Calculate the enthalpy change (∆H) for the reaction- N2(g) + 3 F2(g) –––> 2 NF3(g) given the following bond enthalpies: N≡N 945 kJ/mol F–F 155 kJ/mol N–F 283 kJ/mol

– 844 kJ/mol.

Explanation:

The following data were obtained from the question:

N2(g) + 3 F2(g) –––> 2 NF3(g)

Enthalpy of N≡N (N2) = 945 kJ/mol

Enthalpy of F–F (F2) = 155 kJ/mol

Enthalpy of N–F3 (NF3) = 283 kJ/mol

Enthalpy change (∆H) =?

Next, we shall determine the enthalpy of reactant.

This is illustrated below:

Enthalpy of reactant (Hr) = 945 + 3(155)

Enthalpy of reactant (Hr) = 945 + 465

Enthalpy of reactant (Hr) = 1410 kJ/mol

Next, we shall determine the enthalpy of the product.

This is illustrated below:

Enthalpy of product (Hp) = 2 x 283

Enthalpy of product (Hp) = 566 kJ/mol

Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:

Enthalpy of reactant (Hr) = 1410 kJ/mol

Enthalpy of product (Hp) = 566 kJ/mol

Enthalpy change (∆H) =?

Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)

Enthalpy change (∆H) = 566 – 1410

Enthalpy change (∆H) = – 844 kJ/mol

idk what the question is about