Question

based on the bond energies for the reaction below, what is the enthalpy of the reaction, in kj?
h₂(g) + n₂(g) + 2 c(g) → 2 hcn(g)
bond energy table: single bond (h-h:432, h-c:411, h-n:386, h-o:459; c-c:346, c-n:305, c-o:358; n-n:167, n-o:201; o-o:142); multiple bonds (c=c:602, c≡c:835, c=n:615, c≡n:887; c=o:799, c≡o:1072; o=o:494; n≡n:942) all values in kj/mol

Explanation:

Step1: Identify bonds broken/formed

• Bonds broken: 1 H-H (432 kJ/mol), 1 N≡N (942 kJ/mol), 2 C (no bond, but in reactants, C is atomic; wait, reactants: H₂ (1 H-H), N₂ (1 N≡N), 2 C (g) (atomic, so no bonds to break). Wait, no: H₂ has H-H, N₂ has N≡N, and C is gaseous atoms (so no bonds). Products: 2 HCN. Each HCN has H-C (single) and C≡N (triple). So for 2 HCN: 2 H-C bonds and 2 C≡N bonds.

Step2: Calculate energy for bonds broken

Bonds broken: H-H (1 mol) + N≡N (1 mol)
Energy broken = 432 + 942 = 1374 kJ/mol (for 1 mol reaction, but reaction is 1 H₂, 1 N₂, 2 C → 2 HCN. Wait, moles: H₂ is 1, N₂ is 1, C is 2. Products: 2 HCN.

Step3: Calculate energy for bonds formed

Bonds formed: 2 H-C (single) + 2 C≡N (triple)
Energy per H-C: 411 kJ/mol, per C≡N: 887 kJ/mol
Energy formed = 2×411 + 2×887 = 822 + 1774 = 2596 kJ/mol

Step4: Enthalpy (ΔH) = Bonds broken - Bonds formed

ΔH = 1374 - 2596 = -1222 kJ/mol (wait, no: ΔH = Σ(bonds broken) - Σ(bonds formed). Wait, bonds broken: energy absorbed (positive), bonds formed: energy released (negative). So ΔH = (energy to break bonds) - (energy released from forming bonds) = (432 + 942) - (2×411 + 2×887)
Calculate:
Bonds broken: 432 (H-H) + 942 (N≡N) = 1374 kJ
Bonds formed: 2×(H-C) + 2×(C≡N) = 2×411 + 2×887 = 822 + 1774 = 2596 kJ
ΔH = 1374 - 2596 = -1222 kJ (since reaction forms 2 moles of HCN, but the stoichiometry is 1 H₂, 1 N₂, 2 C → 2 HCN, so per reaction as written, ΔH = -1222 kJ? Wait, wait: wait, the reactants are H₂ (g) + N₂ (g) + 2 C (g) → 2 HCN (g). So C is gaseous atoms, so no bonds to break for C. So bonds broken: H-H (1 mol) and N≡N (1 mol). Bonds formed: in each HCN, H-C (single) and C≡N (triple). So 2 HCN: 2 H-C and 2 C≡N. So yes, above calculation. Wait, but let's check again:

Wait, H₂ has 1 H-H bond (broken: +432), N₂ has 1 N≡N bond (broken: +942). C is atomic, so no bonds. Then, forming 2 HCN: each HCN has H-C (single) and C≡N (triple). So 2 H-C bonds (each 411, so 2×411) and 2 C≡N bonds (each 887, so 2×887). So energy released (bonds formed) is 2×411 + 2×887 = 822 + 1774 = 2596. So ΔH = (bonds broken) - (bonds formed) = (432 + 942) - 2596 = 1374 - 2596 = -1222 kJ. Wait, but is that correct? Wait, maybe I messed up the stoichiometry. Wait, the reaction is 1 H₂, 1 N₂, 2 C → 2 HCN. So moles: H₂ is 1, N₂ is 1, C is 2. Products: 2 HCN. So bonds broken: 1 H-H (432), 1 N≡N (942), and 2 C (but C is atomic, so no bonds). Bonds formed: 2 H-C (each 411) and 2 C≡N (each 887). So yes, that's correct. So ΔH = 432 + 942 - (2×411 + 2×887) = 1374 - (822 + 1774) = 1374 - 2596 = -1222 kJ. Wait, but let's check the bond energies again. H-C single bond: 411 (from table: C row, H column: 411). C≡N: 887 (from multiple bonds: C≡N is 887). N≡N: 942 (correct). H-H: 432 (correct). So yes, calculation seems right.

Wait, but maybe I made a mistake in the direction. Enthalpy of reaction is ΔH = Σ(bonds broken) - Σ(bonds formed). Because breaking bonds requires energy (positive), forming bonds releases energy (negative). So ΔH = (energy absorbed) - (energy released) = (bonds broken) - (bonds formed). So if bonds formed have more energy released than bonds broken absorbed, ΔH is negative (exothermic). So in this case, 1374 (absorbed) - 2596 (released) = -1222 kJ (so exothermic, 1222 kJ released per reaction as written).

Answer:

\boxed{-1222}