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The enthalpy and entropy change for the ...

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 reaction will be in equilibrium is:

A

285.7K

B

273 K

C

450 K

D

300 K

Text Solution

Verified by Experts

The correct Answer is:
A
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`
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