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?
1 Answer
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
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,
#|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
The half-life is done like so, since apparently,
#-(d[A])/(dt) = k[A][B]#
Separation of variables gives:
#-1/([A][B])d[A] = kdt#
Now we integrate the left side for
#-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,
#-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,
#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
#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
Thus, it being near an hour makes sense.