The enthalpy and entropy change for the reaction :
`Br_(2)(l) + Cl_(2)(g) rarr 2BrCl(g)`
are `30 kJ mol^(-1) and 105 JK^(-1)` respectively. The temperature at which the reaction will be in equilibrium is :

Options

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

Answer

D

Explanation

At equilibrium,
`Delta G = Delta H - T Delta S`
or `Delta H = T Delta S`
or `T = (Delta H)/(Delta S)`
`(30 xx 10^(3) J mol^(-1))/(105 J K^(-1) mol^(-1))`
= 285.7 K