The observed rate law for the reaction H₂(g)+ 2 ICl(g) $\to$ I₂(g)+ 2 HCl(g)is first order with respect to each reactant.Draw an energy diagram that shows energy versus reaction progress for the mechanism given.Both steps are exothermic.
Step 1: H₂(g)+ ICl(g) $\to$ HI(g)+ HCl(g)(slow)
Step 2: HI(g)+ ICl(g) $\to$ I₂(g)+ HCl(g)(fast)

Question

The observed rate law for the reaction H₂(g)+ 2 ICl(g)

\to

I₂(g)+ 2 HCl(g)is first order with respect to each reactant.Draw an energy diagram that shows energy versus reaction progress for the mechanism given.Both steps are exothermic.
Step 1: H₂(g)+ ICl(g)

\to

HI(g)+ HCl(g)(slow)
Step 2: HI(g)+ ICl(g)

\to

I₂(g)+ HCl(g)(fast)

Quizplus · Accepted Answer

The Answer of The observed rate law for the reaction H₂(g)+ 2 ICl(g)$\to$I₂(g)+ 2 HCl(g)is first order with respect to each reactant.Draw an energy diagram that shows energy versus reaction progress for the mechanism given.Both steps are exothermic. Step 1: H₂(g)+ ICl(g)$\to$ HI(g)+ HCl(g)(slow) Step 2: HI(g)+ ICl(g)$\to$ I₂(g)+ HCl(g)(fast)

The Observed Rate Law for the Reaction H2(g)+ 2 ICl(g) →\to→

The Observed Rate Law for the Reaction H₂(g)+ 2 ICl(g) $\to$