For a reaction `M_(2)O_((g))to2M_((s))+1/2O_(2(g))`,
`DeltaH=30 k J mol^(-1)` and `DeltaS=0.07 k J mol^(-1)` at `1 atm`. Calculate up to which temperature, the reaction would not be spontaneous.

For a reversible reaction at constant temperature and at constant pressure the equilibrium composition of reaction mixture corresponds to the lowest point on Gibbls energy Vs progress of reaction diagrams as shown. At equilibrium Gibbs energy of reaction is equal to zero. For a reaction M_(2)O_((s))^(3//4) rarr 2 M_((S)) + (1)/(2) O_(2 (g)) Delta H = 30 KJ//mol and Delta S = 0.07 KJ//mol //K at 1 atm. The reaction would not be spontaneous at temperature

For a reaction, DeltaH=10000" kJ "mol^(-1) and DeltaS=25" kJ "k^(-1)mol^(-1) . The minimum temperature, above which the reaction would be spontaneous is

For the reaction X _(2) O _(4 (l)) to 2 XO _(2 (g)) , Delta u = 2.1 kcal mol ^(-1), Delta S = 20 cal K ^(-1) mol ^(-1) at 300 K. Calculate Delta G and predict whether the reaction is spontaneous at 300 K

Delta H and Delta S for the reaction, Ag_(2)O(s)to 2A(s)+(1)/(2)O_(2)(g) , are 30.56 kJ mol^(-1) and 66.0 J mol^(-1) respectively. Calculate the temperature at which this reaction will be at equilibrium. Predict whether the forward reaction will be favoured above or below this temperature.

Delta H and Delta S for the reaction, Ag_(2)O(s)to 2A(s)+(1)/(2)O_(2)(g) , are 30.56 kJ mol^(-1) and 66.0 J mol^(-1) respectively. Calculate the temperature at which this reaction will be at equilibrium. Predict whether the forward reaction will be favoured above or below this temperature.

H_(2)O(g)toH_(2)O(l),DeltaH=-40.4kJ*mol^(-1) and DeltaS= - 108.3J*K^(-1)*mol^(-1) . At which temperature the process will be spontaneous at constant 1 atm?

DeltaH and DeltaS for the reaction Ag_(2)O_((s))rarr2Ag_((s))+1/2O_(2_((g))) " are "30.56 kJ mol^(-1) and 66.0 Jk^(-1) mol^(-1) respectively. Calculate the temperature at which the free energy for this reaction will be zero. What will be the direction of reaction at this temperature and at temperature below this and why ? Given: DeltaH=30.56kJ mol^(-1)=30560 J mol^(-1) DeltaS=66.0JK^(-1)mol^(-1) DeltaG=0