Question

standard entropies for H2 cl2 and HCL is 60 40 and 60 joule Kelvin inverse respectively for reaction H2 + cl2 = 2 HCL Del H = + 30 kj to be at equilibrium, the temperature will be​

Answer 1

Answer: 1500 K

Explanation: The expression for entropy change is,

$\Delta S=\sum [n\times S_(product)]-\sum [n\times S_(reactant)]$

$H_2+Cl_2\rightarrow 2HCl$

$\Delta S=[(n_{HCl}\times S_{HCl})-[(n_{H_2}\times S_{H_2})+(n_{Cl_2}\times S_{Cl_2})]$

$\Delta S=[2\times 60+-[1\times 60+(1\times 40)]=20JK^{-1}$

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

$\Delta G$ = Gibbs free energy = zero at equilibrium

$\Delta H$ = enthalpy change = +30 KJ = 30000J

$\Delta S$ = entropy change = 20 J/K

$\Delta H=T\Delta S$ at eqm

$30000J=T\times 20J/K$

$T=1500K$