e. Which reaction intermediate would be considered a catalyst (if any) and why?

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Step 3: solution b

#(b):

The overall chemical equation can be obtained by adding the three elementary step reactions and canceling the common species like "Pt(s)", "Cl(g)" and "ClCO(g)"

(i): Cl2(g) + Pt(s)  2Cl(g) + Pt(s) 

(ii): Cl(g) + CO(g) + Pt(s)  ClCO(g) + Pt(s)  

(iii): Cl(g) + ClCO(g) Cl2CO(g) 

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Adding the above 3 equations and canceling the common species like "Pt(s)", "Cl(g)" and "ClCO(g)" gives:

Overall chemical equation,  Cl2(g) + CO(g)  Cl2CO(g)   (Answer) 

 

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Step 4: Solution c and solution d

#(c):

Initial enthalpy, H(Initial) = 600 kJ

The final enthalpy can be obtained by adding the enthalpies of all the 3 elementary reactions.

Hence the final enthalpy, H(Final)

= H(i) + H(ii) + H(iii) 

= - 950 kJ + 575 kJ - 825 kJ

= -1200 kJ 

Overall change in enthalpy, H = H(Final) - H(Initial) = - 1200 kJ - 600 kJ = - 1800 kJ 

Overall change in enthalpy = - 1800 kJ  (Answer) 

 

#(d): 

Overall activation energy can be obtained by adding the activation energy(Ea) of all the elementary reactions.

Overall activation energy, Ea = Ea(i) + Ea(ii) +Ea(iii) = 1550 kJ + 2240 kJ + 2350 kJ = 6140 kJ 

Overall activation energy = 6140 kJ  (Answer) 

(i) Cl₂ (g) + Pt (s) → 2Cl (g) + Pt (s)
(ii) Cl (g)+ CO (g) + Pt (s) → CICO (g) + Pt (s)
(iii) Cl (g) + CICO (g) → Cl₂CO (g)
Ea = 1550 kJ
Ea = 2240 kJ
Ea = 2350 kJ
ΔΗ = – 950 kJ
ΔΗ = 575 kJ
ΔΗ = − 825 kJ
Transcribed Image Text:(i) Cl₂ (g) + Pt (s) → 2Cl (g) + Pt (s) (ii) Cl (g)+ CO (g) + Pt (s) → CICO (g) + Pt (s) (iii) Cl (g) + CICO (g) → Cl₂CO (g) Ea = 1550 kJ Ea = 2240 kJ Ea = 2350 kJ ΔΗ = – 950 kJ ΔΗ = 575 kJ ΔΗ = − 825 kJ
e. Which reaction intermediate would be considered a catalyst (if any) and why?
Transcribed Image Text:e. Which reaction intermediate would be considered a catalyst (if any) and why?
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