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}) quad delta h_{mathrm{rxn}}^{circ}=-122.8 mathrm{~kj} )
(2) ( 2 mathrm{o}_{3}(mathrm{~g}) \to 3 mathrm{o}_{2}(mathrm{~g}) quad 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}) quad delta h_{mathrm{rxn}}^{circ}=? )
( delta h_{mathrm{rxn}}^{circ}= ) kj
tools
( \times 10^{y} )

Explanation:

Step1: Reverse reaction (1)

Reverse reaction (1): \(Cl(g)+2O_{2}(g)\to ClO(g) + O_{3}(g)\), and \(\Delta H = 122.8\space kJ\) (the sign of \(\Delta H\) changes when a reaction is reversed)

Step2: Manipulate reaction (2)

Divide reaction (2) by \(2\): \(O_{3}(g)\to\frac{3}{2}O_{2}(g)\), and \(\Delta H=\frac{- 285.3}{2}=-142.65\space kJ\)

Step3: Add the two manipulated reactions

\((Cl(g)+2O_{2}(g)\to ClO(g) + O_{3}(g))+(O_{3}(g)\to\frac{3}{2}O_{2}(g))\) gives \(O_{3}(g)+Cl(g)\to ClO(g)+O_{2}(g)\)
Add the \(\Delta H\) values: \(122.8+(-142.65)\)

Answer:

\(-19.85\space kJ\)