O3 + NO --> O2 + NO2

Write the rate law equation for the reaction. Explain how you obtained your answer.

Experiment 1:
[O3] .0010
[NO] .0010
Rate of formation of NO2 x

Experiment 2:
[O3] .0010
[NO] .0020
Rate of formation of NO2 2x

Experiment 3:
[O3] .0020
[NO] .0010
Rate of formation of NO2 2x

Experiment 4:
[O3] .0020
[NO] .0020
Rate of formation of NO2 x

How do you go about writing this equation?

So far, I think I have figured out the rate order for the O3 and NO is 1, so I have:

Reaction Rate=k[O3]^1[NO]^1

I got this because in exp. 1&2 the [O3] remained the same while the [NO] and the rate of formation of NO2 increased by a factor of 2.

Then in exp 1&3 the [NO] remained the same while the [O3] and the rate of formation of NO2 increased by a factor of 2.

How do I solve for K?

If the superscripts are as you have them, then take any experiment and write the rate expression. For example,

experiment #1,
rate = x = k(O3)^1(NO)^1
Plug in 0.002 for (O3) and 0.001 for (NO) and solve for k. You must solve for k in terms of x if you don't have a value of x for the rate of reaction.

To solve for the rate constant, k, in the rate law equation, you need to look at one of the experiments that have a known value for k.

Let's use experiment 2 as an example:

[O3] = 0.0010 M
[NO] = 0.0020 M
Rate of formation of NO2 = 2x

Plugging these values into the rate law equation, we have:
Rate = k[O3]^1[NO]^1 = 2x

Substituting the concentrations from experiment 2:
2x = k(0.0010)^1(0.0020)^1

Simplifying the equation:
2x = k(0.000002)

Now, let's solve for k:
k = (2x) / (0.000002)

Since we don't have the exact value of x, we can't determine the exact value of k. However, we can say that the rate constant, k, would be equal to 2x divided by 0.000002.

You can repeat this process for any other experiment with a known rate constant to solve for k.

To solve for the rate constant, k, we can use any of the given experiments to substitute the concentrations and rate values into the rate law equation. Let's use Experiment 1, where [O3] = 0.0010 M, [NO] = 0.0010 M, and the rate of formation of NO2 is x.

Plugging these values into the rate law equation, we get:
Rate = k[O3]^1[NO]^1 = k(0.0010 M)(0.0010 M) = k(0.0010^2) M^2

Since the given rate is denoted as x, we can rewrite the equation as:
x = k(0.0010^2) M^2

Now, to solve for k, we can rearrange the equation:
k = x / (0.0010^2) M^2

Using the given rate value for Experiment 1, we can substitute x = x:
k = x / (0.0010^2) M^2

Repeat the same process using Experiment 4 with [O3] = 0.0020 M, [NO] = 0.0020 M, and the rate of formation of NO2 also denoted as x. Plugging these values into the rate law equation, we get:
x = k(0.0020^2) M^2

Solving for k in this scenario, we have:
k = x / (0.0020^2) M^2

Now that we have obtained two equations for k, we can set them equal to each other:
x / (0.0010^2) M^2 = x / (0.0020^2) M^2

Since x is a common term on both sides of the equation, we can cancel it out:
1 / (0.0010^2) M^2 = 1 / (0.0020^2) M^2

Simplifying the equation further, we have:
1 / 0.000001 M^2 = 1 / 0.000004 M^2

Now, we can calculate the values on both sides of the equation to determine the value of the rate constant, k:
1 / 0.000001 M^2 = 1,000,000 M^-2
1 / 0.000004 M^2 = 250,000 M^-2

Therefore, the rate constant, k, is either 1,000,000 M^-2 or 250,000 M^-2, depending on the units of the concentrations used in the experiments.