Question

A rate is equal to 0.0200 M/s. If [A] = 0.100 M and rate = k[A][B]2, what is the new rate if the concentration of [A] is increased to 0.200 M?

Answer 1

The new rate is 0.0400 mol·L⁻¹s⁻¹.

r = k[A][B]²

The reaction is first order in [A].

If you double [A], you double the rate.

The new rate is 2 × 0.0200 mol·L⁻¹s⁻¹ = 0.0400 mol·L⁻¹s⁻¹ .

Answer 2

The new rate is $\boxed{0.0400{\text{ M/s}}}$ if concentration of A is increased to 0.200 M.

Further Explanation:

Rate of reaction:

It is the speed by which any chemical reaction occurs. It is generally defined in terms of concentration of products or concentration of reactants consumed per unit time.

We are provided with the rate of reaction as follows:

${\text{Rate}} = {\text{k}}\left[ {\text{A}} \right]{\left[ {\text{B}} \right]^2}$                                                    …… (1)

Here,

k is the rate constant for the reaction.

[A] is the concentration of A.

[B] is the concentration of B.

Substitute0.100 M for [A] in equation (1) to calculate original rate of reaction.

${\text{Original Rate}} = {\text{k}}\left( {0.100{\text{ M}}} \right){\left[ {\text{B}} \right]^2}$                                 …… (2)

Substitute 0.200 M for [A] in equation (1) to calculate new rate of reaction.

${\text{New rate}} = {\text{k}}\left( {0.200{\text{ M}}} \right){\left[ {\text{B}} \right]^2}$                      …… (3)

Dividing equation (3) by equation (2), we get:

$\dfrac{{{\text{New Rate}}}}{{{\text{Original Rate}}}} = \dfrac{{{\text{k}}\left( {0.200{\text{ M}}} \right){{\left[ {\text{B}} \right]}^2}}}{{{\text{k}}\left( {0.100{\text{ M}}} \right){{\left[ {\text{B}} \right]}^2}}}$  

Solving above expression, we get the relation between new and original rate of reaction as follows:

${\text{New Rate}} = 2\left( {{\text{Original Rate}}} \right)$                                                      …… (4)

Substitute 0.0200 M/s for original rate in equation (4).

$\begin{aligned}{\text{New Rate}}&= 2\left( {{\text{0}}{\text{.0200 M/s}}} \right) \\&= 0.0400{\text{ M/s}} \\\end{aligned}$  

Learn More:  

1. What is the half-life of the reaction? https://brainly.com/question/8907464
2. Rate of chemical reaction: https://brainly.com/question/1569924

Answer Details:  

Grade: Senior School

Subject: Chemistry

Chapter: Chemical Kinetics

Keywords: rate of reaction, speed, chemical reaction, rate, k, A, B, concentration, [A], [B], new rate, original rate, 0.0400 M/s, 0.0200 M/s, 0.100 M, 0.200 M, unit time.