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Standar entropies of x(2), y(2) and xy(3...

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

A

750K

B

1000K

C

1250K

D

500K

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(2) y_(2) + (3)/(2) y_(2) hArr xy_(3)`
`Delta S = Delta S_(xy) - ((1)/(2) sy_(2) + (3)/(2) sy_(2))`
`50 - ((1)/(2) xx 60 + (3)/(2) xx 40) = -40JK^(-1) mol^(-1)`
`Delta G = Delta H - T Delta S, Delta G = 0 = T = (Delta H)/(Delta S) = 750k`
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