Question

Please answer #7 and #8 only, Thank you!

Ozone Decomposition: Ozone formation is faster during the day due to the action of sunlight on NO2 molecules. Thus, Os(g) accumulates in the troposphere and its concentration peaks sometime in the afternoon. During the night, ozone reacts with NO(g) according to the following reaction: NO(g) + Os(g) = NO2(g) + Ox(g). This is a single-step process with an activation energy E. = 12.5 kl/mol and a AH..=-199.5 kJ/mol. 5. Build potential energy diagram that represents the transformation of reactants to products. Indicate on the diagram the activation energy and the heat of the reaction for this process. Hour 6. Derive the rate law for this process and predict the effect on the rate of reaction of doubling og concentrations. 7. Estimate how many times slower this process is at 0 °C than at 25 °C. 8. This reaction is product-favored at 25°C. Estimate the temperature at which it becomes reactant-favored.

Answer 1

Ans

7. Let k₁ be the rate constant of the reaction at 25⁰C and k₂ be the rate constant at temperature 0⁰C.

log k₂/k₁ = [E_a/2.303R] [T₂-T₁]/T₁T₂

=[12.5*1000/2.303*8.314] [25/273*298]

=0.2006

k₂/k₁ = antilog 0.2006 =1.5

From the ratio,it is clear that rate of the reaction at 0⁰C (273 K) is slower by 1.5 fold than at a temperature 25⁰C (298 K).

8.For calculating the temperature at which the reaction reverses,free energy change value must calculate.For this along with enthalpy of the reaction $\Delta$ H and entropy change values are required. But in the question,the entropy change $\Delta$ S is not given. So data is inadequate.