The enthalpy and entropy change for the reaction,
`Br_(2)(l)+Cl_(2)(g)rarr2BrCl(g)`
are `30KJmol^(-1)` and `105JK^(-1)mol^(-1)` respectively. The temperature at which the raction will be in equilibrium is:

Options

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

Answer

C

Explanation

According to thermodynamics,
`Delta G = Delta H - T Delta S`
At equilibrium (by definition), `Delta G = 0`. Thus,
` 0 = 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`