The Haber process for the manufacture of ammonia is
represented by the following equation.
N2(g) + 3H2(g) <--->
2NH3(g) ΔH = –92 kJ
mol–1
Which statement is correct about this reaction when
the temperature is increased?
A) Both forward and backward rates increase.
B) The backward rate only increases.
C) The forward rate only increases.
D) There is no effect on the backward or forward
rate.
Answer
: A
When the temperature is increased, the percentage of reactant
molecules with energy equal or exceeding the activation energies
for both the backward and forward reactions, are increased
(along with temperature). Hence, both the forward and backward
reaction rates increase.
However, the
percentage increase in rate is greater for the reaction in the
direction with the larger activation energy. If the
forward reaction is endothermic (ie. the forward reaction will
accordingly have a larger activation energy, compared to the
backward reaction's smaller activation energy), then increasing
the temperature will increase both the forward and backward
reaction rates, but will increase the forward reaction rate by
a larger magnitude compared to the backward reaction rate. This
results in the position of equilibrium shifting towards the right
(ie. towards the products), even as the rate of both the
forward and backward reactions simultaneously increase.
In an endothermic forward reaction, heat may be regarded as
a reactant (rather than product). Increasing the temperature is
thus akin to increasing the concentration of the reactants, which
will therefore cause the position of equilibrium to shift towards
the right (ie. towards the products), as predicted by Le
Chatelier's principle (ie. the system attempting to remove the
added heat energy by using some of it to convert the reactants to
the products). Mathematically, the Kc (ie. equilibrium constant)
value will increase, while the Qc (ie. reaction quotient) value
remains initially unchanged.
Hence (while increasing the temperature of the system increases
both the forward and backward rate constants), the rate of the
forward reaction (ie. indicated by the new relatively
much larger k-forward value) will now exceed the
rate of the backward reaction (ie. indicated by the new
relatively only slightly larger k-backward
value) until equilibrium is eventually re-established for
the system, ie. Qc = Kc.
Once Qc = Kc and the system has re-established equilibrium,
the rates of the forward and backward reactions will be exactly the
same. This is despite the k-forward rate constant still having a
larger value compared to the k-backward rate constant. To
understand this, recall that the rate of reaction is given as the
rate constant multiplied by the molarities of the reactants, each
raised to the power of their orders. Thus at position of
equilibrium, having smaller reactant molarities and a larger
k-forward rate constant, will be mathematically equal to having
larger product molarities and a smaller k-backward rate
constant.