Chapter 7, Problem 7.61QP

Interpretation Introduction

Interpretation:

Rate law for the given reaction has to be written if the order of all reactant is one and if rate constant in doubled then there will be increase, decrease, or no change in the rate of the reaction has to be given.

    ${\text{CH}}_{\text{4}}(g)+{\text{2O}}_{\text{2}}(g)\to {\text{2H}}_{\text{2}}\text{O}(l)+{\text{CO}}_{\text{2}}(g)$

Concept Introduction:

Rate of the reaction is proportional to the concentration of the reactant if all the other parameters are held as constant.  Rate of a reaction is equal to the product of rate constant and the concentration of the reactant that is raised to exponent that is reaction order.  This is termed as rate law.

Rate law for single reactant converting into product:

    $Rate=k{[A]}^n$

Where,

    $k$ is the rate constant.

    $[A]$ is the molar concentration of reactant.

    $n$ is the order of reaction.

Rate law for two reactants converting into products:

    $Rate=k{[A]}^n{[B]}^{n\text{'}}$

Where,

    $k$ is the rate constant.

    $[A],[B]$ is the molar concentration of reactant.

    $n,n\text{'}$ is the order of reaction.