Ethylene C₂H₄ reacts with diazene H₂N₂ to afford ethane C₂H₆ and nitrogen gas.

1) Draw the Lewis structure of each compound

2) Write a chemical equation for this reaction

3) Calculate the enthalpy change ΔH_rxn for this reaction

Data:

H_bond(H-H) = 436 kJ.mol^-1

H_bond(C-H) = 413 kJ.mol^-1

H_bond(N-H) = 391 kJ.mol^-1;

H_bond(C-C) = 348 kJ.mol^-1

H_bond(C=C) = 614 kJ.mol^-1

H_bond(N-N) = 163 kJ.mol^-1;

H_bond(N=N) = 418 kJ.mol^-1

H_bond(N $\equiv$ N) = 941 kJ.mol^-1

Question

Ethylene C₂H₄ reacts with diazene H₂N₂ to afford ethane C₂H₆ and nitrogen gas.

1) Draw the Lewis structure of each compound

2) Write a chemical equation for this reaction

3) Calculate the enthalpy change ΔH_rxn for this reaction

Data:

H_bond(H-H) = 436 kJ.mol^-1

H_bond(C-H) = 413 kJ.mol^-1

H_bond(N-H) = 391 kJ.mol^-1;

H_bond(C-C) = 348 kJ.mol^-1

H_bond(C=C) = 614 kJ.mol^-1

H_bond(N-N) = 163 kJ.mol^-1;

H_bond(N=N) = 418 kJ.mol^-1

H_bond(N

\equiv

N) = 941 kJ.mol^-1

| General Chemistry 2

Accepted Answer

1) Lewis structures:

2) C₂H₄ + N₂H₂ → C₂H₆ + N₂

3) ΔH_rxn = ΣH_{bond broken} - ΣH_{bond formed}

⇒ ΔH_rxn = ΣH_bond (reactants) - ΣH_bond (products)

⇒ ΔH_rxn = 4 H_bond (C-H) + H_bond (C=C) + 2 H_bond (N-H) + H_bond (N=N) - 6 H_bond (C-H) - H_bond (C-C) - H_bond (N

\equiv

N)

⇒ ΔH_rxn = -301 kJ.mol^-1

Midterm 2 - Question 6 | General Chemistry 2