Chemistry - Thermochemistry Question with Solution | TestHub
Use the data (all values are in kJ per mole at 25°C) given below to answer the following:
Enthalpy of formation of CH3CN = +88.0
Enthalpy of formation of C3H8 = -85.0
Enthalpy of sublimation of graphite = 719.0
Enthalpy of dissociation of nitrogen = 948.0
Enthalpy of dissociation of hydrogen = 435.0
Bond enthalpies: C-H = 414.0; C-N = 378.0; N-H = 426.0
The enthalpy of hydrogenation of CH3CN (kJ/mol) is
Options:
Answer:
Solution:
CH₃CN + 2H₂ → CH₃CH₂NH₂.
To calculate the enthalpy change (Δ H) for this reaction, we consider the energy of bonds broken and bonds formed.
Bonds broken:
* 1 C≡N bond: 899.5 kJ/mol
* 2 H-H bonds: 2 × 435.0 kJ/mol = 870.0 kJ/mol
Total energy for bonds broken: 899.5 + 870.0 = 1769.5 kJ/mol
Bonds formed:
* 1 C-N bond: 378.0 kJ/mol
* 2 N-H bonds: 2 × 426.0 kJ/mol = 852.0 kJ/mol
* 2 new C-H bonds: 2 × 414.0 kJ/mol = 828.0 kJ/mol
Total energy for bonds formed: 378.0 + 852.0 + 828.0 = 2058.0 kJ/mol
The enthalpy change (Δ H) is calculated as:
Δ H = Energy of bonds broken - Energy of bonds formed
Δ H = 1769.5 kJ/mol - 2058.0 kJ/mol = -288.5 kJ/mol
This corresponds to option (a).
