Question

The enthalpy and entropy change for the reaction :
$Br_2(\ell) + Cl_2(g) \rightarrow 2BrCl(g)$

are $30 \,kJmol^{-1}$ and $105\,JK^{-1} mol^{-1}$ respectively. The temperature at which the reaction will be in equilibrium is :

• A. 285.7 K
• B. 373 K
• C. 250 K
• D. 400 K

Answer 1

Answer: (A)

For the reaction

$Br _{2}(\ell)+ Cl _{2}( g ) \rightarrow 2 BrCl ( g )$

$\Delta H =30 kJ / mol$

$\Delta S =105 JK ^{-1} mol ^{-1}$

For at equilibrium $\Delta G =0$

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

$\Delta H = T \Delta S$

$T =\frac{\Delta H }{\Delta S }=\frac{30 \times 1000 J mol ^{-1}}{105 JK ^{-1} mol ^{-1}}$

$=2857 \,K$