The enthalpy of formation of $C O_{(g)}, C O_{2 (g)}, N_{2} O_{(g)}$ and $N_{2} O_{4 (g)}$ is $- 110, - 393, + 811$ and $10 k J / m o l$ respectively. For the reaction, $N_{2} O_{4 (g)} + 3 C O_{(g)} \to N_{2} O_{(g)} + 3 C O_{2 (g)} \cdot △ H_{r} (k J / m o l)$ is

- 212

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+ 212

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+ 48

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- 48

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Solution

The correct option is D $- 48$
$N_{2} O_{4 (g)} + 3 C O_{(g)} \to N_{2} O_{(g)} + 3 C O_{2 (g)}$
$△ H_{r e a c t i o n} = \sum_{Heat of formation of products} - \sum_{Heat of formation of reactants}$
$△ H_{r e a c t i o n} = [△ H_{f} N_{2} O + 3 \times △ H_{f} C O_{2}] - [△ H_{f} N_{2} O_{4} + 3 \times △ H_{f} C O]$
$△ H_{r} = [+ 811 + 3 (- 393)] - [10 + 3 (- 110)]$
$= [811 - 1179] - [- 320] = - 368 + 320$
$= - 48 k J / m o l$

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