In the reaction, `A_(2)(g)+3B_(2)(g)rarr2AB_(3)(g)` the standard entropies in `(JK^(-1)mol^(-1))` of `A_(2)(g),B_(2)(g) and AB_(3)(g)` are respectively 190, 130 and 195 and the standard enthaly change for the reaction is `-95 kJmol^(-1)`. The temperature (in K) at which the reaction attains equilibrium is (assuming both the standard entropy change and standard enthalpy change for this reaction are constant over a wide range of temperature)

Options

A. 500
B. 400
C. 300
D. 600

Answer

A

Explanation

For the reaction,
`A_(2)(g)+3B_(2)(g)rarr2AB_(3)(g)`
Entropy change, `DeltaS^(@)=SigmaS_("product")-SigmaS_("reactants")`
`=2xxS_(AB_(3))-[S_(A_(2))+3S_(B_(2))]`
`=2xx195-[190+3xx130]`
`=390-[190+390]`
`=-190JK^(-1)mol^(-1)`
Given, `DeltaH^(@)=-95kJmol^(-1)`
At equilibrium, `DeltaG^(@)=0`
We know that,
`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`0=DeltaH^(@)-TDeltaS^(@)`
or `DeltaH^(@)=TDeltaS^(@)`
`T=(DeltaH^(@))/(DeltaS^(@))=(-95xx10^(3)Jmol^(-1))/(-190JK^(-1)mol^(-1))`
`=500K`