Questions: Consider the following reaction: N2(g)+2 H2O(g)=2 NO(g)+2 H2(g) The enthalpy change at 298 K for this reaction is ΔH°rnn=664.38 kJ / mol. Under the same conditions the entropy change for the reaction is ΔS°ran=113.24 J / mol-K. Calculate the value of K Eq for 8711 K. Assume that ΔH°r×n and ΔS°ren does not change with temperature. Use R=8.314 J / mol-K for the value of the gas constant.

Solution Steps

The Gibbs free energy change (\(\Delta G^\circ\)) for a reaction can be calculated using the formula:

\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]

Given:

• \(\Delta H^\circ = 664.38 \, \text{kJ/mol} = 664380 \, \text{J/mol}\)
• \(\Delta S^\circ = 113.24 \, \text{J/mol-K}\)
• \(T = 8711 \, \text{K}\)

Substitute these values into the formula:

\[ \Delta G^\circ = 664380 - 8711 \times 113.24 \]

Calculate:

\[ \Delta G^\circ = 664380 - 986,666.44 = -322286.44 \, \text{J/mol} \]

The relationship between the Gibbs free energy change and the equilibrium constant \(K\) is given by:

\[ \Delta G^\circ = -RT \ln K \]

Rearrange to solve for \(K\):

\[ \ln K = -\frac{\Delta G^\circ}{RT} \]

Substitute the known values:

\[ \ln K = -\frac{-322286.44}{8.314 \times 8711} \]

Calculate:

\[ \ln K = \frac{322286.44}{72423.474} \approx 4.4501 \]

Now, solve for \(K\):

\[ K = e^{4.4501} \approx 85.4 \]

Final Answer