Question

A scientist measures the standard enthalpy change for the following reaction to be -2847.0 kJ: 2C2H6(g) + 7 02(g) —4CO2(g) + 6H2O(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CzH6(g) is kJ/mol.

Answer 1

Enthalpy of reaction is

ΔH_reaction=∑ΔH_f(products)−∑ΔH_f(Reactants)

1. 2C2H6(g) + 7O2(g) = 4CO2(g) + 6 H2O(g)

ΔH_reaction = -2847.0 KJ = ∑ΔH_f(products)−∑ΔH_f(Reactants)

-2847.0 KJ= 4 *(-393.5) + 6 (-241.8) - 2*ΔH_f(C6H6) - 7*0

-2*ΔH_f(C6H6) = -2847.0 - 4 *(-393.5) - 6 (-241.8) = 177.8

ΔH_f(C6H6) = -177.8/2 = -88.9 KJ/mol

2. Ca(OH)2 (aq) + 2HCl(aq)  = CaCl2 (s) + 2H2O (l)   

ΔH_reaction = -15.0 KJ = ∑ΔH_f(products)−∑ΔH_f(Reactants)

-15 = -795.0 + 2*(-285.8) - (−1002.82) -2*ΔH_fHCl(aq)

-2*ΔH_fHCl(aq) = -15 +795.0 - 2*(-285.8) +(−1002.82)= 348.78

ΔH_fHCl(aq) = -348.78/2 = -174.3 KJ/mol

3. 4NH3 (g)+5O2 (g) =4NO(g) + 6H2O(g)

ΔH_reaction = -963.4 KJ = ∑ΔH_f(products)−∑ΔH_f(Reactants)

-963.4 = 4(90.4) + 6*ΔH_fH2O(g) -4 (-46.2) -5*0

6*ΔH_fH2O(g) = -963.4-4(90.4)+4 (-46.2) = -1509.8

ΔH_fH2O(g) = -1509.8/6 = -251.6 KJ/mol