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49. The standard Gibb's energy change fo...

49. The standard Gibb's energy change for the formation of propane. CzH3(g) at 298K is [4, HⓇ for propane = -103.85K.J mole", S® C3H2(g)=270.2J.K-mol-1, SemH2(g)=130.68J.K-Imol-1 and somigraphite) =-5.74JK-mol-l] 1)-23.4 KJ 2) -44.4 KJ 3) -54.4 KJ 4)-104.5 KJ.

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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)]

Calculate the Gibb's energy change for the formation of propane, C_(3)H_(8(g)) at 298 K. Given that Delta_(f)H for propane = -103.85 kJ mol^(-1) Delta S for the reaction is - 269.74 JK^(-1) .

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) 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 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 formation of methane at 300K : C("graphite") +2H_(2) (g) rarr CH_(4)(g) The following data are given: Delta_(f)H^(Theta) (kJ mol^(-1)): CH_(4)(g) =- 74.81 Delta_(f)S^(Theta)(kJ mol^(-1)): C("graphite") = 5.70, H_(2)(g) = 130.7 CH_(4)(g) = 186.3