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The standard Gibb's energy change for th...

The standard Gibb's energy change for the formation of propane. `C_(3)H_(8) (g)` at 298 K is
`[Delta_(f)H^(theta)` for propane `=-103.85 kJ" mole"^(-1), S^(theta)._(m)C_(3)H_(8)(g)=270.2J.K^(-1) mol^(-1), S^(theta)._(m)H_(2)(g)=130.68J.K^(-1) mol^(-1)`. and `S^(theta)._(m)C_("(graphite)")=5.74JK^(-1) mol^(-1)]`

A

`-23.4 k.J`

B

`-44.4 k.J`

C

`-54.4 k.J`

D

`-104.5 kJ`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaG^(@)=DeltaH^(@)-T DeltaS^(@)`
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Calculate the standard Gibbs energy change for the formation of propane at 298 K: 3C("graphite") + 4H_(2)(g) to C_(3)H_(8)(g) Delta_(f)H^(@) for propane, C_(3)H_(8)(g) = -103.8 kJ mol^(-1) . Given : S_(m)^(0)[C_(3)H_(8)(g)] = 270.2 J K^(-1) "mol"^(-1) S_(m)^(@)("graphite") = 5.70 J K^(-1) "mol"^(-1) and S_(m)^(0)[H_(2)(g)] = 130.7 J K^(-1) "mol"^(-1) .

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) .

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) .

Calculate the entropy change for the following reaction H_(2)(g) +CI_(2)(g) rarr 2HCI (g) at 298 K Given S^(Theta)H_(2) = 131 J K^(-1) mol^(-1), S^(Theta)CI_(2) = 233 J K^(-1) mol^(-1) , and S^(Theta) HCI = 187 J K^(-1) mol^(-1)

Calculate Delta_(r)S_(m)^(Theta) for the reaction: 4Fe(s)+3O_(2)(g) rarr 2Fe_(2)O_(3)(s) Given that S_(m)^(Theta)(Fe) = 27.3J K^(-1) mol^(-1) , S_(m)^(Theta)(O_(2)) = 205.0J K^(-1)mol^(-1) and S_(m)^(Theta)(Fe_(2)O_(3)) = 87.4 J K^(-1) mol^(-1) .

Using Delta_(f)H^(@) and S_(m)^(@) calculate the standard Gibbs energy of formation, Delta_(f)G^(@) for each following : (a) CS_(2)(l) (b) N_(2)H_(4)(l) Delta_(f)H^(@)(CS_(2)) = 89.70 kJ "mol"^(-1), Delta_(r)S^(@)(CS_(2)) = 151.34 J K^(-1) "mol"^(-1) Delta_(f)H^(@)(N_(2)H_(4)) = 50.63 kJ "mol"^(-1) , Delta_(r)S^(@)N_(2)H_(2) = 121.21 J K^(-1) "mol"^(-1) .

Calculate the change in entropy for the following reaction 2CO(g) +O_(2)(g) rarr 2CO_(2)(g) Given: S_(CO)^(Theta)(g)=197.6 J K^(-1)mol^(-1) S_(O_(2))^(Theta)(g)=205.03 J K^(-1)mol^(-1) S_(CO_(2))^(Theta)(g)=213.6 J K^(-1)mol^(-1)

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