For a demerization reaction ,
` 2A (g) to A_2 (g)`
at 298 K, `Delta U(Theta) = - 20 kJ"mol"^(-1), Delta S^(Theta) = - 30 JK^(-1) "mol"^(-1) ` , then the `Delta G^(Theta)` will be _________J.

What is the equilibrium constant K_(c) for the following reaction at 400K ? 2NOCI(g) hArr 2NO(g) +CI_(2)(g) DeltaH^(Theta) = 77.2 kJ mol^(-1) and DeltaS^(Theta) = 122 J K^(-1) mol^(-1) at 400K .

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 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)

For the reaction, 2A (s) + 2B (g)to C (g) + D(s) at 298K Delta U ^(@) =-10.5KJ and Delta S ^(@)=-44.10 JK^(-1) Calculate Delta G ^(@) for the reaction (in Joules).

For the reaction, 4C(graphite) +5H_(2)(g) rarr nC_(4)H_(10)(g) , DeltaH^(Theta) =- 124.73 kJ mol^(-1), DeltaS^(Theta) =- 365.8 J K^(-1) mol^(-1) 4C(graphite) +5H_(2)(g) rarr iso-C_(4)H_(10)(g) DeltaH^(Theta) =- 131.6 kJ mol^(-1), DeltaS^(Theta) =- 381.079 J K^(-1) mol^(-1) Indicate whther normal butane can be spontaneously converted to iso-butane or not.

For the reaction 2NO_2 (g) hArr N_2O_4(g) , when DeltaS = - 176.0 J K^(-1)mol^(-1) and DeltaH = - 57. 8 kJ mol^(-1) , the magnitude of Delta G at 298 K for the reaction is __________ kJ mol^(-1) . (Nearest integer)

Calculate the dissociation constant of NH_(4)OH at 298k , if DeltaH^(Theta) and DeltaS^(Theta) for the given changes are as follows:- NH_(3) + H^(o+) hArr overset(o+)NH_(4) , DeltaH^(Theta) =- 52.2 KJ mol^(-1), DeltaS^(Theta) = 1.67 J K^(-1)mol^(-1) H_(2)O hArr H^(o+) + overset(Theta)OH, DeltaH^(Theta) = 56.6 kJ mol^(-1) . DeltaS^(Theta) =- 76.53 JK^(-1)mol^(-1)

For the reaction 2A (g) + B(g) to C (g) Delta U^(@) = 20 kcal/mole, Delta S^(@) = 100 cal/kmole at 300 K. Hence Delta G^(@) in kcal is __________ [R = 2 cal mol^(-1) K^(-1)]