Answer
1. $-147\text{ kJ/mol}$
2. $-1165\text{ kJ/mol}$
Work Step by Step
To estimate the change in energy (ΔE) for the given reactions using bond energy values, we need to follow these steps:
1. $H-C\equiv N(g)+2H_{2}(g)\rightarrow H_{2}^{H}-1^{-}$ $\underset {H}{N})N$
- Bonds broken: H-C (413 kJ/mol), 2 × H-H (436 kJ/mol), 1 x C≡N (891 kJ/mol)
- Bonds formed: 2 × H-N (391 kJ/mol), 3 × C-H (412 kJ/mol), 1 x C-N (305 kJ/mol)
- ΔE = Σ(Energy of bonds broken) - Σ(Energy of bonds formed)
- ΔE = (413 + 2 × 436+891) - (2 × 391 + 3x412+305) = -147 kJ/mol
The reaction is exothermic.
2. $N2H_4+2F_{2(g)}\rightarrow N_2(g)+4HF(g)$
- Bonds broken: 1x N-N, 4x N-H, 2 × F-F (159 kJ/mol)
- Bonds formed: 1 × N≡N (945 kJ/mol), 4 × H-F (565 kJ/mol)
ΔE = Σ(Energy of bonds broken) - Σ(Energy of bonds formed)
ΔE =160+ 4 × 391+2x158 - (945 + 4 × 565) = -1165 kJ/mol
The reaction is strongly exothermic.
