Calculate the equilibrium constant for the following reaction at `298K` and `1` atmospheric pressure:
`C(graphite) +H_(2)O(l) rarr CO(g) +H_(2)(g)`
Given `Delta_(f)H^(Theta) at 298 K` for `H_(2)O(l) =- 286.0 kJ mol^(-1)`
for `CO(g) =- 110.5 kJ mol^(-1)`
`DeltaS^(Theta)` at `298K` for the reaction `= 252.6 J K^(-1) mol^(-1)`

Calculate the equilibrium constant for the following reaction at 298 K and 1 atm pressure.C(graphite)+H2O(l)=CO(g)+H2(g) Given : 'Delta_f H^@ , [H_2O (l)] = -286.0 kJ mol^(-1) , Delta_f H^@ [ CO (g)] = - 110.5 kJ mol^(-1) , DeltaS^@ at 298 K for the reaction =252.6 J k^(-1) mol^(-1) . Gas constant R = 8.31 Jk^(-1) mol^(-1) .

Calculated the equilibrium constant for the following reaction at 298K : 2H_(2)O(l) rarr 2H_(2)(g) +O_(2)(g) Delta_(f)G^(Theta) (H_(2)O) =- 237.2 kJ mol^(-1),R = 8.314 J mol^(-1) K^(-1)

Calculated the equilibrium constant for the following reaction at 298K : 2H_(2)O(l) rarr 2H_(2)(g) +O_(2)(g) Delta_(f)G^(Theta) (H_(2)O) =- 237.2 kJ mol^(-1),R = 8.314 J mol^(-1) K^(-1)

Calculate the equilibrium constant for the following reaction at 298 K: 2H_2O (l) to 2H_2(g) + O_2(g) Given : Delta_fG^@ [ H_2O (l)] = - 273.2 kJ mol^(-1) R = 8.314 J mol^(-1) K^(-1) .

Calculate the equilibrium constant for the reaction : NO(g) + 1/2 O_(2)(g) iff NO_(2)(g) Given, Delta_(f)H^(@) at 298 K : NO(g) = 90.4 kJ "mol"^(-1), NO_(2)(g) = 33.8 kJ "mol"^(-1) and DeltaS^(@) at 298 K = -70.8 J K^(-1) "mol"^(-1)

Calculate the equilibrium constant for the reaction : NO(g) + 1/2 O_(2)(g) iff NO_(2)(g) Given, Delta_(f)H^(@) at 298K : NO(g) = 90.4 kJ "mol"^(-1), NO_(2)(g) = 33.8 kJ "mol"^(-1) and DeltaS^(@) at 298 K = -70.8 J K^(-1) "mol"^(-1) , R = 8.31 J K^(-1) "mol"^(-1) .

Calculate the standard Gibbs free energy change from the free energies of formation data for the following reaction: C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr 6CO_(2)(g) +3H_(2)O(g) Given that Delta_(f)G^(Theta) =[C_(6)H_(6)(l)] = 172.8 kJ mol^(-1) Delta_(f)G^(Theta)[CO_(2)(g)] =- 394.4 kJ mol^(-1) Delta_(f)G^(Theta) [H_(2)O(g)] =- 228.6 kJ mol^(-1)

Calculate the standard Gibbs free energy change from the free energies of formation data for the following reaction: C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr 6CO_(2)(g) +3H_(2)O(g) Given that Delta_(f)G^(Theta) =[C_(6)H_(6)(l)] = 172.8 kJ mol^(-1) Delta_(f)G^(Theta)[CO_(2)(g)] =- 394.4 kJ mol^(-1) Delta_(f)G^(Theta) [H_(2)O(g)] =- 228.6 kJ mol^(-1)

Calculate DeltaG^(Theta) for the following reaction at 298K: CO(g) +((1)/(2))O_(2)(g) rarr CO_(2)(g), DeltaH^(Theta) =- 282.84 kJ Given, S_(CO_(2))^(Theta)=213.8 J K^(-1) mol^(-1), S_(CO(g))^(Theta)= 197.9 J K^(-1) mol^(-1), S_(O_(2))^(Theta)=205.0 J K^(-1)mol^(-1) ,