Question

1. given the information

$$ mathrm { n } _ { 2 } mathrm { o } _ { 4 } ( mathrm { g } ) ightarrow 2 mathrm { no } _ { 2 } ( mathrm { g } ) quad delta h _ { mathrm { r } } ^ { circ } = + 57.2 mathrm { kj } $$

is the following correct? briefly explain your reasoning.

$$ mathrm { no } _ { 2 } ( mathrm { g } ) ightarrow \frac { 1 } { 2 } mathrm { n } _ { 2 } mathrm { o } _ { 4 } ( mathrm { g } ) quad delta h _ { mathrm { r } } ^ { circ } = - 57.2 mathrm { kj } $$

Explanation:

Step1: Analyze the first reaction

The reaction \(N_{2}O_{4}(g)\to2NO_{2}(g)\) has \(\Delta H_{r}^{\circ}= + 57.2\space kJ\). This means it is endothermic as \(\Delta H>0\).

Step2: Analyze the second reaction

The reaction \(NO_{2}(g)\to\frac{1}{2}N_{2}O_{4}(g)\) is the reverse of the first reaction divided by 2. When a reaction is reversed, the sign of \(\Delta H\) changes. When a reaction is multiplied or divided by a factor, the \(\Delta H\) is also multiplied or divided by the same factor. For the reverse of \(N_{2}O_{4}(g)\to2NO_{2}(g)\) (i.e., \(2NO_{2}(g)\to N_{2}O_{4}(g)\)), \(\Delta H=- 57.2\space kJ\). Then dividing by 2 gives \(\Delta H=\frac{-57.2}{2}=-28.6\space kJ\) for \(NO_{2}(g)\to\frac{1}{2}N_{2}O_{4}(g)\). But the given \(\Delta H_{r}^{\circ}=-57.2\space kJ\) for \(NO_{2}(g)\to\frac{1}{2}N_{2}O_{4}(g)\) is incorrect.

Answer:

The second reaction's \(\Delta H_{r}^{\circ}\) value is incorrect. The correct \(\Delta H_{r}^{\circ}\) for \(NO_{2}(g)\to\frac{1}{2}N_{2}O_{4}(g)\) is \(-28.6\space kJ\) (not \(-57.2\space kJ\)) because it is the reverse and half - of the first reaction, so \(\Delta H\) should be \(-\frac{57.2}{2}=-28.6\space kJ\)