Standard entropies of `X_(2), Y_(2)` and `XY_(3)` are 60, 30 are `50 JK^(-1) mol^(-1)` respectively. For the reaction `(1)/(2)X_(2)+(3)/(2)Y_(2)hArr XY_(3), Delta H = -30 kJ` to be at equilibrium, the temperature should be :

Options

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

Answer

b

Explanation

`(1)/(2) X_(2) + ( 3)/(2)Y_(2) hArr XY_(3)`
`DeltaS^(@) = SigmaS_(P)^(@) - Sigma _(R)^(@) = 50-( 30+60) = - 40 JK^(-1) mol^(-1)`
`DeltaG^(@) =DeltaH^(@) - T DeltaS^(@) ` . At equilibrium , `DeltaG^(@) = 0`. Hence,`TDeltaS^(@) = DeltaH^(@)`
or `T= (DeltaH^(@))/( DeltaS^(@)) = ( -30 xx 10^(3) J mol^(-1))/( -40JK^(-1) mol^(-1))= 750 K`