For reversible reaction : `X_((g))+3Y_((g))hArr 2Z_((g)), DeltaH=-"40 kJ"` Standard entropies of X, Y and Z are 60, 40and `"50 J K"^(-1)",ol"^(-1)` respectively. The temperature at which the above reaction is in equilibrium is

Options

A. 400K
B. 500 K
C. 273K
D. 373K

Answer

B

Explanation

`X_((g))+ 3Y_((g)) hArr 2Z_((g))`
`Delta S^(@)= 2S^(@) (Z)- {S^(@) (X) + 3S^(@)(Y)}`
`=2 xx 50 - {60 + 3 xx 40} = 100-180= -80 JK^(-1) mol^(-1)`
Given `Delta H^(@)= -40kJ = -40,000J, Delta G^(@)= Delta H^(@)- T Delta S^(@)`
At equilibrium, `Delta G^(@)= 0 therefore Delta H^(@)= T Delta S^(@)`
or `T = (Delta H^(@))/(Delta S^(@))= (40000)/(80) = 500K`