(Given, $\frac{ d (\ln K)}{ d \left(\frac{1}{T}\right)}=-\frac{\Delta H^{0}}{R}$, where the equilibrium constant, $K=\frac{p_{z}}{p^{0}}$ and the gas constant, $R=8314 \,J \,K ^{-1} mol ^{-1}$ )
The value of standard enthalpy, $\Delta H^{\phi}$ (in $kJ\, mol ^{-1}$ ) for the given reaction is ______
Correct Answer: 166.28
Solution and Explanation
The given reaction is: \[ X(s) \rightleftharpoons Y(s) + Z(g) \] The plot shows \( \ln \left( \frac{p_z}{p^\circ} \right) \) versus \( 10^4 / T \), where: - \( p_z \) is the pressure (in bar) of the gas Z at temperature \( T \), - \( p^\circ = 1 \, \text{bar} \), - The temperature range is \( 10^4 / T \), and the slope of the line from the plot is used to calculate the enthalpy change (\( \Delta H^\phi \)) for the reaction.
The formula provided is: \[ \frac{d (\ln K)}{d (1/T)} = -\frac{\Delta H^\phi}{R} \] where: - \( K = \frac{p_z}{p^\circ} \) is the equilibrium constant, - \( \Delta H^\phi \) is the standard enthalpy change of the reaction, - \( R = 8314 \, \text{J/molΒ·K} \) is the gas constant.
Step 2: Extracting the slope from the graph
From the graph, we can determine the change in \( \ln \left( \frac{p_z}{p^\circ} \right) \) and \( 10^4 / T \). The graph shows a line that has a slope between the points where \( 10^4 / T \) is 10 and 12.
- The change in \( \ln \left( \frac{p_z}{p^\circ} \right) \) is: \[ \Delta \left( \ln \left( \frac{p_z}{p^\circ} \right) \right) = -3 - (-7) = 4 \] - The change in \( 10^4 / T \) is: \[ \Delta \left( \frac{10^4}{T} \right) = 12 - 10 = 2 \] Thus, the slope of the line is: \[ m = \frac{4}{2} = 2 \] Step 3: Applying the slope to the formula
Now, we can use the relationship between the slope and \( \Delta H^\phi \): \[ \text{Slope} = -\frac{\Delta H^\phi}{R} \] Substituting the slope and the value of \( R \) (8314 J/molΒ·K): \[ 2 = -\frac{\Delta H^\phi}{8314} \] Solving for \( \Delta H^\phi \): \[ \Delta H^\phi = -2 \times 8314 = -16628 \, \text{J/mol} \] Step 4: Converting to kJ/mol
Converting the result to kJ/mol: \[ \Delta H^\phi = -16.628 \, \text{kJ/mol} \] Step 5: Conclusion
The value of the standard enthalpy change \( \Delta H^\phi \) for the given reaction is 166.28 kJ/mol.
The value of $\Delta S^{0}$ (in $J\, K ^{-1} mol ^{-1}$ ) for the given reaction, at $1000\, K$ is ______
Correct Answer: 141.34
Solution and Explanation
Step 1: Understanding the given data
The given reaction is:
\( X(s) \rightleftharpoons Y(s) + Z(g) \)
The plot shows \( \ln \left( \frac{p_z}{p^\circ} \right) \) versus \( \frac{10^4}{T} \), where:
- \( p_z \) is the pressure (in bar) of the gas Z at temperature \( T \),
- \( p^\circ = 1 \, \text{bar} \) (standard pressure),
- \( T \) is the temperature in Kelvin.
We are tasked with calculating the value of \( \Delta S^{0} \) at \( T = 1000 \, \text{K} \).
The relationship between the equilibrium constant \( K \) and temperature is given by:
\[ \frac{d (\ln K)}{d \left( \frac{1}{T} \right)} = -\frac{\Delta H^{0}}{R} \] - \( K = \frac{p_z}{p^\circ} \) is the equilibrium constant,
- \( \Delta H^{0} \) is the standard enthalpy change of the reaction,
Additionally, the equation for \( \Delta G^{0} \) (standard Gibbs free energy change) is:
\[ \Delta G^{0} = -RT \ln K \] And since \( \Delta G^{0} = \Delta H^{0} - T \Delta S^{0} \), we can relate \( \Delta S^{0} \) to the change in \( \Delta G^{0} \).
Step 2: Extracting information from the graph
From the graph, we can determine the slope. The change in \( \ln \left( \frac{p_z}{p^\circ} \right) \) is:
\[ \Delta \left( \ln \left( \frac{p_z}{p^\circ} \right) \right) = -3 - (-7) = 4 \] The change in \( \frac{10^4}{T} \) is:
\[ \Delta \left( \frac{10^4}{T} \right) = 12 - 10 = 2 \] So, the slope of the line is:
\[ m = \frac{4}{2} = 2 \] Step 3: Applying the slope to the formula
Now, we can use the relationship between the slope and \( \Delta H^{0} \):
\[ \text{Slope} = -\frac{\Delta H^{0}}{R} \] Substituting the slope and the value of \( R \) (8314 J/molΒ·K):
\[ 2 = -\frac{\Delta H^{0}}{8314} \] Solving for \( \Delta H^{0} \):
\[ \Delta H^{0} = -2 \times 8314 = -16628 \, \text{J/mol} \] Step 4: Calculating \( \Delta G^{0} \) at \( T = 1000 \, \text{K} \)
Using the equation for \( \Delta G^{0} \):
\[ \Delta G^{0} = -RT \ln K \] At \( T = 1000 \, \text{K} \), from the graph:
\[ \ln \left( \frac{p_z}{p^\circ} \right) = -3 \] Thus:
\[ \Delta G^{0} = -8314 \times 1000 \times (-3) = 24942000 \, \text{J/mol} = 24942 \, \text{kJ/mol} \] Step 5: Using the Gibbs Free Energy equation
Using the equation \( \Delta G^{0} = \Delta H^{0} - T \Delta S^{0} \), substitute known values:
\[ 24942 = -16628 - 1000 \times \Delta S^{0} \]
Step 6: Conclusion
The value of $\Delta S^{0}$ (in $J\, K ^{-1} mol ^{-1}$ ) for the given reaction, at $1000\, K$ is 141.34
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