Question

In a second order reaction, the concentration of `A` has dropped to `"0.020 M"` in one hour. If the initial concentration of `A` is `"0.050 M"` and that of `B` is `"0.080 M"`, what is the rate constant? What then is the half-life, if `[B]` is held fixed?

Answer 1

RATE LAW

We can write the rate law as:

`\mathbf(r(t) = -(d[A])/(dt) = k[A]^m[B]^n)`

In this case, we have:

`\mathbf(r(t) = k[A][B])`

Since the initial concentration of `B` was given, I'll assume NOT that it's zero order with respect to `B`, but that `[B]` is fixed.

RATE CONSTANT

To get the rate constant, we need the initial rate, which we don't actually have yet. For now, let's solve for the final equation.

`color(green)(k = (r_0(t))/([A]_0[B]_0))`

Now, `[A] = 2/5 [A]_0` after `"1 hour"` (`"3600 s"`), so we can acquire a rate by recognizing that the concentration decreased by `3/5 [A]_0` in `"3600 s"`:

`|r(t)| = |-(d[A])/(dt)| = |(-3/5 [A]_0)/("3600 s")| = 8.bar33xx10^(-6) "M/s"`

So, now we get:

`color(blue)(k) = (8.bar33xx10^(-6) cancel"M""/s")/(("0.050 M")("0.080" cancel"M"))`

`= color(blue)(2.08bar3xx10^(-3))` `color(blue)("1/""M"*"s")`

THIS REACTION'S HALF-LIFE IS KNOWN IF `\mathbf(Delta[B] = 0)`

The half-life is done like so, since apparently, `[B]` doesn't change during that hour, which is nice. It simplifies our calculations.

`-(d[A])/(dt) = k[A][B]`

Separation of variables gives:

`-1/([A][B])d[A] = kdt`

Now we integrate the left side for `Delta[A]` and the right side for `Deltat`. `Delta[B] = 0`, so we can pull that out of the integral.

`-1/([B])int_([A]_0)^([A]) 1/([A])d[A] = int_0^t kdt`

`-1/([B])(ln[A] - ln[A]_0) = kt`

Now, for the half-life, `[A] = [A]_0/2`, which simplifies this down to:

`-1/([B])ln\frac([A])([A]_0) = kt_"1/2"`

Use the properties of logarithms to turn their subtraction into the logarithm of a fractional argument:

`-1/([B])ln\frac(cancel([A]_0))(2cancel([A]_0)) = kt_"1/2"`

Here, `ln(1/2) = -ln2`, thus cancelling out the negative sign out front.

`1/([B])ln2 = kt_"1/2"`

And solving for the half-life, we get:

`color(blue)(t_"1/2" = 1/([B])ln2/k)`

This makes sense, because it looks like the half-life of a first-order reaction involving only `[A]`, which would have been `t_"1/2" = (ln2)/k`. In fact, here, the units become:

`1/"s" = 1/"M" * 1/(1/("M/s"))`

`1/"s" = 1/cancel"M" * cancel"M"/"s"`

`color(green)(1/"s" = 1/"s")`

So, the half-life is therefore:

`t_"1/2" = 1/("0.080" cancel"M")(ln2)/(2.08bar3 xx 10^(-3) 1/(cancel"M"*"s")`

`=` `"4158.88 s"`

`~~` `color(blue)("1.155 hours")`

Naturally, if it took an hour to get to `2/5` the concentration of `A`, incorporating `B` into the reaction in such a way that `Delta[B] = 0` should increase the half-life by some amount because it is relative to both reactants at once.

Thus, it being near an hour makes sense.