Question

Given the average bond enthalpies listed below, calculate the ΔH for the reaction in kJ: N2 + 3 H2 → 2 NH3

Be (H–H) = 436 kJ/mol
Be (N≡N) = 946 kJ/mol
Be (N–N) = 163 kJ/mol
Be (N–H) = 389 kJ/mol
a. -1323 kJ
b. -196 kJ
c. 196 kJ
d. 1323 kJ

Answer 1

To calculate the ΔH for the reaction, you need to determine the sum of the bond enthalpies of the reactants and subtract the sum of the bond enthalpies of the products. The ΔH for the reaction N2 + 3 H2 → 2 NH3 is 1476 kJ/mol.

Step-by-step explanation:

To calculate the ΔH for the given reaction, we need to determine the sum of the bond enthalpies of the reactants and subtract the sum of the bond enthalpies of the products. Here's how to do it:

1. Calculate the total bond enthalpy of N2: 1 mol N2 x 946 kJ/mol = 946 kJ/mol
2. Calculate the total bond enthalpy of H2: 3 mol H2 x 436 kJ/mol = 1308 kJ/mol
3. Calculate the total bond enthalpy of NH3: 2 mol NH3 x 389 kJ/mol = 778 kJ/mol
4. Subtract the sum of the bond enthalpies of the products from the sum of the bond enthalpies of the reactants: 946 kJ/mol + 1308 kJ/mol - 778 kJ/mol = 1476 kJ/mol

So, the ΔH for the reaction is 1476 kJ/mol.