Example
Answer
Q 1: The rate of a chemical reaction. A. is always dependent of the concentration of all reactants.
B. is always increased with increasing temperatures. C. is directly proportional to the value of
∆E.
Answer: (B) The collision model indicates that molecular collision is necessary for reaction.
Because an increase in temperature raises molecular velocities and the percentage of effective collisions, the reaction rate increases.
Q 2: For the first-order reactions of different substances A and X.
A → B, at t½ = 30.0 min. X → Y, at t½ = 60.0 min. This means that:
A. a certain number of grams of A will react twice as fast as the same number of grams of X.
B. a certain number of grams of X will react twice as fast as the same number of grams of A.
C. 3 moles of A will react more rapidly than 3 moles of X.
Answer: (C) It takes half as much time for A to form B as for X to form Y, as seen by the smaller half-life. Note that option “B” would be incorrect as the grams of A and the grams of X are not the same number of moles.
Q 3: A reaction is first order with respect to [X] and second order with respect to [Y]. When [X] is
0.20 M and [Y] = 0.20 M the rate is 8.00 × 10-3 M/min. The value of the rate constant, including correct units, is
A. 1.00 M/ min. B. 1.00 M/ min. C. 2.00 M/ min.
Answer: (B) From these data, it follows that the rate law is Rate = k[X]Y]2. Solving for the rate constant and substituting data for this reaction: K = Rate / [X][Y]2 = 8.00 × 10-3 M/min/(0.200 M)(0.200 M)2 = 0.008 M/min/0.008 M3 = 1.00 M/min.
Q 4: What is meant by the half-life of a reactant?
Answer: The half-life of a reactant is the time required for half of that reactant to be converted into products. For a first order reaction, the half-life is independent of concentration so that the same time is required to consume half of any starting amount or concentration of the reactant. On the other hand, the half-life of a second-order reaction does depend on the starting amount of the reactant.
Q 5: Calculate the concentration of H2O2 ([H2O2]) at 4000 second after the start of the reaction. By knowing that, the initial concentration Of H2O2 = [H2O2]o = 1.00M and k = 8.3 x 10-4 s-1.
Answer: To determine [H2O2] at 4000 second, use the integrated rate law where at t = 0, [H2O2]0 = 1.M. ln [H2O2]t - ln [H2O2]0 = − k t. ln [H2O2]t = ln [H2O2]0 − k t.
ln [H2O2]t - ln [1.00 M]0 − (- 8.3 x 10-4 s-1) x 4000 s = 3.3. [H2O2]t = e-3.3 = 0.037 M.
Q6: A certain first-order reaction is 45.0% complete in 65 s. What are the rate constant and the half-life for this process?
Answer: If [A]0 = 100.0, then after 65 s, 45.0: of A has reacted or [A]t = 55.0. For first order reactions:
ln[A]t - ln[A]0 = - k t. ln[55.0] - ln[100.0] = - k (65 s). k = 9.2 × 10-3 s-1. t½ = 0.693 / k = 0.693 / 9.2 × 10-3 s-1 = 75 s.
3- Second Order Reaction
(1) A + A (2A) → B + C. or (2) A + B → C + D a a, at t = 0. a b 0 0
(a –x)2 2x, at t = t1, (a – x) (b – x ) x x
(1) A + A → B + C. or 2A → B + C.
. 1
1 1
1
1 1
2 2
x) a(a . x , t x) a(a . x t , k x)
a(a x k t
a (a - x) a - x - a kt
(a - x) a
.kt
a kt (a - x)
, dt k
x) (a , dx x) k(a dt
dx
k
-
- -
= −
= −
= −
=
=
=
− =
−
=
Half live time (t1/2). When t = (t1/2), then x = a/2.
(2) A + B → C + D
(1)
Using the partial fractions, to resolving Eq. 1.
Accordingly, dx = A(b-x) + B(a-x). (3). If, a = x in Eq. 3, then:
Also if b = x in Eq. 3, then: , Substituting the values of A and B in Eq. 2.
From Eq. (1).
At, t =0, x = 0. Accordingly,
=
−
−
−
−
= k dt
x) x)(b (a
dx x) , x)(b
k(a dt
dx
1 1
2 1
=
s
−k
a t
, x)
a(b x) b(a
b) t(a .
k −
−
= 2 303 − log
Example: Hydrolysis of ester in basic medium.
CH3COOCH3 + H2O + NaOH → CH3COONa + CH3OH.
Q: The rate constant for the decomposition of nitrogen dioxide: 2NO2 + 2NO + O2, with a laser beam is 1.70 M-1 min-1. Find the time, in seconds, needed to decrease 2.00 mol/L of NO2 to 1.25 mol/L.
Answer: Since the units of the rate constant are M-1· min-1, we know that the reaction is second order overall. Since the reaction has only one reactant, NO2, the reaction is second order with respect to NO2. The integrated rate equation for a reaction that is second order with respect to NO2 as the only reactant is:
Q: What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds?
Answer: a. rate of a chemical reaction. The units for rate are always mol/L•s.
b. rate constant for a zero-order rate law. Rate=k; k must have units of mol/L•s c. rate constant for a first-order rate law. Rate=k[A], k must have units of s-1.
d. rate constant for a second-order rate law. Rate=k[A]2, k must have units of L/mol• s-1. Q: A certain bimolecular reaction in which a = b, 10% completed at 10 min. How long does it take for 50% completion?
Answer: For 2nd order reaction, a = 100 & x = 10 so:
4- Third Order Reaction
1- A + A + A → Product. 2- 2A + B → Product. 3- A + B + C → Product.
A + A + A → Product (B + C + D). Or 3A → Product.
A + A + A → B + C + D. or (3A) → B + C + D.
a a a, at t = 0. (a –x)3 3x, at t = t1,
. k a
t
, a (a - x)
a - x) . x (
t , k a (a - x)
x ) a - x (
a - (a - x)
k t
. ,
dt k
x) (a , dx k(a - x) dt
dx
2 2 / 1
2 2 2 2
2 2
3 3
