The enthalpy of formation of ammonia when calculated from the following bond energy data is (B.E. of N - H, H - H, N ≡ N is 389 kj mol-1, 435 kj mol-1, 945.36 kj mol-1 respectively)

Question

The enthalpy of formation of ammonia when calculated from the following bond energy data is (B.E. of N - H, H - H, N ≡ N is 389 kj mol-1, 435 kj mol-1, 945.36 kj mol-1 respectively) (A) - 41.82 kj mol-1 (B) + 83.64 kj mol-1 (C) - 945.36 kj mol-1 (D) - 833 kj mol-1

Tardigrade Admin · Accepted Answer

N2 +3H2 arrow 2NH3 Δ H = Δ H(N ≡ N)+3×Δ H (H-H)-2 × 3Δ H(N-H) = 945.36 + 3 × 435.0 - 6 × 389.0 = - 83.64 kj Heat of formation of NH3= (-83.64/2) = -41.82 kj/mol

Q. The enthalpy of formation of ammonia when calculated from the following bond energy data is ( $B . E .$ of $N - H$ , $H - H$ , $N \equiv N$ is $389 kj m o l^{- 1}$ , $435 kj m o l^{- 1}$ , $945.36 kj m o l^{- 1}$ respectively)

$- 41.82 kj m o l^{- 1}$

$+ 83.64 kj m o l^{- 1}$

$- 945.36 kj m o l^{- 1}$

$- 833 kj m o l^{- 1}$

$N_{2} + 3 H_{2} \to 2 N H_{3}$
$Δ H = Δ H (N \equiv N) + 3 \times Δ H (H - H) - 2 \times 3Δ H (N - H)$
$= 945.36 + 3 \times 435.0 - 6 \times 389.0 = - 83.64 kj$
Heat of formation of $N H_{3} = \frac{- 83.64}{2} = - 41.82 kj / m o l$

Q. The enthalpy of formation of ammonia when calculated from the following bond energy data is (B.E. of N−H, H−H, N≡N is 389kjmol−1, 435kjmol−1, 945.36kjmol−1 respectively)

−41.82kjmol−1

+83.64kjmol−1

−945.36kjmol−1

−833kjmol−1