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 ) + O 3 ( g ) ⟶ Cl ( g ) + 2 O 2 ( g ) Δ H ∘ rxn = − 122.8 kJ ( 2 ) 2 O 3 ( g ) ⟶ 3 O 2 ( g ) Δ H ∘ rxn = − 285.3 kJ ( 3 ) O 3 ( g ) + Cl ( g ) ⟶ ClO ( g ) + O 2 ( g ) Δ H ∘ rxn = ?

ΔH°rxn = - 162.5 kJ

Explanation:

Hello,

In this case, we use the Hess law to compute the required enthalpy of the reaction for the chlorine with ozone.

1. At first, we invert the original (1) equation in order to place the chlorine at the reactants, so the enthalpy sign is inverted to positive:

Cl ( g ) + 2O2 ( g ) ⟶ClO ( g ) + O3 ( g ) ΔH°rxn = 122.8 kJ

2. Then we do not modify the second reaction:

2O3 ( g ) ⟶ 3O2 ( g ) ΔH°rxn = −285.3 kJ

Next, the add the aforementioned reactions:

Cl ( g ) + 2O2 ( g ) + 2O3 ( g )  ⟶ ClO ( g ) + O3 ( g ) + 3O2 ( g )

In order to obtain (3):

O3 ( g ) + Cl ( g ) ⟶ ClO ( g ) + O2 ( g )

And the enthalpy of reaction results:

ΔH°rxn = 122.8 kJ − 285.3 kJ

ΔH°rxn = - 162.5 kJ

Best regards.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.