Question

The enthalpy and entropy change for the reaction :
$Br_2( l )+ Cl _2( g ) \rightarrow 2 BrCl ( g )$
are $30\, kJ\, mol ^{-1}$ and $105\, J \,K ^{-1} \,mol ^{-1}$ respectively. The temperature at which the reaction will be in equilibrium is

• A. 300 K
• B. 285.7 K
• C. 273 K
• D. 450 K

Answer 1

Answer: (B)

$Br _2(l)+ Cl_2( g ) \rightarrow 2 BrCl ( g )$
$\Delta H =30\, kJ\, mol { }^{-1}, \Delta S =105 \,J\, K ^{-1} \,mol ^{-1}$
$\Delta S =\frac{\Delta H }{ T }$ i.e., $105=\frac{30}{ T } \times 1000$
$\therefore T =\frac{30 \times 1000}{105}=285.7\, K$