Question

What is an example of a rate of reactions practice problem?

Answer 1

Here's an example of a rate of reaction problem:

What is the rate constant k for the first-order decomposition of `N_2O_(5(g))` at `25^@C` if the half-life of `N_2O_(5(g))` at that temperature is `4.03 * 10^4` seconds? Under these conditions, what percentage of `N_2O_5` molecules will not have reacted after 1 day?

We know that half-life represents the time needed for a reactant's concentration to reach half of its initial value. The equation for the half-life of a reactant in a first-order reaction is given by

`t_("1/2") = 0.6931/k`, where `k` represents the rate constant.

Since we know that `N_2O_5` has a half-life of `4.03 * 10^4s` under these conditions, the rate constant will be

`k = 0.6931/t_("1/2") = 0.6931/(4.03 * 10^4s) = 1.72*10^(-5)s^(-1)`

Since this is a first-order reaction, the equation we'll use to determine what percentage is left unreacted is

`ln(([N_2O_5])/([N_2O_5]_0)) = -k*t`, where

`[N_2O_5]` - the amount that does not react;
`[N_2O_5]_0` - the initial amount;
`t` - the time of the reaction - in this case, 1 day;

Since `[N_2O_5]` represents the amount that does not react, we can say that `([N_2O_5])/([N_2O_5]_0) * 100%` will be the percentage of `N_2O_5` that does not react.

Converting 1 day to seconds will give

`1` `"day" * ("24 hours")/("1 day") * ("60 minutes")/("1 hour") * ("60 seconds")/("1 minute") = 86400s`

`ln(([N_2O_5])/([N_2O_5]_0)) = - 1.72*10^(-5)s^(-1) * 86400s = -1.49`

This means that

`([N_2O_5])/([N_2O_5]_0) = e^(-1.49) = 0.225`

Therefore, the percentage left unreacted is

`0.225 * 100% = 22.5%`