Question

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) ClO(g) +O₃(g) ----> Cl(g) +2O₂(g); Hrxn = -122.8 kJ/mol
2) 2O₃(g) ---> 3O₂(g); Hrxn = -285.3 kJ/mol
3) O₃(g) + Cl(g) -----> ClO(g) + O₂(g); Hrxn = ????

Answer 1

Answer: The $\Delta H^o_{rxn}$ for the reaction is -162.5 kJ/mol

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$O_3(g)+Cl(g)\rightarrow ClO(g)+O_2$      $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $ClO(g)+O_3(g)\rightarrow Cl(g)+2O_2(g)$    $\Delta H_1=-122.8kJ/mol$

(2) $2O_3(g)\rightarrow 3O_2(g)$     $\Delta H_2=-285.3kJ/mol$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (-\Delta H_1)]+[1\times \Delta H_2]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-(-122.8))+(1\times (-285.3))=-162.5kJ/mol$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -162.5 kJ/mol