Question

the reaction
\\ a + b \longrightarrow c + d \quad \text{rate} = kab^2 \\
has an initial rate of \\(0.0910\text{ m/s}\\).

what will the initial rate be if \\(a\\) is halved and \\(b\\) is tripled?

initial rate:

what will the initial rate be if \\(a\\) is tripled and \\(b\\) is halved?

initial rate:

Explanation:

Identify the rate law and initial conditions

We are given the rate law:

$$ \text{rate} = k[\text{A}][\text{B}]^2 $$

The initial rate is:

$$ \text{rate}_0 = 0.0910\text{ M/s} $$

This is related to the Rate Law concept.

Calculate the rate when [A] is halved and [B] is tripled

Let the new concentrations be \([\text{A}]_1 = 0.5[\text{A}]_0\) and \([\text{B}]_1 = 3[\text{B}]_0\).
The new rate is:

$$ \text{rate}_1 = k[\text{A}]_1[\text{B}]_1^2 = k(0.5[\text{A}]_0)(3[\text{B}]_0)^2 $$

$$ \text{rate}_1 = 0.5 \times 9 \times k[\text{A}]_0[\text{B}]_0^2 = 4.5 \times \text{rate}_0 $$

$$ \text{rate}_1 = 4.5 \times 0.0910\text{ M/s} = 0.4095\text{ M/s} $$

Calculate the rate when [A] is tripled and [B] is halved

Let the new concentrations be \([\text{A}]_2 = 3[\text{A}]_0\) and \([\text{B}]_2 = 0.5[\text{B}]_0\).
The new rate is:

$$ \text{rate}_2 = k[\text{A}]_2[\text{B}]_2^2 = k(3[\text{A}]_0)(0.5[\text{B}]_0)^2 $$

$$ \text{rate}_2 = 3 \times 0.25 \times k[\text{A}]_0[\text{B}]_0^2 = 0.75 \times \text{rate}_0 $$

$$ \text{rate}_2 = 0.75 \times 0.0910\text{ M/s} = 0.06825\text{ M/s} $$

Answer:

Question 1

What will the initial rate be if [A] is halved and [B] is tripled?

initial rate: <blank>\(0.410\)</blank> M/s

Question 2

What will the initial rate be if [A] is tripled and [B] is halved?

initial rate: <blank>\(0.0683\)</blank> M/s