Question

For the given reaction:
\[ \mathrm{X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)} \] The following data table is provided at \(600 \, \mathrm{K}\):
& \(\Delta_f H \, (\mathrm{kJ/mol})\) & \(S_m \, (\mathrm{J\,K^{-1}\,mol^{-1}})\)
\(\mathrm{X_2}\) & 80 & 140
\(\mathrm{Y_2}\) & 8 & 250
\(\mathrm{XY}\) & 42 & 200
\endtabular Calculate \(\Delta_r G\) at \(600 \, \mathrm{K}\).

• A. \(-10 \, \mathrm{kJ/mol}\)
• B. \(-100 \, \mathrm{kJ/mol}\)
• C. \(-2 \, \mathrm{kJ/mol}\)
• D. \(+2 \, \mathrm{kJ/mol}\)

Hint:

For thermodynamics questions, always use \(\Delta G = \Delta H - T\Delta S\). Be careful with units: if \(\Delta H\) is in \(\mathrm{kJ/mol}\) and entropy is in \(\mathrm{J\,K^{-1}\,mol^{-1}}\), convert \(T\Delta S\) into \(\mathrm{kJ/mol}\) before subtraction.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Gibbs Free Energy change for a reaction (\(\Delta_rG\)) combines enthalpy and entropy contributions to tell us the spontaneity of a reaction at a given temperature.
The Gibbs-Helmholtz equation is: \(\Delta_rG = \Delta_rH - T\Delta_rS\).
Both \(\Delta_rH\) and \(\Delta_rS\) are calculated from standard molar enthalpies of formation and standard molar entropies, respectively.
Step 2: Key Formula or Approach:
1. \(\Delta_rH = \sum \Delta_fH(\text{products}) - \sum \Delta_fH(\text{reactants})\)
2. \(\Delta_rS = \sum S_m(\text{products}) - \sum S_m(\text{reactants})\)
3. Convert \(\Delta_rS\) from J/K to kJ/K by dividing by 1000.
4. Substitute into \(\Delta_rG = \Delta_rH - T\Delta_rS\).
Step 3: Detailed Explanation:
Calculation of \(\Delta_rH\):
From the data table (using values: \(\Delta_fH(XY) = 42\) kJ/mol, \(\Delta_fH(X_2) = 80\) kJ/mol, \(\Delta_fH(Y_2) = 8\) kJ/mol):
\[ \Delta_rH = [2 \times \Delta_fH(XY)] - [\Delta_fH(X_2) + \Delta_fH(Y_2)] \]
\[ \Delta_rH = [2 \times 42] - [80 + 8] = 84 - 88 = -4 \text{ kJ/mol} \]
Calculation of \(\Delta_rS\):
From the data table (using values: \(S_m(XY) = 200\) J/K$\cdot$mol, \(S_m(X_2) = 140\) J/K$\cdot$mol, \(S_m(Y_2) = 250\) J/K$\cdot$mol):
\[ \Delta_rS = [2 \times S_m(XY)] - [S_m(X_2) + S_m(Y_2)] \]
\[ \Delta_rS = [2 \times 200] - [140 + 250] = 400 - 390 = +10 \text{ J/K}\cdot\text{mol} \]
\[ \Delta_rS = \frac{10}{1000} = 0.01 \text{ kJ/K}\cdot\text{mol} \]
Calculation of \(\Delta_rG\) at 600 K:
\[ \Delta_rG = \Delta_rH - T\Delta_rS = -4 - (600 \times 0.01) \]
\[ \Delta_rG = -4 - 6 = -10 \text{ kJ/mol} \]
Step 4: Final Answer:
The value of \(\Delta_rG\) at 600 K is --10 kJ/mol.