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The rate of a reaction is usually expressed as the change in
concentration of a species divided by the time during which that
change took place. If a graph of concentration against time is
plotted, the rate is the gradient of the graph, which is found by
drawing a tangent to the curve:
The graph above shows the concentration of a chemical species
against time, and since the concentration of the species is falling
over time, the gradient of the graph is negative. If the
concentration of the product was plotted against time, the
gradient would be positive since the concentration
increases with time.
From the graph it can be seen that the rate (i.e. the gradient) is
changing with time and, of course, with concentration. For many
reactions it is found that the rate is proportional to the
concentration raised to some power. If the rate depends on the
concentration raised to the power of one, the reaction is called
first order:
$$\text{first order:}\ \ rate = k_1[A]$$
In this equation, [A] represents the concentration of the species A
and $k_1$ is the first order rate constant which has units of
time$^{-1}$. If a graph of rate against concentration is plotted, a
straight line of gradient $k_1$ would be observed.
However, if the rate is proportional to the square of the
concentration, the reaction is called second order: $$\text{second
order:}\ \ rate = k_2[A]^2$$
$k_2$ is the second order rate constant which has units of
concentration$^{-1}$ time$^{-1}$. For such a reaction, a plot of
rate against concentration squared should yield a straight line of
slope $k_2$. Note that the order of a reaction cannot be determined
simply from the coefficients in the balanced chemical
equation.
Consider the equation below. What
is the order of reaction with respect to A? B? C? Overall? What
would such orders mean physically?
$$ rate =
k\frac{[A]^{\frac{3}{2}}[B]^{2}}{[C]^{\frac{1}{2}}} $$
NOBr decomposes in the gas phase according to the following
equation:
$$2NOBR_{(g)} \rightarrow 2NO_{(g)} + Br_{2\ (g)}$$
The reaction is second order in NOBR:
$$rate = k_2[NOBR]^2$$
The following data show the concentration of NOBr, measured as a
function of time:
(i) Plot a graph of concentration
against time; draw a smooth curve through the points.
(ii) For about five different
concentration draw a tangent to the curve and determine the
gradient of the tangent. This gradient is
that rate at the chosen concentration.
(iii) Plot a second graph of the
rates you determined in (ii) against the square of the
concentration. You should find this to be a straight line, and its
gradient will be the second order rate constant, $k_2$.
Find the gradient as best you can, as there will be a
degree of scatter!
(iv) You may see in higher study
that a more convenient way of analysing a second order reaction is
to plot 1/concentration against time; it will be shown that such a
graph is a straight line with gradient $k_2$. Using the
data above, plot a third graph of 1/[NOBr] against time. Find the
gradient, $k_2$, and compare it with the value found in
(iii).
When considering a reaction, we can draw a plot showing how the
energy of molecules involves varies as we pass from reagents to
products; such a plot is called an energy profile, and a typical
example is shown below:
In this plot the products are lower in energy than the reactants,
so energy is given out when going from reactants to products - the
reaction is exothermic. As shown above, on this diagram the heat of
reaction is the difference in energy between the reagents and
products. For most reactions there is also an energy barrier which
has to be overcome in order to go from reagents to products. This
barrier is called the activation energy, $E_a$, and as shown on the
diagram is the energy gap between the reagents and the top of the
barrier.
Draw energy profiles for the
reactions which have an activation energy and which are (a)
endothermic and (b) exothermic. Mark each plot with the activation
energy and the heat of reaction.
(ii) Also mark on your plot the
activation energy of the reverse reaction i.e. the one where
products go back to reactants.
(iii) For a reaction which is
endothermic when going from reagents to products, what is the
relationship between the activation energy of the forward reaction,
the activation energy of the back reaction and the heat of
reaction?
(iv) Some reactions, such as
those between atoms i.e $I + I \rightarrow I_2$, have zero
activation energy. Draw energy profiles for reactions which are (a)
endothermic and have zero activation energy, (b) are exothermic and
have zero activation energy and (c) have zero heat of reaction but
a finite activation energy.
[Reproduced with the kind permission of James Keeler and Peter
Wothers, Department of Chemistry, University of Cambridge]