Question

Consider the following data for the reaction
$X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$
at 600 K. The $\Delta_r G^\ominus$ (in kJ $\text{mol}^{-1}$) for the reaction is :}

• A. $-21000$
• B. $-10$
• C. $-1000$
• D. $-9.012$

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Gibbs Free Energy change for a reaction can be calculated using the relation $\Delta_r G^\ominus = \Delta_r H^\ominus - T \Delta_r S^\ominus$. We first need to calculate the standard enthalpy and entropy change for the reaction using the provided formation data.
: Key Formula or Approach:
\[ \Delta_r H^\ominus = \sum \Delta_f H^\ominus (\text{products}) - \sum \Delta_f H^\ominus (\text{reactants}) \]
\[ \Delta_r S^\ominus = \sum S^\ominus (\text{products}) - \sum S^\ominus (\text{reactants}) \]
\[ \Delta_r G^\ominus = \Delta_r H^\ominus - T \Delta_r S^\ominus \]
Step 2: Detailed Explanation:
1. Calculation of $\Delta_r H^\ominus$:
Reaction: $X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)$
$\Delta_r H^\ominus = [2 \times \Delta_f H^\ominus(XY)] - [\Delta_f H^\ominus(X_2) + \Delta_f H^\ominus(Y_2)]$
$\Delta_r H^\ominus = [2 \times 42] - [8 + 80] = 84 - 88 = -4 \text{ kJ mol}^{-1}$.

2. Calculation of $\Delta_r S^\ominus$:
$\Delta_r S^\ominus = [2 \times S^\ominus(XY)] - [S^\ominus(X_2) + S^\ominus(Y_2)]$
$\Delta_r S^\ominus = [2 \times 200] - [140 + 250] = 400 - 390 = 10 \text{ J mol}^{-1}\text{ K}^{-1}$.

3. Calculation of $\Delta_r G^\ominus$:
Convert $\Delta_r S^\ominus$ to kJ units: $10 \text{ J} = 0.010 \text{ kJ}$.
$T = 600 \text{ K}$.
$\Delta_r G^\ominus = (-4 \text{ kJ mol}^{-1}) - (600 \text{ K} \times 0.010 \text{ kJ mol}^{-1}\text{ K}^{-1})$
$\Delta_r G^\ominus = -4 - 6 = -10 \text{ kJ mol}^{-1}$.
Step 3: Final Answer:
The $\Delta_r G^\ominus$ for the reaction at 600 K is $-10$ kJ $\text{mol}^{-1}$.