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: Relate \(\Delta H^\circ\) and \(\Delta U^\circ\).
For gaseous reactions: \[ \Delta H^\circ=\Delta U^\circ+\Delta n_gRT. \] Here: \[ \Delta n_g=\text{moles of gaseous products}-\text{moles of gaseous reactants}. \]

Step 2: Calculate \(\Delta n_g\).
Reaction: \[ 2A(g)+B(g)\rightarrow 2D(g) \] Moles of gaseous products: \[ 2 \] Moles of gaseous reactants: \[ 2+1=3 \] Therefore: \[ \Delta n_g=2-3=-1. \]

Step 3: Calculate \(\Delta H^\circ\).
\[ \Delta H^\circ=\Delta U^\circ+\Delta n_gRT. \] \[ \Delta U^\circ=-10\ kJ\ mol^{-1}. \] Now: \[ RT=8.31\times298=2476.38\ J\ mol^{-1}. \] \[ RT=2.47638\ kJ\ mol^{-1}. \] So: \[ \Delta n_gRT=(-1)(2.47638) \] \[ =-2.47638\ kJ\ mol^{-1}. \] Therefore: \[ \Delta H^\circ=-10-2.47638. \] \[ \Delta H^\circ=-12.47638\ kJ\ mol^{-1}. \]

Step 4: Convert entropy into kJ unit.
Given: \[ \Delta S^\circ=-44\ J\ K^{-1}mol^{-1}. \] Convert into kJ: \[ \Delta S^\circ=-0.044\ kJ\ K^{-1}mol^{-1}. \]

Step 5: Use Gibbs free energy equation.
\[ \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ. \] \[ \Delta G^\circ=-12.47638-298(-0.044). \] \[ 298\times0.044=13.112. \] So: \[ \Delta G^\circ=-12.47638+13.112. \] \[ \Delta G^\circ=+0.63562\ kJ\ mol^{-1}. \] Approximately: \[ \Delta G^\circ=+0.63568\ kJ\ mol^{-1}. \]

Step 6: Decide spontaneity.
If: \[ \Delta G^\circ<0, \] the reaction is spontaneous. If: \[ \Delta G^\circ>0, \] the reaction is non-spontaneous. Here: \[ \Delta G^\circ>0. \] Therefore, the reaction is: \[ \text{non-spontaneous}. \]