Question

The enthalpy and entropy change for the reaction : ${Br2(\ell) + Cl2(g) -> 2BrCl(g)}$ 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)

285.7 K

Answer 2

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 $\Delta G =0$ $\therefore \Delta G =\Delta H - T \Delta S$ $\Delta H = T \Delta S$ $T =\frac{\Delta H }{\Delta S }=\frac{30 \times 1000 J mol ^{-1}}{105 JK ^{-1} mol ^{-1}}$ $=2857 \,K$