The following is a concerted, bimolecular reaction:CH3+NaCN→CH3CN+NaBr.

a. What is the rate equation for this reaction?

b. What happens to the rate of the reaction if[CH3Br] is doubled?

c. What happens to the rate of the reaction if [NaCN] is halved?

d. What happens to the rate of the reaction if [CH3Br] and [NaCN] are both increased by a factor of five?

Short Answer

Expert verified

Answer

a. The rate equation for the given reaction is:

Rate=kCH3BrNaCN

b. The rate gets doubled when the concentration ofCH3Brgets doubled.

c. The rate gets halved when the concentration of NaCN gets halved.

d. The rate increases by a factor of 25 when the concentration of both reactants increases by a factor of five.

Step by step solution

01

Step-by-Step SolutionStep 1: Rate Equation

The rate equation joins the rate of the forward reaction with the concentration and pressure of the reactants. The rate equation is written as:

Rate=kreactantAxreactantBy

Where x and y are the orders of the reaction with respect to the reactants A and B, respectively.

02

Order of reaction

The order of the reaction is the relationship between the amounts of the reactants and the speed of the reaction.

It determines the active participation of each reactant and helps in determining that to what extent the rate is affected if the concentration of any one of the reactants is to be changed.

03

Explanation

a. The rate equation for the given reaction is:

Rate=kCH3BrNaCN

b. The rate gets doubled when the concentration of CH3Brgets doubled.

c. The rate gets halved when the concentration of NaCN gets halved.

d. To determine the rate, increase the concentration of both of the reactants by a factor of five. So the rate equation becomes:

Rate=k55=25k

So, the rate increases by a factor of 25.

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