Question

Write the rate law for this reaction, and give the numerical value of the rate constant?

What will the observed initial rate be if the initial concentration
[Y]=0.400 and [Z]=0.300?

Answer 1

The rate law in general just relates the rate `r(t)` with the rate constant `k` and the concentration `[Y]` of reactant `Y` and `[Z]` of reactant `Z`:

`r(t) = k[Y]^m[Z]^n`

where `m` is the order of reactant `Y` and `n` is the order of reactant `Z`. We do not know `m` or `n` yet, so we must find those to finish writing the rate law.

RATE LAW ORDERS, AND RATE LAW

To make our lives easier, let us set `[Z]` as a constant in the rate law. That way, when `[Y]` changes, we know it must influence `r(t)`. Thus, we focus on experiments 1 and 2 and the change in the initial `Y` concentration in relation to the change in initial rate:

`r_(i,1)(t) = k[Y]_(i,1)^m[Z]_(i,1)^n`
`r_(i,2)(t) = k[Y]_(i,2)^m[Z]_(i,2)^n`

But since `[Z]` is constant, `[Z]_(i,2) = [Z]_(i,1)`.

`(r_(i,2)(t))/(r_(i,1)(t)) = (cancel(k)[Y]_(i,2)^mcancel([Z]_(i,2)^n))/(cancel(k)[Y]_(i,1)^mcancel([Z]_(i,1)^n))`

`(r_(i,2)(t))/(r_(i,1)(t)) = ([Y]_(i,2)^m)/[Y]_(i,1)^m`

`(1.6xx10^(-4))/(4.0xx10^(-5)) = ("0.200 M"/"0.100 M")^m`

If you do the math, you'd get a comparison:

`4 = 2^m`

Thus, `color(green)(m = 2)`, and `Y` is second order, or the reaction is "second order with respect to `Y`".

For `Z`, we set `[Y]` constant and follow the same process, comparing experiments 1 and 3, such that `[Y]_(i,1) = [Y]_(i,3)`:

`(r_(i,3)(t))/(r_(i,1)(t)) = ([Z]_(i,3)^n)/[Z]_(i,1)^n`

`(8.0xx10^(-5))/(4.0xx10^(-5)) = ("0.200 M"/"0.100 M")^m`

If you do the math, you'd get a comparison:

`2 = 2^m`

Thus, `color(green)(n = 1)`, and `Z` is first order, or the reaction is "first order with respect to `Z`".

Therefore, the overall rate law is:

`color(blue)(r(t) = k[Y]^2[Z])`

RATE CONSTANT

The rate constant is specific to the reaction, not to the experiment. That means we can pick any rate from any trial, and find the rate constant for the entire reaction.

`r_(i,1)(t) = k_1[Y]_(i,1)^2[Z]_(i,1)`

`4.0xx10^(-5) = k_1("0.100 M")^2("0.100 M")`

`=> color(blue)(k_1) = (4.0xx10^(-5) "M/s")/(("0.100 M")^2("0.100 M"))`

`= color(blue)("0.04 M"^(-2)cdot"s"^(-1))`

To prove that `k` is the same for any chosen trial of the same reaction...

`r_(i,2)(t) = k_2[Y]_(i,2)^2[Z]_(i,2)`

`1.6xx10^(-4) = k_2("0.200 M")^2("0.100 M")`

`=> color(blue)(k_2) = (1.6xx10^(-4) "M/s")/(("0.200 M")^2("0.100 M"))`

`= color(blue)("0.04 M"^(-2)cdot"s"^(-1))`

Thus, `k_1 = k_2` and the rate constant is the same for any trial we choose, since it is merely the same reaction at a different test run.

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At this point you should be able to finish the problem yourself. Once you have the rate constant, you can use your given concentrations in the last part of the problem to find the initial rate of the reaction that corresponds to those new concentrations.