Enthalpies of formation of CO ( g ), CO _2(g), N _2 O ( g ), and … - Vidyadip

MCQ

Enthalpies of formation of $CO ( g ), CO _2(g), N _2 O ( g )$, and $N _2 O _4(g)$ are $-110,-393,81$, and $9.7 kJ mol ^{-1}$ respectively. Find the value of $\Delta_{ r } H$ for the reaction: $N _2 O _4(g)+3 CO ( g ) \rightarrow N _2 O ( g )+3 CO _2(g)$

A
$-850 kJ$
B
$-600 kJ$
✓
$-778 Kj$
D
$-802 kJ$

✓

Answer

Correct option: C.

$-778 Kj$

(c) -778 kJ
Explanation: Heat of reaction, $\Delta_r H =\sum \triangle_r H_{\text {products }}-\sum \Delta_r H_{\text {reactants }}$
$
\begin{aligned}
& \Rightarrow \Delta_r H=\left[\Delta_f H\left(N_2 O\right)+3 \Delta_f H\left(CO_2\right)\right]-\left[\Delta_f H\left(N_2 O_4\right)+3 \Delta_f H(CO)\right] \\
& \left.\Rightarrow \Delta_r H=[81+\{3 \times(-393)\}]-99.7+\{3 \times(-110)\}\right] kJ \\
& \Rightarrow \Delta_r H=-777.7 kJ \approx-778 kJ
\end{aligned}
$

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