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

Options

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

Answer

A

Explanation

For the reaction
`Br_(2)(l)+Cl_(2)(g)to2BeCl(g)`
`DeltaH=30KJ//mol
DeltaS=105jk^(-1)mol^(-1)`
For at equilibrium`DeltaG`=0
`thereforeDeltaG=DeltaH-TDeltaS
DeltaH=TDeltaS`
`T=(DeltaH)/(DeltaS)=(30xx1000J mol^(-1))/(105 JK^(-1))mol^(-1)=285.7k`