Delta S\), at equilibrium \(\Delta G = 0\), so \(T=\frac{\Delta H}{\Delta S}\). But let's check the units again. \(\Delta H\) is in kJ, \(\Delta S\) in J/K. So convert \(\Delta H\) to J: \(42.4\space kJ/mol=42400\space J/mol\). Then \(T=\frac{42400}{562.3}\approx75.4\space K\)? But that is below 25°C (298 K). Wait, maybe I made a mistake? Wait, no, the formula is correct. Wait, let's recalculate: \(42400\div562.3\approx75.4\) Kelvin. But let's check the problem again. The reaction is given, \(\Delta H^{\circ}=42.4\space kJ/mol\) (endothermic), \(\Delta S^{\circ}=562.3\space J/mol\cdot K\) (positive, increase in disorder). So for \(\Delta G = 0\), \(T=\Delta H/\Delta S\). So \(42400\space J/mol\div562.3\space J/mol\cdot K\approx75.4\space K\). But that is -197.75°C. Wait, but maybe the problem is correct. Let's proceed.

Answer:

The temperature at equilibrium is approximately \(\boxed{75.4\space K}\) (or \(- 197.75^{\circ}C\)). Wait, but let's check the calculation again. \(42400\div562.3\): 562.3*70 = 39361, 42400 - 39361 = 3039, 3039/562.3≈5.4, so total≈75.4 K. Yes. So the temperature is approximately 75.4 K.