Consider these three reactions as the elementary steps in the mechanism for a chemical reaction.(i) Cl2 (g) + Pt (s) à 2Cl (g) + Pt (s) Ea = 1550 kJ ∆H = – 950 kJ(ii) Cl (g)+ CO (g) + Pt (s) à ClCO (g) + Pt (s) Ea = 2240 kJ ∆H = 575 kJ(iii) Cl (g) + ClCO (g) à Cl2CO (g) Ea = 2350 kJ ∆H = – 825 kJe. Which reaction intermediate would be considered a catalyst (if any) and why?f. If you were to add 2700kJ of activation energy to the reaction, would you be able to make thereaction reverse itself (i.e. have the products become reactants)? Justify your answer.g. If you were to added a positive catalyst to step (iii) what would the end result be? Justify yourprediction. h. Your friend is looking at your graph and states that she believes that step (ii) is the ratedetermining step. Do you agree with her? Justify your reasoning. d. What is the overall amount of activation energy for the above chemical reaction?e. Which reaction intermediate would be considered a catalyst (if any) and why?f. If you were to add 2700kJ of activation energy to the reaction, would you be able to make thereaction reverse itself (i.e. have the products become reactants)? Justify your answer.g. If you were to added a positive catalyst to step (iii) what would the end result be? Justify yourprediction.h. Your friend is looking at your graph and states that she believes that step (ii) is the ratedetermining step. Do you agree with her? Justify your reasoning. Consider these three reactions as the elementary steps in the mechanism for a chemical reaction.(i) Cl₂ (g) + Pt (s) → 2Cl (g) + Pt (s)Ea = 1550 kJEa = 2240 kJ(ii) Cl (g)+ CO (g) + Pt (s) → CICO (g) + Pt (s)(iii) Cl (g) + CICO (g) → Cl₂CO (g)Ea = 2350 kJΔΗ = – 950 kJΔΗ = 575 kJΔΗ = − 825 kJ

d.overall amount of activation energy = 2350 KJ e. Pt will be the catalyst, becasue it remained…

Answered: e. Which reaction intermediate would be…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Consider these three reactions as the elementary steps in the mechanism for a chemical reaction.
(i) Cl2 (g) + Pt (s) à 2Cl (g) + Pt (s) Ea = 1550 kJ ∆H = – 950 kJ
(ii) Cl (g)+ CO (g) + Pt (s) à ClCO (g) + Pt (s) Ea = 2240 kJ ∆H = 575 kJ
(iii) Cl (g) + ClCO (g) à Cl2CO (g) Ea = 2350 kJ ∆H = – 825 kJ

e. Which reaction intermediate would be considered a catalyst (if any) and why?
f. If you were to add 2700kJ of activation energy to the reaction, would you be able to make the
reaction reverse itself (i.e. have the products become reactants)? Justify your answer.
g. If you were to added a positive catalyst to step (iii) what would the end result be? Justify your
prediction.
h. Your friend is looking at your graph and states that she believes that step (ii) is the rate
determining step. Do you agree with her? Justify your reasoning.

d. What is the overall amount of activation energy for the above chemical reaction?
e. Which reaction intermediate would be considered a catalyst (if any) and why?
f. If you were to add 2700kJ of activation energy to the reaction, would you be able to make the
reaction reverse itself (i.e. have the products become reactants)? Justify your answer.
g. If you were to added a positive catalyst to step (iii) what would the end result be? Justify your
prediction.
h. Your friend is looking at your graph and states that she believes that step (ii) is the rate
determining step. Do you agree with her? Justify your reasoning.

Consider these three reactions as the elementary steps in the mechanism for a chemical reaction.
(i) Cl₂ (g) + Pt (s) → 2Cl (g) + Pt (s)
Ea = 1550 kJ
Ea = 2240 kJ
(ii) Cl (g)+ CO (g) + Pt (s) → CICO (g) + Pt (s)
(iii) Cl (g) + CICO (g) → Cl₂CO (g)
Ea = 2350 kJ
ΔΗ = – 950 kJ
ΔΗ = 575 kJ
ΔΗ = − 825 kJ

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