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Calculate the standard Gibbs enegry chan...

Calculate the standard Gibbs enegry change for the combustion of `alpha-D` glucose at `300K`.
`C_(6)H_(12)O_(6)(s) +6O_(2)(g) rarr 6CO_(2)(g) +6H_(2)O(l)`
Given the standard enthalpies of formation `(kJ mol^(-1))`
`C_(6)H_(12)O_(6) =- 1274.5, CO_(2) =- 393.5, H_(2)O =- 285.8`.
Entropies `(J K mol^(-1))`
`C_(6)H_(12)O_(6) = 212.1, O_(1) = 205.0, CO_(2) =213, H_(2)O = 69.9`

Text Solution

Verified by Experts

`DeltaG^(@) = DeltaH^(@) -TDeltaS^(@)`
`DeltaH^(@) = [6Delta_(f)H^(@) (CO_(2)) + 6Delta_(f)H^(@) (H_(2)O)] - [Delta_(f)H^(@)(C_(6)H_(12)O_(6) + 6 Delta_(f)H^(@)(O_(2))]`
`= [6 xx(-393.5) + 6 xx (-285.8)] - [(-1274.5) + 6 xx 0]`
`[because Delta_(f)H^(@)(O_(2)) = 0]`
` = -2361.0 - 1714.8 + 1274.5 = -2801.3 kJ "mol"^(-1)`
`DeltaS^(@) = [6S^(@)(CO_(2))+ 6S^(@)(H_(2)O)] - [S^(@)(C_(6)H_(12)O_(6)) + 6^(@)(O_(2))]`
` = [6 xx (213.6) + 6 xx (69.9)] - [(212.1) + 6 xx (205.0)]`
` = [1281.6 + 419.4] - [212.1 + 1230]`
` = 258.9 J K^(-1) "mol"^(-1)`
`DeltaG^(@) = DeltaH^(@) - TDeltaS^(@)`
`DeltaH^(@) = -2801.3 kJ = 2801300 J mol^(-1), DeltaS^(@) = 258.9`
T = 298 K
`therefore DeltaG^(@) = -2801300 - 298 xx (258.9)`
`= -2878452 or = -2878.4 kJ "mol"^(-1)`.
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