Question

Three part rate law question:) Will always thumbs up if the answer is right. Thanks again!

Part a)

Consider the overall reaction: AB(g) + CB(g) ⟶ A(g) + CB2(g)

A proposed mechanism for this reaction is

2 AB(g) ⇌  AB2(g) + A(g) (fast)

AB2(g) + CB(g) ⟶ AB(g) + CB2(g) (slow)

Based on this mechanism, determine the rate law for the overall reaction.

Part b)

Consider the following three-step mechanism for a reaction:

Br2 (g) ⇌ 2 Br (g) Fast

Br (g) + CHBr3 (g) → HBr (g) + CBr3 (g) Slow

Br (g) + CBr3 (g) → CBr4 (g) Fast

Based on this mechanism, determine the rate law for the overall reaction.

Part c)

Phosgene (Cl2CO), a poison gas used in World War I, is formed by the reaction of Cl2 and CO. The proposed mechanism for the reaction is
Cl2 ⇌ 2 Cl (fast,equilibrium)
Cl + CO ⇌ ClCO (fast,equilibrium)
ClCO + Cl2 → Cl2CO + Cl (slow)

Based on this mechanism, determine the rate law for the overall reaction.

Express your answers in terms of k and the necessary concentrations (ex: k[X][Y]^2). Use brackets around the concentrations and the caret symbol prior to the exponent. Do not include any multiplication signs.

Answer 1

Slowest step in a reaction is a rate determining step. Thus, Rate will be described by that step.

Rate law: It states that rate of a reaction is directly proportional to the concentration of reactants with each concentration term raise to some power equals to their stoichiometry co-efficients.

A. second stepis the slowest step

Thus, rate lasw is:

Rate = k [AB₂] [CB]

Where k ; rate constant

B. middle step is a slowest step.

THus, rate law is:

Rate = k [Br] [CHBr₃]

C. Last step is the slowest step

Thus, Rate law is:

Rate = k [ClCO] [Cl₂]