Question

Question : Given the following data:

2${\mathbf{O}}_{\mathbf{3}}$ (g)$\underset{}{\mathbf{\to }}$3role="math" localid="1648705821921" ${\mathbf{O}}_{\mathbf{2}}$ (g) ∆H = -427 kJ

O₂(g)$\stackrel{}{\mathbf{\to }}$2O(g) ∆H = 495 kJ

NO(g) + ${\mathbf{O}}_{\mathbf{3}}$(g)$\underset{}{\mathbf{\to }}$N (g) + (g) ∆H = -199 kJ

calculate ∆H for the reaction

NO(g) + O(g)$\underset{}{\mathbf{\to }}$NO₂ (g)

Short answer

∆H for the reaction = -133.5 kJ

Step-by-step solution

Numbering of equations 

$$\begin{gathered} 2\mathrm{O}\left(\mathrm{g}\right)\stackrel{}{\to }3{\mathrm{O}}_2\left(\mathrm{g}\right)....\mathrm{eq}.1∆\mathrm{H}=-427\mathrm{kJ} \\ {\mathrm{O}}_2\stackrel{}{\to }2\mathrm{O}\left(\mathrm{g}\right)....\mathrm{eq}.2∆\mathrm{H}=495\mathrm{kJ} \\ \mathrm{NO}+{\mathrm{O}}_3\left(\mathrm{g}\right)\stackrel{}{\to }{\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)...\mathrm{eq}.3∆\mathrm{H}=-199\mathrm{kJ} \end{gathered}$$

Reverse equation 1

$3{\mathrm{O}}_2\stackrel{}{\to }2{\mathrm{O}}_3∆\mathrm{H}=427\mathrm{kJ}$

Reverse equation 2  

$2\mathrm{O}\underset{}{\to }{\mathrm{O}}_2∆\mathrm{H}=-495\mathrm{kJ}$

Multiply equation 3 by 2

$2\mathrm{NO}+2{\mathrm{O}}_3\to 2{\mathrm{NO}}_2+2{\mathrm{O}}_2∆\mathrm{H}=-199\mathrm{kJ}$

Calculation of ∆H by the Hess Law

$3{\mathrm{O}}_2\stackrel{}{\to }2{\mathrm{O}}_3∆\mathrm{H}=427\mathrm{kJ}$

$2\mathrm{O}\underset{}{\to }{\mathrm{O}}_2∆\mathrm{H}=-495\mathrm{kJ}$

$2\mathrm{NO}+2{\mathrm{O}}_3\to 2{\mathrm{NO}}_2+2{\mathrm{O}}_2∆\mathrm{H}=-199\mathrm{kJ}$

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role="math" localid="1648707094930" $$\begin{gathered} 2\mathrm{NO}+2\mathrm{O}\stackrel{}{\to }2{\mathrm{NO}}_2...\mathrm{eq}.4∆\mathrm{H}=-267\mathrm{kJ} \\ \mathrm{Divide}\mathrm{equation}4\mathrm{by}2 \\ \mathrm{NO}+\mathrm{O}\stackrel{}{\to }{\mathrm{NO}}_2\mathrm{H}=-133.5\mathrm{kJ} \end{gathered}$$