X is a product, because its concentration increases with time.
The average rate of reaction between any two points on the graph is the slope of the line
connecting the two points. The average rate is greater between points 1 and 2 than
between points 2 and 3 because they are different stages in the overall process. Points 1
and 2 are earlier in the reaction when more reactants are available, so the rate of forma-
tion of products is greater. As reactants are used up, the rate of X production decreases,
and the average rate between points 2 and 3 is smaller.
. Using the relationship rate = k[A]
, determine the value of x that produces a rate
law to match the described situation.
x = 0. The rate of reaction does not depend on [A], so the reaction is zero-order
x = 2. When [A] increases by a factor of 3, rate increases by a factor of (3)
x = 3. When [A] increases by a factor of 2, rate increases by a factor of (2)
. Given concentrations of reactants and products at two times, as represented in the dia-
gram, find t
for this first-order reaction.
. For a first order reaction, t
= 0.693/k; t
depends only on k. Use equation [14.12] to
solve for k.
Since reactants and products are in the same container, use number of particles as a meas-
ure of concentration. The red dots are reactant A, and the blue are product B. [A]
= 8, [A]
= 2, t = 30 min.
= –kt. ln(2/8) = –k(30 min);
k = 0.046210 = 0.0462 min
= 0.693/k = 0.693/0.046210 = 15 min
By examination, [A]
= 8, [A]
= 2. After 1 half-life, [A] = 4; after a second half-life,
[A] = 2. Thirty minutes represents exactly 2 half-lives, so t
= 15 min. [This is more
straightforward than the calculation, but a less general method.]
After 4 half-lives, [A]
× 1/2 × 1/2 × 1/2 × 1/2 = [A]
/16. In general, after n half-
lives, [A] = [A]
This is the profile of a two-step mechanism, A
B and B
C. There is one intermediate, B. Be-