Question

Consider the following reaction: \[ 2A(g)+B(g)\rightarrow 2D(g) \] \[ \Delta U^\circ=-10\ kJ\ mol^{-1} \] \[ \Delta S^\circ=-44\ J\ K^{-1}mol^{-1} \] at \(298K\). Identify the correct option with \(\Delta G^\circ\) for the reaction and spontaneity of the reaction at \(298K\). \[ R=8.31\ J\ mol^{-1}K^{-1} \]

• A. \(-0.63568\ kJ\ mol^{-1}\), spontaneous
• B. \(+0.63568\ kJ\ mol^{-1}\), non-spontaneous
• C. \(-1.635\ kJ\ mol^{-1}\), spontaneous
• D. \(+1.635\ kJ\ mol^{-1}\), non-spontaneous

Hint:

Use \(\Delta H=\Delta U+\Delta n_gRT\) first, then calculate \(\Delta G=\Delta H-T\Delta S\).

Answer 1

The Correct Option is B

Step 1: Understanding the Topic:
The question is based on Chemical Thermodynamics and the criteria for the spontaneity of a chemical reaction. The standard Gibbs free energy change ($\Delta G^o$) is the ultimate predictor of whether a reaction will proceed forward under standard conditions at a specific temperature. To calculate $\Delta G^o$, we need to relate the internal energy change ($\Delta U^o$) to the enthalpy change ($\Delta H^o$), and then use the Gibbs-Helmholtz equation involving entropy.
Step 2: Key Formulas and Approach:
The following fundamental thermodynamic relations are used:
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
$\Delta H^o = \Delta U^o + \Delta n_g RT$
$\Delta G^o = \Delta H^o - T \Delta S^o$
Calculations must be done carefully regarding units, converting everything to Joules before the final step.
Step 3: Detailed Explanation:

Find $\Delta n_g$: For the reaction $2A(g) + B(g) \rightarrow 2D(g)$, the gaseous product moles = 2 and gaseous reactant moles = $2+1 = 3$. Thus, $\Delta n_g = 2 - 3 = -1$.
Calculate $\Delta H^o$: We are given $\Delta U^o = -10 \text{ kJ} = -10000 \text{ J}$. \[ \Delta H^o = -10000 + (-1 \times 8.31 \times 298) \] \[ \Delta H^o = -10000 - 2476.38 = -12476.38 \text{ J} \]
Calculate $\Delta G^o$: Now substitute $\Delta H^o$, $T=298\text{ K}$, and $\Delta S^o = -44 \text{ J/K}$. \[ \Delta G^o = -12476.38 - (298 \times -44) \] \[ \Delta G^o = -12476.38 + 13112 = +635.62 \text{ J} \]
Spontaneity Check: Converting to kJ, $\Delta G^o \approx +0.63568 \text{ kJ/mol}$. Since the value is positive ($>0$), the reaction is non-spontaneous at 298 K under standard conditions.
Step 4: Final Answer:
The calculated value is approximately $+0.63568 \text{ kJ mol}^{-1}$ and the reaction is non-spontaneous. This matches option (C).