Consider the reaction:

2O_{3}(g) 3O_{2}(g) rate = k[O_{3}]^{2}[O_{2}]^{-1}

What is the overall order of the reaction and the order with
respect to [O_{3}]?

1 and 2

* The exponent of [O_{3}] in the rate law is 2.
So, the reaction is second order with respect to [O_{3}].
Adding the exponents together, we get:*

* 2 + (-1) = 1*

* So, the reaction is first order overall.*

When the reaction below:

3NO(g) N_{2}O(g) + NO_{2}(g)

is proceeding under conditions such that 0.015 mol/L of
N_{2}O is being formed each second, the rate of the overall
reaction is ________ and the rate of change for NO is ________.

0.015 M/s; -0.045 M/s

** Notice that three moles of NO are required to make one
mole of N_{2}O in this reaction. So, the rate of
disappearance of NO (which has a negative sign) is THREE times
the rate of appearance N_{2}O. As the coefficient of
N_{2}O is a ONE, the rate of its appearance matches the
rate of the overall reaction.**

What is the rate law for the reaction below:

A + B + C D

if the following data were collected?

Exp [A]0 [B]0 [C]0 Initial Rate

1 0.4 1.2 0.7 2.32x10-3

2
1.3 1.2 0.9 7.54x10-3

3 0.4 4.1 0.8 9.25x10-2

4 1.3 1.2 0.2 7.54x10-3

rate = 3.36x10-3 [A] [B]3

* The general form of the rate law has to be:*

* rate = k [A]x [B]y [C]z*

* Using the experiments given to set up several equations,
we get:*

* 2.32x10-3 = k(0.4)x(1.2)y(0.7)z*

* 7.54x10-3 = k(1.3)x(1.2)y(0.9)z*

* 9.25x10-2 = k(0.4)x(4.1)y(0.8)z*

* 7.54x10-3 = k(1.3)x(1.2)y(0.2)z*

* With algebra, we find x=1, y=3, and z=0. Plugging these
values in, we get the rate law as:*

* rate = k [A] [B]3 [C]0 = k [A] [B]3*

* We can now plug in any experiment into the rate law
above to solve for k. Plugging in experiment 1:*

* 2.32x10-3 = k(0.4)(1.2)3*

* k = 3.36x10-3*

* The complete rate law:*

* rate = 3.36x10-3 [A] [B]3*

A chemical reaction is expressed by the balanced chemical equation:

A + 2B C

Consider the data below:

exp [A]0 [B]0 initial rate (M/min)

1 0.15 0.15
0.00110363

2 0.15 0.3 0.0044145

3 0.3 0.3 0.008829

Find the rate law for the reaction.

rate = k [A] [B]2

* The general form of the rate law has to be:*

* rate = k [A]x [B]y*

* Using the experiments given to set up several equations,
we get:*

* 0.00110363 = k(0.15)x(0.15)y*

* 0.0044145 = k(0.15)x(0.3)y*

* 0.008829 = k(0.3)x(0.3)y*

* With algebra, we find x=1 and y=2. Plugging these values
in, we get the rate law as:*

* rate = k [A]1 [B]2 = k [A] [B]2*

Calculate the value of the rate constant (k) for the reaction in question above

0.327

rate = k [A] [B]2

* Any of the three experimental data sets can then be
plugged in to solve for k. Plugging in the first experiment:*

* 0.00110363 = k(0.15)(0.15)2*

* k = 0.327*

If the initial concentrations of both A and B are 0.31 M for the reaction in questions 4 and 5, at what initial rate is C formed?

rate = 0.327 [A] [B]2

r** ate = 0.327(0.31)(0.31)2 = 0.00974 M/min**

We know that the rate expression for the reaction below:

2NO + O2 2NO2

at a certain temperature is rate = [NO]2 [O2]. We carry out two experiments involving this reaction at the same temperature, but in the second experiment the initial concentration of NO is doubled while the initial concentration of O2 is halved. The initial rate in the second experiment will be how many times that of the first?

2

* Let's say that [NO] = x and [O2] = y in the first
reaction. Therefore, in the second reaction, [NO] = 2x and [O2] = 0.5y.*

* If rate = k [NO]2 [O2]...*

* First reaction: rate = kx2y*

* Second reaction: rate = k(2x)2(0.5y) = 2kx2y*

* Therefore, the initial rate in the second experiment
will be two times that of the first.*

Consider the data collected for a chemical reaction between compounds A and B that is first order in A and first order in B:

rxn [A]0 [B]0 rate (M/s)

1 0.2 0.05 0.1

2 ? 0.05
0.4

3 0.4 ? 0.8

From the information above for 3 experiments, determine the missing concentrations of A and B. Answers should be in the order [A] then [B].

