Standard entropy of `X_(2)` , `Y_(2)` and `XY_(3)` are `60, 40 ` and `50JK^(-1)mol^(-1)` , respectively. For the reaction, `(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3),DeltaH=-30KJ` , to be at equilibrium, the temperature will be:

Options

A. 1250 K
B. 500 K
C. 750 K
D. 1000 K

Answer

C

Explanation

`Delta S^(@)=Sigma(S^(@))_("Products")-Sigma (S^(@))_("Reactants")`
`Delta S^(@)=50-((1)/(2)xx60+(3)/(2)xx40)=-40 JK^(-1)mol^(-1)`
`Delta H^(@)=-30 kJ mol^(-1)`
at equalibrium `-Delta G = 0`
`T=(Delta H)/(Delta S)=(-30xx1000 J mol^(-1))/(40JK^(-1)mol^(-1))=750K`