# Calculate rate constant k?

A + B --> C

Consider the above reaction. The rate law can be written as:

Rate = k[A]^m[B]^n

where k is the rate constant for the reaction.

Under the conditions: [B]0 >> [A]0 , a linear regression fit was obtained for 1/[A] versus t. The equations of these were found to be:

Equation of regression line; [B]0

y = 3.00×10-2 x + 0.12; 1.70×10-2 M

y = 3.00×10-2 x + 0.71; 2.80×10-1M

What is the value of m?

What is the value of n?

What is the rate constant k?

I calculated that m = 2 and n = 0. How do I find rate constant k?

Relevance

The rate constant is

k = 3.00×10⁻² M⁻¹s⁻¹

For the unit of the rate constant I assumed that [A] was measured in M and time was measured in s.

Explanation:

Because B is present in large excess, change of [B] is negligible small. That means [B] remains constant at initial level and you can describe rate law as pseudo nth order reaction:

rate = k∙[A]^m∙[B]^n ≈ k∙[A]^m∙[B]₀^n = k'∙[A]^m

where k' = k∙[B]₀^n is the pseudo nth order rate constant.

A plot of 1/[A] versus t yields a straight line. So the pseudo nth order reaction is 2nd order reaction, i.e. n=2. The integrated second order rate law is given by:

1/[A] = k'∙t + 1/[A]₀

As you can see the slope of your regression represents the pseudo second order rate constant.

From the two run with different [B]₀ you find that k' does not depend on [B]₀. SO reaction is zeroth order with respect to B, i.e. n=0.

Hence,

k' = k∙[B]₀⁰ = k

=>

k = k' = 3.00×10⁻²