InterviewSolution
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InterviewSolution
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Calculate `Delta_(r)G^(Theta)` for the following reactions using `Delta_(f)G^(Theta)` values and and predict which reactions are spontaneous. a. `Ca(s) +CI_(2)(g) rarr CaCI_(2)(s)` b. `H_(2)O(s) rarr Hg(l) +(1)/(2)O_(2)(g)` c. `NH_(3)(g) +2O_(2)(g) rarr HNO_(3)(l) +H_(2)O(l)` `Delta_(f)G^(Theta)` value `(kJ mol^(-1))` are: `CaCI_(2)(s) =- 748.1, HgO(s) =- 58.84` `NH_(3)(g) =- 16.45, HNO_(3)(l) = - 80.71`, `H_(2)O(l) =- 237.13` |
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Answer» `Delta_(r)G^(Theta)`can be calculated by using the relation. `Delta_(r)G^(Theta) = sum Delta_(r)G^(Theta) ("products") -sum Delta_(r)G^(Theta) ("reactants")` a. `Ca(s) +CI_(2) (g) rarr CaCI_(2)(s)` `Delta_(r)G^(Theta) = Delta_(f)G^(Theta) [CaCI_(2)(s)]` `-{Delta_(f)G^(Theta)[Ca(s)] +Delta_(f)G^(Theta) [CI_(2)(g)]}` `=- 748.1 -(0+0) =- 748.1 kJ` Reaction is spontaneous. b. `HgO(s) rarr Hg(l) +(1)/(2)O_(2)(g)` `Delta_(r)G^(Theta) =DeltaG^(Theta) [Hg(l) +(1)/(2)DeltaG^(@) [O_(2)(g)] -Delta_(f)G^(Theta) [HgO(s)]` `=0+0 -(-58.54) = 58.84 kJ` Reaction is not spontaneous c. `NH_(3)(g) +2O_(2)(g) rarr HNO_(3)(l) +H_(2)O(l)` `Delta_(r)G^(Theta) =Delta_(f)G^(Theta) [HNO_(3)(l)] +Delta_(f)G^(Theta) [H_(2)O(l)]` `-{Delta_(f)G^(Theta) [NH_(3)(g)] +2Delta_(f)G^(Theta) [O_(2)(g)]}` `=- 80.71 +(-237.13) -{-16.45+0}` `=- 80.71 -237.13 +16.45 =- 301.39kJ` Reaction is spontanoeus. a. `DeltaG^(Theta) =- 2.303 RT logK` |
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