0.80M : 0.20M

** Use trial 1 to determine the rate constant k.**

** rate = k[A][B] = k(0.2)(0.05) = 0.1**

** k = 10 M-1·s-1**

** Using trial 2, solve for [A]:**

** rate = 10(x)(0.05) = 0.4**

** x = 0.8 M**

** Using trial 3, solve for [B]:**

** rate = 10(0.4)(y) = 0.8**

** y = 0.2 M**

For a reaction that is zero-order overall...

the rate does not change during the reaction.

* Remember that for a zero-order reaction, the reactant is
still used up. Its concentration will decrease as time passes, but
the rate will not be affected. The rate will also not change for a
zero-order reaction if you double or triple the concentration of
that reactant*.

Consider the reaction below:

A + B C

If it is 1st order in A and 0th order in B, a plot of ln[A] vs time will have a slope that is...

Constant

** The first order integrated rate equation shows a linear
relationship between the natural log of concentration and time.**

Consider the reaction below:

H2CO3(aq) CO2(aq) + H2O(l)

If it has a half-life of 1.6 sec, how long will it take a system with [H2CO3]0 of 2M to reach [H2CO3] of 125mM?

4

At a certain fixed temperature, the reaction below:

A(g) + 2B(g) AB2(g)

is found to be first order in the concentration of A and zeroth order in the concentration of B. The reaction rate constant is 0.05s-1. If 2.00 moles of A and 4.00 moles of B are placed in a 1.00 liter container, how many seconds will elapse before the concentration of A has fallen to 0.30 moles/liter?

37.94

* rate = k[A]*

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

* [A]0 = 2.00 mol / 1.00 liter = 2 M*

* [A] = 0.3 M*

* \ln\left(0.3\:M\right)\:=\:\ln\left(2M\right)-\left(0.05\:s^{-1}\right)t*

* t = 37.94 sec*

The reaction below:

A >>>>> products

is observed to obey first-order kinetics. Which of the following plots should give a straight line?

ln[A] vs t

** The first order integrated rate equation is**

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

** Since k and ln[A]0 are constants, this equation is in the
form y = b + mx, where y = ln[A] and x = t. A straight line is
produced by this equation.**

For the reaction below:

cyclobutane(g) >>>>>>>>> 2ethylene(g)

at 800K, a plot of ln[cyclobutane] vs t gives a straight line with a slope of -1.6 s-1. Calculate the time needed for the concentration of cyclobutane to fall to 1/16 of its initial value.

1.7 sec

The initial concentration of the reactant A in a first-order reaction is 1.2 M. After 69.3 sec, the concentration has fallen to 0.3 M. What is the rate constant k?

0.02 s-1

A reaction is found to be first order with respect to one of the reactant species, A. When might a plot of ln[A] vs time NOT yield a straight line?

All the answers could be correct

**
If a reaction is first order with respect to species A, a plot
of ln[A] vs time will yield a straight line only if the reactions
is zeroth order with respect to all other species (first order
overall) or if the concentrations of the other species is
effectively constant (pseudo-first order). Additionally, the
reaction should not come to equilibrium, as this will mean that
the forward rate and backward rates are equal. At this point, the
reaction rate will level off to a constant value. This is the same
as having a significant backwards reaction**.

Consider the following elementary reactions:

a) NO + O3 NO2 + O2

b) CS2 CS + S

c) O + O2 + N2 O3 + N2

Identify the molecularity of each reaction respectively.

bimolecular, unimolecular, termolecular

*
The elementary reactions given have two reactants, one
reactant, and three reactants respectively. Therefore, they are
bimolecular, unimolecular, and termolecular respectively.*

A and B react to form C according to the single step reaction below:

A + 2B C

Which of the following is the correct rate equation for [B] and the correct units for the rate constant of this reaction?

...

Consider the mechanism below:

NO2 + F2 NO2F + F k1, slow

F + NO2 NO2F k2, fast

What is the rate law?

...

Determine the overall balanced equation for a reaction having the following proposed mechanism:

Step 1: B2 + B2 E3 + D slow

Step 2: E3 + A B2 + C2 fast

and write an acceptable rate law.

...

Consider the reaction below:

H2(g) + I2(g) 2HI(g)

The proposed mechanism of this reaction is:

I2 ⇌ 2I k1, k-1(reverse rxn), fast

2I + H2 2HI k2, slow

What is the rate of the overall reaction?

...

A reaction rate increases by a factor of 655 in the presence of a catalyst at 37°C. The activation energy of the original pathway is 106 kJ/mol. What is the activation energy of the new pathway, all other factors being equal?

...