Consider the following reaction : 2A (g) + B (g) arrow 2D (g) Δ U^...

The change in the number of moles of gaseous species is:

$\Delta n_g = n_p - n_r = 2 - (2 + 1) = -1$

The standard enthalpy change $\Delta H^{\ominus}$ is given by:

$\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g R T$

Substituting the given values:

$\Delta H^{\ominus} = -10 + (-1) \times (8.31 \times 10^{-3}) \times 298$

$\Delta H^{\ominus} = -10 - 2.47638 = -12.47638 \text{ kJ mol}^{-1}$

The standard Gibbs free energy change $\Delta G^{\ominus}$ is:

$\Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S^{\ominus}$

$\Delta G^{\ominus} = -12.47638 - 298 \times (-44 \times 10^{-3})$

$\Delta G^{\ominus} = -12.47638 + 13.112 = +0.63562 \text{ kJ mol}^{-1}$

Since $\Delta G^{\ominus} > 0$, the reaction is non-spontaneous.

Answer: $+0.63568$ kJ mol$^{-1}$, non-spontaneous