Answer to Question #191869 in Physical Chemistry for Abdullah Latif

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.