okay let's practice looking at initial rates and rate laws. given the initial
rates data below determine the rate law and the value of the rate constant for
this reaction. so the reaction we are looking at is 2NO + Cl2 yields 2NOCl.
and here is the initial rates data that we are going to use to find the rate law
and the rate constant. so if this seems confusing go ahead and check out my
tutorial on kinetics and when you're ready give this a try.
ok so we're gonna put this reaction at the top and then here is all of our initial rates data
now in order to find the rate law we're going to have to compare two different
trials and see how the data changes in terms of concentration and rate so the
first thing we want to do is compare trials 1 & 2 so we're looking at trials
1 & 2 there. now we first notice that the concentration of NO2 stays the same, point 1 molar
and then point 1 molar however the concentration of Cl2
increases from 0.10 to 0.15 so that is a factor of 1.5 so for the
second trial the Cl2 concentration is a time and a half that for trial 1. then
we look at the rate of NO consumption and that goes from point zero zero three
to point zero zero four five so the rate also increases by a factor of one point
five now a lot of times when we're using initial rates data we are seeing
concentrations double and that makes the arithmetic nice and tidy but we won't
always see that it's not always so convenient so we need a way to determine
the rate law when these concentrations are not
doubling but are instead varying by some other factor so we can use this
technique all we have to do is take the rate from trial to over the rate from
trial one and that is going to be equal to the concentration of Cl2 from the
second trial raised to the X power over the concentration of Cl2 from the
first trial also raised to the X power and if we solve this equation for x we
will find the order with respect to Cl2 so this side will simplify to one
point five and then on the right since these are raised to the same exponent we
can take 0.15 and divide by point one oh and that will be one point five to the X
now in order for one point five to the X to be equal to one point five x must be
one so an x value of one means that the reaction is first-order in Cl2 so
first order because x equals one okay so next let's compare trials one and three
so first we will notice that this time the concentration of Cl2 stays the
same however the concentration of NO2 of one-point-five again from point one to
point one five however this time the rate increases by a different factor and
to find out what that factor is that is going to be point zero zero six seven
five divided by point zero zero three and so that is a factor of two point two
five we would have to multiply that first rate by two point two five in
order to get the second rate so this is a little bit different let's use the
same equation so we've got our rate from trial three over the rate from trial one
and then we have the concentrations raised to the X power this time those
rates that will simplify to two point two five as we found here once again we
get one point 5 to the X however now in order to solve for X we're going to have
to do a little bit of algebra it's not as simple as it was before so
we can do is we can take the natural log of both sides the reason that's useful
is that due to some rules regarding logarithms we can take that exponent X
and bring it down as a coefficient in front of that natural log term so now we
have natural log of 2.25 equals x times the natural log of 1.5 now to solve for
X we divide both sides by the natural log of 1.5 and then we just plug that
into the calculator and we get 2 so because x equals 2 the reaction is
second order in n-o so this technique is useful because we can use any kind of
data where the concentrations and the rates are increasing by any factor it
doesn't even have to be an integer factor it can be any random data doesn't
have to be the perfect situation where things are doubling we can take any data
and we will be able to calculate the order with respect to any of these
reactants however if that seemed a little bit confusing there's also
another way to rationalize it where we don't have to do quite as much
arithmetic we can just say that the Cl2 concentration increases by a factor of
1.5 the rate increases by a factor of one point five and one point five equals
one point five to the 1 so it is first-order in Cl2 then we would say
that the NO concentration increases by a factor of 1.5 the rate increases by a
factor of two point two five and two point two five equals one point five
squared so because of that too right there we know it is second order in NO
so in other words if the rate increases by a factor that is equivalent to the
factor by which the concentration increases that is first order if the
concentration increases by some factor and the rate increases by that factor
squared that is second order if the concentration were to increase by some
factor and the rate were to increase by that factor cubed that would be third
order so we're beginning to see the arithmetic relationship here and we can
probably apply this to really any situation with initial rates not just
the oversimplified ones so now that we know that the reaction is second order
with respect to n oh and first order with respect to Cl2 we can find this
rate law and the rate law is going to be as follows
the rate equals K times the concentration of NO squared because it is second order
with respect to NO, times the concentration of Cl2 with no exponent
because it is first order with respect to Cl2. now it is very important to note
that any correlation with stoichiometric coefficients is coincidental it is true
that NO is raised to the second power and it has a coefficient of 2, and Cl2
is not raised to any power and it has a coefficient of 1 so those do correspond
to the stoichiometric coefficients in the balanced equation but that is
coincidental it is not the case that those coefficients become the exponents
in the rate law and you will see many examples where they do not correlate so
we must remember that is purely coincidental so now we want to solve for
the rate constant K now that is going to be very easy so first let's solve for K
in the rate law and to do that we just divide by those concentrations so the
rate constant is going to be equal to the rate divided by NO concentration
squared times Cl2 concentration and now in order to solve for that rate constant
we can just use the data from any trial we wish so let's just arbitrarily choose
trial 1 and we just plug everything in so k is equal to the rate or point 0 0 3
0 0 molar per second divided by NO concentration squared so that's point 1
Oh molar squared times Cl2 concentration which is point 1 molar
now we can't forget to square that NO concentration so that becomes
0.01 molar squared and then that all simplifies to 3 point 0 molar to the
minus 2 seconds to the minus 1 so all together this will be our rate law with the rate constant.