Question

Calculate$\mathbf{∆}{\mathbf{H}}^{\mathbf{o}}$ for each of the following reactions, which occur in the atmosphere.

1. ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{3}}\left(g\right)\mathbf{\to }{\mathbf{CH}}_{\mathbf{3}}\mathbf{CHO}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)$
   
2. ${\mathbf{O}}_{\mathbf{3}}\left(g\right)\mathbf{+}\mathbf{NO}\left(g\right)\mathbf{\to }{\mathbf{NO}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)$
   
3. ${\mathbf{SO}}_{\mathbf{3}}\left(g\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\left(\mathrm{aq}\right)$
   
4. $\mathbf{2}\mathbf{NO}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}{\mathbf{NO}}_{\mathbf{2}}\left(g\right)$
   

Short answer

The enthalpy of reaction is

a)$\mathbf{-}\mathbf{361}\mathbf{}\mathbf{kJ}$

b) $\mathbf{-}\mathbf{199}\mathbf{}\mathbf{kJ}$

c)$\mathbf{-}\mathbf{227}\mathbf{}\mathbf{kJ}$

d)$\mathbf{-}\mathbf{112}\mathbf{}\mathbf{kJ}$

Step-by-step solution

Step 1: Calculation of standard enthalpy of reaction.

$$\begin{gathered} {\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+{\mathrm{O}}_3\left(\mathrm{g}\right)\to {\mathrm{CH}}_3\mathrm{CHO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \\ ∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}=\sum ∆{\mathrm{H}}_{\mathrm{P}}^{\mathrm{o}}-\sum ∆{\mathrm{H}}_{\mathrm{R}}^{\mathrm{o}} \end{gathered}$$

Substitute the values of enthalpies of reactant and product.

$\begin{array}{rcl}∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}& =& \left[∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}{\mathrm{CH}}_3\mathrm{CHO}\left(\mathrm{g}\right)\right]-\left[\left(∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}{\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)\right)+\left(∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}{\mathrm{O}}_3\left(\mathrm{g}\right)\right)\right]\\ & =& \left(-166\mathrm{kJ}\right)-\left(143\mathrm{kJ}+52\mathrm{kJ}\right)\\ & =& -361\mathrm{kJ}\end{array}$

Step 2: Calculation of standard enthalpy of reaction.

$$\begin{gathered} {\mathrm{O}}_3\left(\mathrm{g}\right)+\mathrm{NO}\left(\mathrm{g}\right)\to {\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \\ ∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}=\sum ∆{\mathrm{H}}_{\mathrm{P}}^{\mathrm{o}}-\sum ∆{\mathrm{H}}_{\mathrm{R}}^{\mathrm{o}} \end{gathered}$$

Substitute the values of enthalpies of reactant and product.

$\begin{array}{rcl}∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}& =& \left[∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}{\mathrm{NO}}_2\left(\mathrm{g}\right)\right]-\left[\left(∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}{\mathrm{O}}_3\left(\mathrm{g}\right)\right)+\left(∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}\mathrm{NO}\left(\mathrm{g}\right)\right)\right]\\ & =& 34\mathrm{kJ}-\left(143\mathrm{kJ}+90\mathrm{kJ}\right)\\ & =& -199\mathrm{kJ}\end{array}$

Step 3: Calculation of standard enthalpy of reaction.

$$\begin{gathered} {\mathrm{SO}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right) \\ ∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}=\sum ∆{\mathrm{H}}_{\mathrm{P}}^{\mathrm{o}}-\sum ∆{\mathrm{H}}_{\mathrm{R}}^{\mathrm{o}} \end{gathered}$$

Substitute the values of enthalpies of reactant and product.

$\begin{array}{rcl}∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}& =& \left[∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\right]-\left[\left(∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}{\mathrm{SO}}_3\left(\mathrm{g}\right)\right)+\left(∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)\right]\\ & =& -909\mathrm{kJ}-\left[-396\mathrm{kJ}+\left(-286\mathrm{kJ}\right)\right]\\ & =& -227\mathrm{kJ}\end{array}$

Step 4: Calculation of standard enthalpy of reaction.

$$\begin{gathered} 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{NO}}_2\left(\mathrm{g}\right) \\ ∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}=\sum ∆{\mathrm{H}}_{\mathrm{P}}^{\mathrm{o}}-\sum ∆{\mathrm{H}}_{\mathrm{R}}^{\mathrm{o}} \end{gathered}$$

Substitute the values of enthalpies of reactant and product.

$\begin{array}{rcl}∆{\mathrm{H}}_{\mathrm{r}}^{\mathrm{o}}& =& \left[2∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}{\mathrm{NO}}_2\left(\mathrm{g}\right)\right]-\left[2∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\mathrm{for}\mathrm{NO}\left(\mathrm{g}\right)\right]\\ & =& \left(2\times 34\mathrm{kJ}\right)-\left(2\times 90\mathrm{kJ}\right)\\ & =& -112\mathrm{kJ}\end{array}$

Step 2:

As the energy of 3 moles of acetylene is more than benzene, acetylene will release more energy when burned.

Result:

Hence, the enthalpy of the reaction is $632\mathrm{kJ}/\mathrm{mol}$. Due to the higher energy of the formation of acetylene, it will produce more heat when burned.