Standard entropies of $\mathrm{X}_2, \mathrm{Y}_2$ and $\mathrm{XY}_5$ are 70, 50 and $110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature in Kelvin at which the reaction

$$\frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 \Delta \mathrm{H}^{\ominus}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}$$

will be at equilibrium is __________ (Nearest integer)

<p>To determine the temperature at which the given reaction is at equilibrium, we need to apply the concept of Gibbs free energy change $(\Delta G^0)$ at equilibrium, where $\Delta G^0$ is equal to zero. </p> <p>The given reaction is:</p> <p>$ \frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 $</p> <p>We are given the standard entropies and enthalpy change ($\Delta H^{\ominus}$):</p> <p><p>Standard entropies: </p></p> <p><p>$\mathrm{X}_2 = 70 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $</p></p> <p><p>$\mathrm{Y}_2 = 50 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $</p></p> <p><p>$\mathrm{XY}_5 = 110 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $</p></p> <p><p>Standard enthalpy change: </p></p> <p><p>$\Delta H^{\ominus} = -35 \, \mathrm{kJ/mol}$</p></p> <p>First, calculate the standard entropy change ($\Delta S_{Rxn}^0$) of the reaction:</p> <p>$ \Delta S_{Rxn}^0 = 110 - \left(\frac{1}{2} \times 70 + \frac{5}{2} \times 50\right) $</p> <p>$ = 110 - \left(35 + 125\right) $</p> <p>$ = 110 - 160 $</p> <p>$ = -50 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $</p> <p>At equilibrium, $\Delta G^0 = 0$, and the relation $\Delta G^0 = \Delta H^0 - T\Delta S^0$ applies. Thus:</p> <p>$ 0 = -35000 \, \mathrm{J/mol} - T(-50 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}) $</p> <p>Solving for $T$:</p> <p>$ T \cdot 50 = 35000 $</p> <p>$ T = \frac{35000}{50} $</p> <p>$ T = 700 \, \mathrm{K} $</p> <p>Therefore, the temperature at which the reaction is at equilibrium is 700 Kelvin.</p>