Temperature at 298 K: 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)ΔfH/kJmol−1−46.2 90.4−241.8S∘/JK−1mol−1192.5205.0210.6188.7ΔfG′/kJmol−1−16.6 −228.6The standard enthalpy change of the above reaction is

Temperature at 298 K:

4 {NH}_{3} (g) + 5 O_{2} (g) \to 4 NO (g) + 6 H_{2} O (g)

$\begin{array}{lcccc} Δ_{f} H / kJ {mol}^{- 1} & - 46.2 & 90.4 & - 241.8 \\ S^{\circ} / {JK}^{- 1} {mol}^{- 1} & 192.5 & 205.0 & 210.6 & 188.7 \\ Δ_{f} G^{'} / kJ {mol}^{- 1} & - 16.6 & - 228.6 \end{array}$

The standard enthalpy change of the above reaction is

A
$- 90.44 kJ {mol}^{- 1}$
B
$90.44 kJ {mol}^{- 1}$
C
$904.4 kJ {mol}^{- 1}$
D
$- 904.4 kJ {mol}^{- 1}$

Solution:

$\begin{array}{l} Δ H^{\circ} = 4 Δ_{f} H^{\circ} (NO, g) + 6 Δ_{1} H^{\circ} (H_{2} O, g) - 4 Δ_{i} H^{\circ} ({NH}_{3}, g) = [4 \times 90.4 + 6 (- 241.8) - 4 (- 46.2)] kJ {mol}^{- 1} \\ = - 904.4 kJ {mol}^{- 1} \end{array}$ Thus the option D is correct.

Solution:

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