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Calculate the standard free energy chang...

Calculate the standard free energy change for the reaction :
`H_(2)(g) + I_(2)(g) to 2HI(g) DeltaH^(@) = 51.9 kJ "mol"^(-1)` Given : `S^(@)(H_(2)) = 130.6 J K^(-1) "mol"^(-1)`
`S^(@)(I_(2)) = 116.7 J K^(-1) "mol"^(-1)` and `S^(@)(HI) = 206.3 J K^(-1) "mol"^(-1)`.

Text Solution

Verified by Experts

The correct Answer is:
`DeltaG^(@) = + 2640. 6 Jmol^(-1)` . Not feasible
`Delta_(r)S^(@)= 2xxS_((HI))^(@) - [ S_((H_(2)))^(@) +S_((I_(2)))^(@)]= 2 xx 206.3 - ( 130.6+ 116.7) = 165.3 J mol^(-1)`
`Delta_(r)G^(@) = Delta_(r)H^(@) - T Delta_(r)S^(@) = 51900 - 298 xx 165.3 = 2640.6 J mol^(-1)`
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Calculate the standard free energy change for the reaction: H_(2)(g) +I_(2)(g) rarr 2HI(g), DeltaH^(Theta) = 51.9 kJ mol^(-1) Given: S^(Theta) (H_(2)) = 130.6 J K^(-1) mol^(-1) , S^(Theta) (I_(2)) = 116.7 J K^(-1) mol^(-1) and S^(Theta) (HI) =- 206.8 J K^(-1) mol^(-1) .

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Knowledge Check

  • Calculate the free energy change for the following reaction at 300 K. 2CuO_((s)) rarr Cu_(2)O_((s))+(1)/(2)O_(2(g)) Given Delta H = 145.6 kJ mol^(-1) and Delta S = 116.JK^(-1) mol^(-1)

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