The enthalpy and entropy change for the reaction
`Br_(2)(l) + Cl_(2)(g) rightarrow 2BrCl(g)` are `40 kJ mol^(-1)` and `110 JK^(-1) mol ^(-1)` respectively. The temperature at which the reaction will be in equilibrium is

Options

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

Answer

C

Explanation

For the reaction `Br_(2)(l) + Cl_(2)(g) rarr 2BrCl(g)`
`Delta H= 30kJ`/mol
`Delta S= 105JK^(-1) mol^(-1)`
at equilibrium `Delta G= 0`
`therefore Delta G= Delta H- T Delta S`
`Delta H= T Delta S`
`T=(Delta H)/(Delta S)= (30 xx 1000)/(105)= 285.7K`