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, c, n, o) with values (h-h: 432, c-h: 411, c-c: 346, n-h: 386, n-c: 305, n-n: 167, o-h: 459, o-c: 358, o-n: 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 and formed

• Bonds broken: \(1\) \(H - H\) (energy \(432\) kJ/mol), \(1\) \(N \equiv N\) (energy \(942\) kJ/mol), \(2\) \(C\) (free atoms, no bond breaking energy for free C, but wait, the reactants are \(H_2\), \(N_2\), and \(2C(g)\). Wait, \(C(g)\) is already gaseous atoms, so we only break bonds in \(H_2\) and \(N_2\).
• Bonds formed: In \(2\) \(HCN\) molecules, each \(HCN\) has \(1\) \(H - C\) bond and \(1\) \(C \equiv N\) bond. So for \(2\) \(HCN\), we have \(2\) \(H - C\) bonds and \(2\) \(C \equiv N\) bonds.

Step2: Calculate energy for bonds broken

Energy to break bonds: \(E_{broken} = E(H - H) + E(N \equiv N)\)
\(E_{broken} = 432 + 942 = 1374\) kJ/mol (since we have 1 mol of \(H_2\) and 1 mol of \(N_2\), and 2 mol of \(C(g)\) which has no bonds to break)

Step3: Calculate energy for bonds formed

Energy released when forming bonds: \(E_{formed} = 2 \times [E(H - C) + E(C \equiv N)]\)
From the table, \(E(H - C) = 411\) kJ/mol, \(E(C \equiv N) = 887\) kJ/mol
So \(E_{formed} = 2 \times (411 + 887) = 2 \times 1298 = 2596\) kJ/mol

Step4: Calculate enthalpy change (\(\Delta H\))

\(\Delta H = E_{broken} - E_{formed}\) (because \(\Delta H =\) energy required to break bonds - energy released when forming bonds)
\(\Delta H = 1374 - 2596 = -1222\) kJ/mol (for the reaction as written, which is producing 2 mol of \(HCN\))

Wait, wait, let's check again. Wait, the reactants: \(H_2(g)\) (1 mol, break 1 H-H bond), \(N_2(g)\) (1 mol, break 1 N≡N bond), and 2 C(g) (no bonds, so no energy for breaking). Products: 2 HCN(g). Each HCN has H-C (single bond) and C≡N (triple bond). So for 2 HCN: 2 H-C bonds and 2 C≡N bonds.

Calculating \(E_{broken}\): 432 (H-H) + 942 (N≡N) = 1374 kJ (per 1 mol of reaction, since coefficients are 1,1,2 for reactants, 2 for product)

\(E_{formed}\): 2(411 (H-C) + 887 (C≡N)) = 2(1298) = 2596 kJ

\(\Delta H = E_{broken} - E_{formed} = 1374 - 2596 = -1222\) kJ/mol. So the enthalpy of the reaction is -1222 kJ (for the reaction as written, producing 2 mol of HCN).

Answer:

\(\boxed{-1222}\)