Enthalpy of formation of 2 mol of NH3 (g) is - 90 kJ, and Δ HH-H andΔ HN-Hare respectively 435 kJ mol-1 and 390 kJ mol-1. The value of Δ HN-N is

Question

Enthalpy of formation of 2 mol of NH3 (g) is - 90 kJ, and Δ HH-H andΔ HN-Hare respectively 435 kJ mol-1 and 390 kJ mol-1. The value of Δ HN-N is (A) -472.5 kJ (B) -945 kJ mol-1 (C) 472.5 kJ (D) 945 kJ mol-1

Tardigrade Admin · Accepted Answer

N2(g)+3H2(g)→2NH3(g);Δ H=-90kJ Now -90=Δ HN=N+3Δ HH-H+6Δ HN-H or Δ HN=N=6Δ HN-H-3Δ HH-H-90 = 6 × 390 -3 × 435-90 = 945 kJ

Q. Enthalpy of formation of $2$ mol of $N H_{3} (g)$ is $- 90 k J,$ and $Δ H_{H - H}$ and $Δ H_{N - H}$ are respectively $435 k J m o l^{- 1}$ and $390 k J m o l^{- 1}$ . The value of $Δ H_{N - N}$ is

$- 472.5 k J$

$- 945 k J m o l^{- 1}$

$472.5 k J$

$945 k J m o l^{- 1}$

$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); Δ H = - 90 k J$
Now $- 90 = Δ H_{N = N} + 3Δ H_{H - H} + 6Δ H_{N - H}$
or $Δ H_{N = N} = 6Δ H_{N - H} - 3Δ H_{H - H} - 90$
$= 6 \times 390 - 3 \times 435 - 90$
$= 945 k J$

Q. Enthalpy of formation of 2 mol of NH3​(g) is −90kJ, and ΔHH−H​ andΔHN−H​are respectively 435kJmol−1 and 390kJmol−1. The value of ΔHN−N​ is

−472.5kJ

−945kJmol−1

472.5kJ

945kJmol−1