For a reversible reaction : `X_((g))+3Y_((g))hArr 2Z_((g)), Delta H=-40 kJ`, the standard entropies of X,Y and Z are 60, 40 and `50 JK^(-1)mol^(-1)` respectively. The temperature at which the above reaction attains equilibrium is about

Options

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

Answer

B

Explanation

`X_((g))+3Y_((g))hArr 2Z_((g))`
`Delta S^(@)=2S^(@)(Z)-{S^(@)(X)+3S^(@)(Y)}`
`=2xx50-{60+3xx40}=100-180`
`=-80J K^(-1)mol^(-1)`.
Given `Delta H^(@)=-40 kJ =-40,000 J`
`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)=500 K`.