Question
The enthalpy and entropy change for the reaction, $$B{r_2}\left( l \right) + C{l_2}\left( g \right) \to 2\,BrCl\left( g \right)$$ are $$30\,kJ\,mo{l^{ - 1}}$$ and $$105\,J{K^{ - 1}}\,mo{l^{ - 1}}$$ respectively. The temperature at which the reaction will be in equilibrium is
A.
285.7$$\,K$$
B.
273$$\,K$$
C.
450$$\,K$$
D.
300$$\,K$$
Answer :
285.7$$\,K$$
Solution :
At equilibrium Gibbs free energy change $$\left( {\Delta G} \right)$$ is equal to zero.
$$\Delta G = \Delta H - T\Delta S$$
$$0 = 30 \times {10^3}\left( {J\,mo{l^{ - 1}}} \right)$$ $$ - T \times 105\left( {J\,{K^{ - 1}}mo{l^{ - 1}}} \right)$$
$$\eqalign{
& \therefore \,\,T = \frac{{30 \times {{10}^3}}}{{105}}K \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = 285.71\,K \cr} $$