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

Answer

A

Explanation

Given reaction is `1/2 X_2+ 3/2 Y_2 = XY_3`
We know, `Delta S^@=Sigma S_("Products")^@- SigmaS_("ractants")^@`
=50 -(30+60)=-40`K^(-1) " mol "^(-1)`
At equilibrium `DeltaH^@ = T DeltaS^@`
`T=(Delta H^@)/(Delta S^@)=(-30 xx 10^3 " J mol "^(-1))/(-40 JK^(-1) " mol "^(-1)) =750 K`