# 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 #\mathbf(Delta[B] = 0)#**

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.