Question

For the following reaction, 2HI(g) – H (9) +1,9 the equilibrium constant k, = 2.60X10-2 at 1110 K and 2.68x10-2 at 1140 K. Calculate AHⓇ for the reaction, assuming no change in AH between 1110 K and 1140 K. AH = kJ mol-1

Answer 1

Answer:

10.63 kJ/mol is the $\Delta H$ of the reaction.

Explanation:

To calculate $\Delta H$ of the reaction, we use Van't Hoff's equation, which is:

$\ln (\frac{K_1}{K_2})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{1}$ = equilibrium constant at $T_1$

$K_{2}$ = equilibrium constant at $T_2$

$\Delta H$ = Enthalpy change of the reaction

R = Gas constant = 8.314 J/mol K

Given : $K_1=2.60\times 10^{-2}$

$K_2=2.68\times 10^{-2}$

$T_1$ = initial temperature = $1110 K$

$T_2$ = final temperature = $1140 K$

$\Delta H$ = ?

Putting values in above equation, we get:

$\ln(\frac{2.68\times 10^{-2}}{2.60\times 10^{-2}})=\frac{\Delta H}{8.314J/mol.K}[\frac{1}{1110 K}-\frac{1}{1140 K}]$

$\Delta H=10,627.617 J/mol=10.63 kJ/mol$

10.63 kJ/mol is the $\Delta H$ of the reaction.