Page 71 - Basic physical chemistry for the atmospheric sciences

Chemical kinetics 'i7

0
Solution. Substituting k = 1 . 0 x 1 0 - 1 s - 1 . Ea = 1 1 1 kJ mol - 1 , '/'

300°K, and R = * 8 . 3 14 x 1 0 - 3 kJ mo1 - 1 K - 1 into Eq. (3.9) gives
l .O x t o - 10
A = ( - 1 1 ) s - 1 = 2 . l x l 0 9s - 1
1
exp 14
(8 3 . x 1 0 - 3 )300
u
If we now assume (as is s ually done) that A and Ea do not vary with
1
temperature, we can substitute A = 2 . 1 x l 09 s - 1 , Ea = 1 1 kJ mol - 1 ,
1
R = * 8 . 3 1 4 x 1 0 - 3 kJ mo1 - K - 1 , and T = 273K into Eq. (3.9) to obtain
the rate constant at 273 . Hence,
K
( - l l l )
1 0
12
k27 K 3 - 2 . l x 1 0 exp (S . 3 14 x _ 3 273 s - l . 2 x 10 - s - I
- I _
_ 9 )
3.6 Catalysis
A catalyst is a substance that increases the rate of a chemical reaction
without itself undergoing a permanent chemical change. Thus, the
catalyst does not appear in the net chemical equation for the reaction,
although its presence may be indicated above the arrow in the equa­
tion. Generally, a catalyst increases the chemical reaction rate by
providing an alternative pathway that has a lower activation energy.
If the reactants and the catalyst are in the same phase, the catalysis
is said to be homogeneous. A heterogeneous catalyst exists in a differ­
ent phase from the reactant molecules.5
To illustrate a homogeneous catalyst consider the reaction

In the absence of a catalyst this reaction proceeds in the forward
n
directio , but very slowly. However, it can be catalyzed by bromine
(Br 2 ) . This catalysis occurs in two steps. First

then,
2Br ( aq) + H202(aq) + 2H + (aq)� Brz(aq) + 2H20(1)
-
Adding the last two equations we get for the net reaction