Calculate `Deltah` for the eaction
`BaCO_(3)(s)+2HCI(aq) rarr BaCI_(2)(aq)+CO_(2)(g)+H_(2)O(l)`
`Delta_(f)H^(Theta) (BaCO_(3)) =- 290.8 kcal mol^(-1), Delta_(f)H^(Theta) (H^(o+)) =0`
`Delta_(f)H^(Theta) (Ba^(++)) = - 128.67 kcal mol^(-1)`,
`Delta_(f)H^(Theta)(CO_(2)) =- 94.05 kcal mol^(-1)`,
`Delta_(f)H^(Theta) (H_(2)O) =- 68.32 kcal mol^(-1)`

Consider the reaction: 4NH_(3)(g) +5O_(2)(g) rarr 4NO(g) +6H_(2)O(l) DeltaG^(Theta) =- 1010.5 kJ Calculate Delta_(f)G^(Theta) [NO(g)] if Delta_(f)G^(Theta) (NH_(3)) = -16.6 kJ mol^(-1) and Delta_(f)G^(Theta) [H_(2)O(l)] =- 237.2 kJ mol^(-1) .

Consider the reaction: 4NH_(3)(g) +5O_(2)(g) rarr 4NO(g) +6H_(2)O(l) DeltaG^(Theta) =- 1010.5 kJ Calculate Delta_(f)G^(Theta) [NO(g)] if Delta_(f)G^(Theta) (NH_(3)) = -16.6 kJ mol^(-1) and Delta_(f)G^(Theta) [H_(2)O(l)] =- 237.2 kJ mol^(-1) .

Calculate the resonance enegry of toulene (use Kekule structure form the following data C_(7)H_(8)(l) +9O_(2)(g) rarr 7CO_(2)(g) +4H_(2)O(l)+ DeltaH, DeltaH^(Theta) =- 3910 kJ mol^(-1) C_(7)H_(8)(l) rarr C_(7)H_(8)(g), DeltaH^(Theta) = 38.1 kJ mol^(-1) Delta_(f)H^(Theta) (water) =- 285.8 kJ mol^(-1) Delta_(f)H^(Theta) [CO_(2)(g)] =- 393.5 kJ mol^(-1) Heat of atomisaiton of H_(2)(g) = 436.0 kJ mol^(-1) Heat of sulimation of C(g) = 715.0 kJ mol^(-1) Bond energies of C-H, C-C , and C=C are 413.0, 345.6 , and 610.0 kJ mol^(-1) .

Calculate the resonance energy of toluene (use Kekule structure from the following data C_(7)H_(8)(l) +9O_(2)(g) rarr 7CO_(2)(g) +4H_(2)O(l)+ DeltaH, DeltaH^(Theta) =- 3910 kJ mol^(-1) C_(7)H_(8)(l) rarr C_(7)H_(8)(g), DeltaH^(Theta) = 38.1 kJ mol^(-1) Delta_(f)H^(Theta) (water) =- 285.8 kJ mol^(-1) Delta_(f)H^(Theta) [CO_(2)(g)] =- 393.5 kJ mol^(-1) Heat of atomisation of H_(2)(g) = 436.0 kJ mol^(-1) Heat of sublimation of C(g) = 715.0 kJ mol^(-1) Bond energies of C-H, C-C , and C=C are 413.0, 345.6 , and 610.0 kJ mol^(-1) .

Find DeltaH of the process NaOH(s) rarr NaOH(g) Given: Delta_(diss)H^(Theta)of O_(2) = 151 kJ mol^(-1) Delta_(diss)H^(Theta) of H_(2) = 435 kJ mol^(-1) Delta_(diss)H^(Theta) of O-H = 465 kJ mol^(-1) Delta_(diss)H^(Theta) of Na -O = 255 kJ mol^(-1) Delta_(soln)H^(Theta)of NaOH = - 46 kJ mol^(-1) Delta_(f)H^(Theta) of NaOH(s) =- 427 kJ mol^(-1) Delta_("sub")H^(Theta) of Na(s) = 109 kJ mol^(-1)

Find DeltaH of the process NaOH(s) rarr NaOH(g) Given: Delta_(diss)H^(Theta)of O_(2) = 151 kJ mol^(-1) Delta_(diss)H^(Theta) of H_(2) = 435 kJ mol^(-1) Delta_(diss)H^(Theta) of O-H = 465 kJ mol^(-1) Delta_(diss)H^(Theta) of Na -O = 255 kJ mol^(-1) Delta_(soln)H^(Theta)of NaOH = - 46 kJ mol^(-1) Delta_(f)H^(Theta) of NaOH(s) =- 427 kJ mol^(-1) Delta_("sub")H^(Theta) of Na(s) = 109 kJ 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)

From the following data, calculate the standard enthalpy of formation of propane Delta_(f)H^(Theta) CH_(4) = - 17 kcal mol^(-1) Delta_(f)H^(Theta)C_(2)H_(6) =- 24 kcal mol^(-1), BE (C-H) = 99 kcal mol^(-1) (C- C) = 84 kcal mol^(-1) .

From the following data, calculate the standard enthalpy of formation of propane Delta_(f)H^(Theta) CH_(4) = - 17 kcal mol^(-1) Delta_(f)H^(Theta)C_(2)H_(6) =- 24 kcal mol^(-1), BE (C-H) = 99 kcal mol^(-1) (C- C) = 84 kcal mol^(-1) .