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Chem

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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?

  • Chem - ,

    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.

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