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
From the expression (derived below) for the second-order half-life, assuming the stoichiometric coefficient of
t_"1/2" = 1/(k[A]_0)
And so, if
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
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 ofA att = "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
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
=> akint_(0)^(t)dt = -int_([A]_0)^([A])1/([A]^2)d[A]
The integral of
akt = 1/([A]) - 1/([A]_0)
This would rearrange to give a general integrated rate law for any stoichiometric coefficient
barul(|stackrel(" ")(" "1/([A]) = akt + 1/([A]_0)" ")|)
For a half-life, the initial concentration
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
t_"1/2" = 1/(k[A]_0)