Question

1. Synthesis of compound in the gaseous phase reaction has been studied at two temperatures. At 298 K, the kp value is 1.22×10-3 and at 498 K the kp value is 2.16. Calculate the standard enthalpy change of this chemical reaction. Where value of R is 1.987Cal K-1mol-1.

Answer 1

1.987 Cal$\times$K^-1$\times$mol^-1 = 8.314 J$\times$mol^-1$\times$K^-1

$\Delta G = -RTlnK_p$

$\Delta G_1 = -RT_1lnK_{p1}=-8.314\times298\times ln(1.22\times10^{-3}) = 16621.8$

$\Delta G_2 = -RT_2lnK_{p2}=-8.314\times498\times ln(2.16) = -3188.5$

$\Delta G = \Delta H - T\Delta S$

$\Delta H = \Delta G + T\Delta S$

$\Delta S = \frac{\Delta H - \Delta G}{T} = \frac{\Delta H - \Delta G_1}{T_1}$

$\Delta H = \Delta G_2 + T_2(\frac{\Delta H - \Delta G_1}{T_1}) = \Delta G_2 + \frac{T_2\Delta H}{T_1} - \frac{T_2\Delta G_1}{T_1}$

$T_1 \Delta H = T_1\Delta G_2 + T_2\Delta H - T_2\Delta G_1$

$T_1 \Delta H -T_2\Delta H = T_1\Delta G_2 - T_2\Delta G_1$

$\Delta H(T_1-T_2) = T_1\Delta G_2 - T_2\Delta G_1$

$\Delta H = \frac{T_1\Delta G_2 - T_2\Delta G_1}{(T_1-T_2)}$

$\Delta H = \frac{298\times(-3188.5)-498\times(16621.8)}{298-498}=46139 \ (J)$

Answer: $\Delta$H = 46139 J.