Question

Question #4ce8a

Answer 1

The answer given in your book is correct.

Explanation:

The rate of a reaction tells you the rate of change per unit of time in the concentration of the reactants or in the concentration of the products.

In your case, the decomposition of dinitrogen pentoxide, `"N"_2"O"_5` is said to be a first-order reaction, which means that the rate of the reaction varies linearly with the concentration of the reactant.

So, take a look at the first reaction

`"N"_2"O"_text(5(g]) -> 2"NO"_text(2(g]) + 1/2"O"_text(2(g])`

The rate of the reaction as described as the change in the concentration of the reactant per unit of time, `d/dt`, is described as

`"rate" = -(d["N"_2"O"_5])/dt`

The rate law is given as

`"rate" = k * ["N"_2"O"_5]`

This of course means that you have

`(d["N"_2"O"_5])/dt = - k * ["N"_2"O"_5]`

So, what does this mean?

The rate of reaction depends linearly on the concentration of the reactant and on a rate constant `k`. Notice that you have a `1` stoichiometric coefficient in front of dinitrogen pentoxide in the balanced chemical equation.

This is important because it tells you that the reaction proceeds with one molecule of reactant undergoing decomposition. Keep this in mind.

Now take a look at this reaction

`color(red)(2)"N"_2"O"_text(5(g]) -> 4"NO"_text(2(g]) + 2"O"_text(2(g])`

This time, you are told that

`(d["N"_2"O"_5])/dt = - k^' * ["N"_2"O"_5]`

Once again, the rate of reaction depends linearly on the concentration of the product and on a rate constant `k^'`.

But what is different between the two reactions?

This time you have a `color(red)(2)` stoichiometric coefficient in from of dinitrogen pentoxide. This is extremely important because it tells you that the reaction proceeds with two molecules of dinitrogen pentoxide undergoing decomposition, possibly after a successful collision.

For a given concentration of dinitrogen pentoxide, the reactant is consumed twice as fast in this second case, since now two molecules of dinitrogen pentoxide are required in order for the reaction to take place, as opposed to just one in the first case.

You can say that, compared with the first reaction, you have

`1/color(red)(2) * (d["N"_2"O"_5])/dt = -k^' * ["N"_2"O"_5]`

But since you also know that

`(d["N"_2"O"_5])/dt = - k * ["N"_2"O"_5]`

You can coclude that you need to have a rate constant that is half that of the first reaction

`1/2 * (-k * color(red)(cancel(color(black)(["N"_2"O"_5])))) = - k^' * color(red)(cancel(color(black)("N"_2"O"_5])))`

`k/2 = k^' implies color(green)(k = 2 * k^')`