A and B react to form C and D in a reaction that was found to be second order in A. The rate constant at 30^@ "C" is "0.622 M"^(-1)cdot"min"^(-1). What is the half-life of A if 4.10xx10^(-2) "M" is mixed with excess B?

1 Answer
Nov 27, 2017

From the expression (derived below) for the second-order half-life, assuming the stoichiometric coefficient of A is a = 1,

t_"1/2" = 1/(k[A]_0)

And so, if 4.10 xx 10^(-2) "M" of A is added, that is [A]_0 and:

color(blue)(t_"1/2") = 1/(0.622 cancel("M"^(-1))cdot"min"^(-1) cdot 4.10 xx 10^(-2) cancel"M")

= color(blue)ul("39.2 min")


The idea here is that since A is mixed with excess B, the reaction is primarily controlled by the concentration of A (pseudo-second order).

So, the rate law is of the form:

r(t) ~~ k[A]^2[B]^0 = k[A]^2

where r(t) is the rate in "M/min", k is the rate constant in "M"^(-1)cdot"min"^(-1), and [A] is the molar concentration of A at t = "0 min".

The rate is also related to the change in concentration over time:

aA + bB -> cC + dD

r(t) = -1/a(Delta[A])/(Deltat) = -1/b(Delta[B])/(Deltat)

= +1/c(Delta[C])/(Deltat) = +1/d(Delta[D])/(Deltat)

You aren't told what the coefficient a of A is, but since this is all general, we don't assume a = 1. So we keep that in.

Equating the two expressions for the rate:

k[A]^2 = -1/a(Delta[A])/(Deltat)

We mainly care about the rate of reaction near the initial time, so we can look at infinitesimally small changes in concentration such that

(Delta[A])/(Deltat) ~~ (d[A])/(dt).

This then becomes an equation to rearrange and integrate over an interval.

k[A]^2 = -1/a(d[A])/(dt)

Rearrange to get:

akdt = -1/([A]^2)d[A]

Integrate the left side over time 0->t and the right side over [A]_0 -> [A], where [A]_0 is the initial concentration and [A] is the current concentration.

a and k are constants which float outside the integral sign.

=> akint_(0)^(t)dt = -int_([A]_0)^([A])1/([A]^2)d[A]

The integral of -1/x^2 is 1/x, so:

akt = 1/([A]) - 1/([A]_0)

This would rearrange to give a general integrated rate law for any stoichiometric coefficient a:

barul(|stackrel(" ")(" "1/([A]) = akt + 1/([A]_0)" ")|)

For a half-life, the initial concentration [A]_0 drops to half, so we call t = t_"1/2" at the point where [A] = 1/2[A]_0. Thus,

2/([A]_0) = akt_"1/2" + 1/([A]_0)

Solve for the half-life to get:

color(blue)barul(|stackrel(" ")(" "t_"1/2" = 1/(ak[A]_0)" ")|)

And if a = 1 for the stoichiometric coefficient of A, then

t_"1/2" = 1/(k[A]_0)