Question

question 12 of 15
the 1995 nobel prize in chemistry was shared by paul crutzen, f. sherwood rowland, and mario molina for their work concerning the formation and decomposition of ozone in the stratosphere. rowland and molina hypothesized that chlorofluorocarbons (cfcs) in the stratosphere break down upon exposure to uv radiation, producing chlorine atoms. chlorine was previously identified as a catalyst in the breakdown of ozone into oxygen gas.
using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone.
(1) ( mathrm{clo}(mathrm{g})+mathrm{o}_{3}(mathrm{~g}) \to mathrm{cl}(mathrm{g})+2 mathrm{o}_{2}(mathrm{~g}) ) ( delta h_{mathrm{rxn}}^{circ}=-122.8 mathrm{~kj} )
(2) ( 2 mathrm{o}_{3}(mathrm{~g}) \to 3 mathrm{o}_{2}(mathrm{~g}) ) ( delta h_{mathrm{rxn}}^{circ}=-285.3 mathrm{~kj} )
(3) ( mathrm{o}_{3}(mathrm{~g})+mathrm{cl}(mathrm{g}) \to mathrm{clo}(mathrm{g})+mathrm{o}_{2}(mathrm{~g}) ) ( delta h_{mathrm{rxn}}^{circ}=? )
( delta h_{mathrm{rxn}}^{circ}= ) kj

Explanation:

Step1: Reverse reaction (1)

$$\text{Cl(g)} + 2\text{O}_{2}(g)\to\text{ClO}(g)+\text{O}_{3}(g)\quad\Delta H_{1}= + 122.8\,\text{kJ}$$

Step2: Use reaction (2)

$$2\text{O}_{3}(g)\to3\text{O}_{2}(g)\quad\Delta H_{2}=-285.3\,\text{kJ}$$

Step3: Manipulate equations

Let's call the reversed reaction (1) as equation (A) and reaction (2) as equation (B).
We want to find $\Delta H$ for $\text{O}_{3}(g)+\text{Cl}(g)\to\text{ClO}(g)+\text{O}_{2}(g)$.
If we consider equation (A) and equation (B), we can rewrite equation (A) as:
$$\text{Cl(g)}+\text{O}_{2}(g)\to\text{ClO}(g)+\text{O}_{3}(g)-\text{O}_{2}(g)$$
Multiply equation (A) by 1 and equation (B) by $\frac{1}{2}$
For equation (A): $\text{Cl(g)} + 2\text{O}_{2}(g)\to\text{ClO}(g)+\text{O}_{3}(g)\quad\Delta H_{A}= + 122.8\,\text{kJ}$
For equation (B): $\text{O}_{3}(g)\to\frac{3}{2}\text{O}_{2}(g)\quad\Delta H_{B}=-142.65\,\text{kJ}$
Add the two equations:
$$\text{O}_{3}(g)+\text{Cl}(g)\to\text{ClO}(g)+\text{O}_{2}(g)$$
$$\Delta H=\Delta H_{A}+\Delta H_{B}$$
$$\Delta H = 122.8+(- 142.65)$$

Answer:

$-19.85$