Standard entropies of `X_(2), Y_(2) and XY_(3) are 60, 40 and 50 JK^(–1) mol^(–1)` respectively. For the reaction `(1)/(2)X_(2)+(3)/(2)Y_(2)hArr XY_(3)DeltaH=-30KJ `to be at equilibrium, the temperature should be:

Options

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

Answer

C

Explanation

`1//2 X_(2) + 3//2 Y_2 to XY_3`
`Delta S_("rection") = 50 -(3/2 xx 40 + 1/2 xx 60) = -40 J mol^(-1)`
`Delta G = Delta H - T Delta S`
At equilibrium `Delta G = 0`
`Delta H = T Delta S`
`Delta H = -3 kJ` (Given)
`= -30 xx 1000 J = -30000 J`
`:. " " T = (Delta H)/(Delta S) = (-30,000)/(-40) = 750 K`.