1.

Standard entropies of X_(2), Y_(2) and XY_(3) are 60, 40 and 50 JK^(-1) respectively (1)/(2) X_(2) + (3)/(2) Y, Delta H = -30 kJ to be at equilibrium, the temperature should be :

Answer»

1250 K
500 K
750 K
1000 K

Solution :`(1)/(2) X_(2) + (3)/(2) Y_(2) hArr XY_(3)`
`Delta S^(@) = Sum S_(P)^(@) - Sum S_(R)^(@)`
= 50 - (30 + 60)
`Delta S^(@) = - 40 JK^(-1) mol^(-1)`
`Delta G = Delta H - T Delta S`,
`0 = Delta H - T Delta S`
`Delta H = - 30 xx 10^(3) J mol^(-1)`
`T = (Delta H^(@))/(Delta S^(@))`
`= (-30 xx 10^(3) J mol^(-1))/(-40 JK^(-1) mol^(-1))`
= 750 K


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