Question

For the reaction at 25° C, X₂O₄ (l) → 2XO₂(g), U and S are 2.1 K.Cal and 20 Cal/K respectively. what is G for the reaction at the same temperature? (R = 2 CAL K^-1MOL^-1)

• A. -2.67 k.Cal
• B. 2.67 k.Cal
• C. -1.67 k.Cal
• D. 3.67k.Cal

Answer 1

The Correct Option is A

To solve the problem, we need to calculate the change in Gibbs free energy ($\Delta G$) using the given internal energy change ($\Delta U$) and entropy change ($\Delta S$) at a constant temperature.

1. Recall the formula for Gibbs free energy change:
The relation is given by:
$ \Delta G = \Delta U + P\Delta V - T\Delta S $
However, for a reaction involving gases, we use:
$ \Delta G = \Delta H - T\Delta S $, and since $ \Delta H = \Delta U + \Delta n_g R T $,
we modify the formula as:
$ \Delta G = \Delta U + \Delta n_g R T - T\Delta S $

2. Gather the given data:
$\Delta U = 2.1 \, \text{K.Cal}$
$\Delta S = 20 \, \text{Cal/K}$
$T = 25^\circ C = 298 \, \text{K}$
$R = 2 \, \text{Cal mol}^{-1} \text{K}^{-1}$
From the reaction: $X_2O_4 (l) \rightarrow 2XO_2 (g)$
So, change in moles of gas, $\Delta n_g = 2 - 0 = 2$

3. Apply the formula:
$ \Delta G = \Delta U + \Delta n_g R T - T\Delta S $
Substituting the values:
$ \Delta G = 2.1 + 2 \times 2 \times 298 - 298 \times 20 \div 1000 $ (since we convert Cal to K.Cal)
$ \Delta G = 2.1 + 1192 - 5.96 $
$ \Delta G = 2.1 + 1192 - 5.96 = 1188.14 \, \text{K.Cal}$
Wait — this is too high for given options. Let’s re-express all values in **consistent units**:

• $\Delta U = 2.1 \, \text{K.Cal} = 2100 \, \text{Cal}$
• $T = 298 \, \text{K}$
• $\Delta S = 20 \, \text{Cal/K}$
• $R = 2 \, \text{Cal mol}^{-1} \text{K}^{-1}$

Now: $ \Delta G = 2100 + (2)(2)(298) - (298)(20) $
$ \Delta G = 2100 + 1192 - 5960 = 3292 - 5960 = -2668 \, \text{Cal} = -2.668 \, \text{K.Cal} $

4. Final Answer:
The value of $\Delta G$ is $-2.67 \, \text{K.Cal}$