Standard entropies of `x_(2) , y_(2) and xy_(3)` are `60,40 and 50 "J/K mol"^(-1)` respectively for the reaction `(1)/(2) x_(2) + (3)/(2) y_(2) to xy_(3) , Delta H = - 30 "kJ"` to be at equilibrium the temperature should be.

Options

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

Answer

B

Explanation

`Delta S = 50 - ((60)/(2) + (3)/(2) xx 40)`
`= 50 (30+ 60) = - 40 "J/K mol"`
`Delta G = 0`
`therefore Delta H = T Delta S therefore T = (-30 xx 10^(3) )/( - 40) = 750 K`