The problem provides experimental data for the reaction:

2A_{(aq)} + B_2_{(aq)} \rightarrow C_{(s)} + D_{(aq)}

You are asked to determine the reaction order with respect to each reactant, write the rate law equation, and calculate the rate constant $k$ along with other specifics. Here's how we can solve this step by step:

Data from the Table:

a) Determining the Order with Respect to A

• Between Experiment 1 and 2, $[B_2]$ is constant at 0.1 M, but $[A]$ changes from 0.1 M to 0.2 M. The rate decreases from 80 to 40 cm/s.
• To determine the order with respect to A, we use the rate law equation:

[ \text{Rate} = k [A]^m [B_2]^n ]

Taking the ratio of the rates:

[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k [A_2]^m [B_2]^n}{k [A_1]^m [B_2]^n} = \frac{40}{80} = \frac{(0.2)^m}{(0.1)^m} ]

This simplifies to:

$\frac{1}{2} = \left(\frac{0.2}{0.1}\right)^m = 2^{-m}$

Therefore, $m = -1$.

b) Determining the Order with Respect to B2

• Between Experiment 2 and 3, $[A]$ is constant at 0.2 M, but $[B_2]$ changes from 0.1 M to 0.2 M. The rate decreases from 40 to 10 cm/s.

Taking the ratio of the rates:

[ \frac{\text{Rate}_3}{\text{Rate}_2} = \frac{k [A]^m [B_2]^n}{k [A]^m [B_2]^n} = \frac{10}{40} = \frac{(0.2)^n}{(0.1)^n} ]

This simplifies to:

$\frac{1}{4} = \left(\frac{0.2}{0.1}\right)^n = 2^{-n}$

Thus, $n = 2$.

c) Writing the Rate Law Equation

The general form of the rate law is:

[ \text{Rate} = k [A]^m [B_2]^n ]

Substituting the values of $m = -1$ and $n = 2$, the rate law becomes:

[ \text{Rate} = k [A]^{-1} [B_2]^2 ]

d) Determining the Total Reaction Order

The total order of the reaction is the sum of the orders with respect to $A$ and $B_2$:

$\text{Total order} = -1 + 2 = 1$

e) Calculating the Rate Constant (k)

Using the rate law equation and data from Experiment 1:

$80 = k (0.1)^{-1} (0.1)^2$

Simplifying:

$80 = k \times 10 \times 0.01$

$80 = k \times 0.1$

$k = \frac{80}{0.1} = 800 \, \text{cm/s} \cdot \text{M}^{-1}$

f) Calculating the Rate When $[A] = 0.3\, M$ and $[B_2] = 0.3\, M$

Substituting the concentrations into the rate law:

$\text{Rate} = 800 \times (0.3)^{-1} \times (0.3)^2$

Simplifying:

$\text{Rate} = 800 \times \frac{1}{0.3} \times 0.09$

$\text{Rate} = 800 \times 3.33 \times 0.09 = 240 \, \text{cm/s}$

Let me know if you'd like more details or have any questions! Here are some related questions to expand your understanding:

1. How does a negative reaction order physically affect the concentration's impact on rate?
2. Can a reaction order be fractional, and what would that imply in a rate law?
3. Why is it important to experimentally determine reaction orders rather than inferring from the balanced chemical equation?
4. How would temperature affect the value of the rate constant $k$?
5. How could you experimentally determine the rate constant at different temperatures to establish the Arrhenius equation?

Tip: Always check if there is an inverse relationship in the reaction orders by analyzing how the rate changes when concentrations are altered.