Calculate the free energy change for the following reaction at 300 K.
`2CuO_((s)) rarr Cu_(2)O_((s))+(1)/(2)O_(2(g))` Given `Delta H = 145.6 kJ mol^(-1)` and `Delta S = 116.JK^(-1) mol^(-1)`

Predict whether the following reaction is possible or not at 300k . 2CuO(s) rarr Cu_(2)O(s) +1//2O_(2)(g) DeltaH =- 144.6 kJ mol^(-1), DeltaS = 0.116 kJ mol^(-1)

Predict whether the following reaction is possible or not at 300k . 2CuO(s) rarr Cu_(2)O(s) +1//2O_(2)(g) DeltaH =- 144.6 kJ mol^(-1), DeltaS = 0.116 kJ mol^(-1)

Calculate the standard free energy change for the following reaction at 27^@C . H_2(g) + I_2(g) to 2HI(g) , DeltaH^@ = + 51.9 kJ [Given : DeltaS_(H_2)^@ = 130.6 JK^(-1) mol^(-1) DeltaS_(I_2)^@ = 116.7 JK^(-1) mol^(-1) DeltaS_(HI)^@ = 206.3 JK^(-1) mol^(-1)] . Predict whether the reaction is feasible at 27°C or not.

Calculate the standard free energy change for the following reaction Zn(s) + Cu^(2+)(aq) to Zn^(2+) (aq) + Cu(s) Given : Delta_fG^@ [Cu^(2+)(aq)] = 65.0kJ mol^(-1) Delta_f G^@ [ Zn^(2+) (aq)] = -147.2 kJ mol^(-1)

Calculate the temperature above which the reaction of lead oxide to lead in the following reaction become spontaneous. PbO(s) + C(s) rarr Pb(s) + CO(g) Given Delta H = 108.4 kJ mol^(-1), Delta S = 190 JK^(-1) mol^(-1)

Calculate the temperature above which the reaction of lead oxide to lead in the following reaction become spontaneous. PbO(s) + C(s) rarr Pb(s) + CO(g) Given Delta H = 108.4 kJ mol^(-1), Delta S = 190 JK^(-1) mol^(-1)

Calculate the standard free energy change Delta G^(@) for the reaction, 2HgO(s) rarr 2Hg(l) + O_(2)(g) Delta H^(@) = 91 kJ mol^(-1) at 298 K, S_((HgO))^(@) = 72.0 JK^(-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) 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) .