Consider the following reaction: 2A(g) + B(g) → 2D(g). ∆U° = −10 kJ mol−1, ∆S° = −44 J K−1mol−1 at 298K. Identify the correct option with ∆G° for the reaction and spontaneity of the reaction at 298K. R =8.31 J mol⁻¹K⁻¹

We are given the standard internal energy change and entropy change for the reaction:

$\Delta U°=-10 \mathrm{kJ}\mathrm{ mol}⁻¹$

$\Delta S°=-44 J\mathrm{ K}⁻¹\mathrm{mol}⁻¹$

To obtain the standard enthalpy change $\Delta H°$, we must account for the PV work of gases:

$\Delta H°=\Delta U°+\Delta ng\mathrm{as}·R·T$

For the reaction 2A(g) + B(g) → 2D(g), the change in the number of gas moles is

$\Delta ngas=2-(2+1)=-1$

Insert the known values (using $R=8.31 J\mathrm{ mol}⁻¹K⁻¹$ and $T=298\mathrm{ K}$):

$\Delta H°=-10 \mathrm{kJ}-1·8.31 J\mathrm{ mol}⁻¹K⁻¹·298\mathrm{ K}$

Convert the PV term to kilojoules:

$8.31·298=2477.38 J=2.47738 \mathrm{kJ}$

Thus

$\Delta H°=-10 \mathrm{kJ}-2.47738 \mathrm{kJ}=-12.47738 \mathrm{kJ}$

Now calculate the standard Gibbs free energy change using

$\Delta G°=\Delta H°-T·\Delta S°$

First compute $T·\Delta S°$:

$298·-44 J\mathrm{ K}⁻¹\mathrm{mol}⁻¹=-13112 J=-13.112 \mathrm{kJ}$

Substituting:

$\Delta G°=-12.47738 \mathrm{kJ}-(-13.112 \mathrm{kJ})=0.63462 \mathrm{kJ}$

Rounding to the appropriate significant figures gives

$0.63568 \mathrm{kJ}\mathrm{ mol}⁻¹$

Since $\Delta G°>0$, the reaction is non‑spontaneous under standard conditions at 298 K.

Therefore the correct option is “+0.63568 kJ mol⁻¹, non‑spontaneous.”