HCl can be produced by the exothermic reaction H2(g)+2ICl(g)→2HCl(g)+I2(g) . A proposed mechanism for the reaction has two eleme
ntary steps as shown below.
Step 1: H2(g)+ICl(g)→HI(g)+HCl(g) (slow)
Step 2: HI(g)+ICl(g)→HCl(g)+I2(g) (fast)
(d) Write a rate law for the overall reaction that is consistent with the proposed mechanism.
(e) On the incomplete reaction energy diagram below, draw a curve that shows the following two details.
The relative activation energies of the two elementary steps
The enthalpy change of the overall reaction
Step 1: H2(g)+ICl(g)→HI(g)+HCl(g) (slow)
Step 2: HI(g)+ICl(g)→HCl(g)+I2(g) (fast)
(d) Write a rate law for the overall reaction that is consistent with the proposed mechanism.
(e) On the incomplete reaction energy diagram below, draw a curve that shows the following two details.
The relative activation energies of the two elementary steps
The enthalpy change of the overall reaction
1 answer:
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Answer:
The bands are due to:
λmax = 289 nm n→π* transition (E = 12)
λmax = 182 nm π→π* transition (E=10000)
Explanation:
The two types of acetaldehyde transition are as follows:
n→π* and π→π*
From the attached diagram we have to:
ΔEn→π* < ΔEπ→π*
ΔEα(1/λ)
Thus:
λn→π* > λπ→π*
In n→π* spin forbidden, the intensity is low. Thus, the molar extinction E for n→π* is very low.
The same way, for π→π* spin allowed the intensity is high. Thus, the molar extinction coefficient E for π→π* is high too.
The bands are due to:
λmax = 289 nm n→π* transition (E = 12)
λmax = 182 nm π→π* transition (E=10000)
