Question

The equilibrium constant of the reaction $2C_3{H}_6\left(g\right)\rightleftharpoons C_2{H}_4\left(g\right)+C_4{H}_8\left(g\right)$ is found to fit the expression:
$lnK=-1.04-\frac{1088}{T}$

Calculate the standard reaction enthalpy and entropy at 400 K :

• A. $\Delta H^o=4.5kJ/mol;\Delta S^o=4.32J/mo{l}^1{K}^1$
• B. $\Delta H^o=9.04kJ/mol;\Delta S^o=-8.64J/mo{l}^1{K}^1$
• C. $\Delta H^o=18.08kJ/mol;\Delta S^o=17.28J/mo{l}^1{K}^1$
• D. None of these

Answer 1

The correct option is B $\Delta H^o=9.04kJ/mol;\Delta S^o=-8.64J/mo{l}^1{K}^1$

$\Delta G^0=-RTlnK=\Delta H^0-T\Delta S^0$

$lnK=\frac{\Delta S^0}{R}-\frac{\Delta H^0}{RT}$......(1)

But $lnK=-1.04-\frac{1088}{T}$......(2)

From (1) and (2),

$\frac{\Delta S^0}{R}=-1.04$ and $\frac{\Delta H^0}{RT}=\frac{1088}{T}$

$\Delta S^0=-1.04\times 8.314=-8.64J/mol/K$

$\Delta H^0=1088\times 8.314=9040J/mol=9.04kJ/mol$