Question

1. A scientist measures the standard enthalpy change for the following reaction to be -2913.0 kJ:

2C₂H₆(g) + 7 O₂(g)->4CO₂(g) + 6 H₂O(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H₂O(g) is  kJ/mol.

2. A scientist measures the standard enthalpy change for the following reaction to be -138.6 kJ :

H₂(g) + C₂H₄(g)->C₂H₆(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of C₂H₄(g) is  kJ/mol.

3. A scientist measures the standard enthalpy change for the following reaction to be 84.3 kJ :

NH₄Cl(aq)->NH₃(g) + HCl(aq)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of NH₃(g) is  kJ/mol.

Answer 1

2C₂H₆(g) + 7 O₂(g)->4CO₂(g) + 6 H₂O(g)

H⁰rxn   = H⁰f products - H⁰freactants

H⁰rxn   = 6H⁰f H2O(g) + 4H⁰f CO2(g) - [2H⁰f C2H6(g) + 7H⁰f O2(g)]

-2913      = 6H⁰f H2O(g) +4*-393.5 - 2*-84.684 + 7*0

H⁰f H2O(g)   = -251.4KJ/mole >>>>.answer

2.

H₂(g) + C₂H₄(g)->C₂H₆(g)

H⁰rxn   = H⁰f products - H⁰freactants

H⁰rxn   = H⁰f C2H6(g) - [H⁰f H2(g)(g) + H⁰f C2H4(g)]

-138.6      = -84.684 - 0 -H⁰f C2H4(g)

-138.6      = -84.684 -H⁰f C2H4(g)

H⁰f C2H4(g)   = 53.92KJ/mole>>>>answer

3.

NH₄Cl(aq)->NH₃(g) + HCl(aq)

H⁰rxn   = H⁰f products - H⁰freactants

H⁰rxn   = H⁰f NH3(g) +H⁰fHCl(g) - H⁰f NH4Cl(aq)

84.3         = H⁰f NH3(g) -92.3 -*-299.66

84.3         = H⁰f NH3(g) -92.3 +299.66

H⁰f NH3(g)   = -123.06KJ/mole >>>>answer