Topic 16: Chemical Kinetics (HL)

Question 1

rate equation

Answer 1

rate = k[A]^m[B]^n

m = exponent in rate equation described as the order with respect to reactant A
n = exponent in rate equation described as the order with respect to reactant B

Rate equations can only be determined experimentally because the orders can only be deduced empirically.

Question 2

Contact process

Answer 2

Production of sulfuric acid, where sulfur is first reacted with oxygen gas, O2
(g) , to form sulfur dioxide gas, SO2 (g) :

S(s) + O2 (g) → SO2(g)

Sulfur dioxide then reacts with oxygen gas to produce sulfur trioxide, SO3(g) : 2SO2(g) + O2 (g) 2SO3(g)

The catalyst used is vanadium(V) oxide, V2O5(s) , which is a heterogeneous catalyst.

Sulfur trioxide is next absorbed into concentrated sulfuric acid, H2SO4(l) . This produces oleum, H2
S2O7(l) . Oleum reacts with water to
produce aqueous sulfuric acid, H2SO4(aq):

SO3(g) + H2SO4(l) → H2S2O7(l)

H2S2O7(l) + H2
O(l) → 2H2SO4(aq)

Question 3

Reaction mechanism

Answer 3

the sequence of events in a reaction.

Any individual step is called an elementary step.

Question 4

List three classifications of an elementary step

Answer 4

Elementary step is classified by its molecularity:

1) unimolecular: single molecule involved in the step
2) bimolecular: two molecules or atoms are involved in collision
3) termolecular: three molecules or atoms are involved in collision.

Each elementary step has its own rate constant, k and its own activation energy, Ea.

Question 5

Reaction intermediate

Answer 5

a molecule that is formed in step 1 of the reaction mechanism but consumed in step 2.

Question 6

Rate equation for zero order

Answer 6

[A] = -kt + [A]0

[A] = concentration of reactant A 
k = rate constant 
t = time 
[A]0 = initial concentration.

Question 7

Rate equation for 1st order

Answer 7

ln[A] = -kt + ln[A]0

Question 8

Rate equation for 2nd order

Answer 8

1/[A] = kt + 1/[A]0