Worksheet #2
Week #5: 7/30
STEPs Chemistry: 2025
Team: The Reaction Chain
Name:
Instructions:
This worksheet is designed to be solved in class, if it was not finished the rest is homework.
EQUILIBRIUM CONSTANT EXPRESSION IS NOT THE SAME AS RATE LAW!
Problem 1. Write the equilibrium constant expressions for the following reversible reactions:
(a) 2 SO2 (g) + O2 (g) ⇌ 2SO3 (g)
(b) 2NH3 (g) ⇌ N2 (g) + 3H2 (g)
(c) Cu2+ (aq) + 4NH3 (aq) ⇌ [Cu(NH3 )4 ]2+ (aq)
Problem 2. For an elementary reaction aX + bY ⇌ cXY The law of mass action states:
d[X]
= -kf [X]a [Y ]b + kb [XY ]c
dt
What does this equation mean? How does it describe the rate of reaction? What do kf and kb
represent?
It relates the rate of change of concentration for a reactant to the concentrations of the reactants and products for a
reversible reaction.
The forward reaction reduces the concentration of the reactant, and the backwards increases the concentration of
the reactant
kf is the forward rate constant, kb is the backward rate constant.
Problem 3. Answer the following questions related to the conceptual understanding of chemical
kinetics. Each part is independent.
(a) The rate of a reaction increases with temperature. Does that necessarily mean the rate constant
has increased? Explain your reasoning based on kinetic theory.
Yes. Increasing temperature increases the number of particles at or above activation energy, which allows more effective
collisions to occur, increasing the rate of reaction
(b) Two reactions have the same rate constant, activation energy and initial reactant concentration
of 1.5 M. One is second order and the other is zeroth order. Which one has a faster initial rate?
Justify your answer based on reactant concentration.
The second order reaction has a higher initial rate. Since the square of the concentration is related to the
rate, then having an initial molarity above one ensures the second order reaction will always be faster.
(c) Why can the rate law be written directly from the balanced chemical equation in some cases,
but not in others?
The rate law can only be written directly for elementary reactions.
Also, some reactions need a catalyst or a special condition to occur, and the rate would mainly depend on the abundance of
that condition.
(d) Some reactions proceed rapidly even though they have high activation energy. How is this
possible?
If the temperature is high enough, a reaction with a high activation energy can proceed very rapidly.
(e) Can a reaction with a large equilibrium constant Keq still be slow? Explain your answer with
reference to kinetics and thermodynamics.
The equilibrium constant only tells us the ratio of products to reactants at equilibrium, not the speed of a reaction.
The rate constant tells us the speed of reaction. So yes, reactions with a large equilibrium constants can still be slow.
(f) Two reactions have the same reactants and products, but proceed through different mechanisms. Can they have different rates? Why or why not?
Yes. Different mechanisms often have different rate-determining steps.
(g) In a proposed reaction mechanism, an intermediate species appears in both a fast formation
step and a slow consumption step:
k
(slow)
k
(fast)
Step 1:
1
A −→
I
Step 2:
2
I −→
B
A student applies the steady-state approximation to derive the rate law for the formation of
B, assuming d[I]
dt ≈ 0.
Why is the steady-state approximation valid in this case?
Because the intermediate is used up as soon as it is made.
Problem 4. The ammonium salt of isocyanic acid is a product of the decomposition of urea
CO(NH2 )2 , represented below:
CO(NH2 )2 (aq) −−→ NH4 + (aq) + OCN – (aq)
Ahmed AlSaggaf is studying the decomposition reaction runs the reaction at 90 °C. The student
collects data on the concentration of urea as a function of time, as shown in the data table and the
graph below.
Time (hours)
[CO(NH2 )2 ] (M)
0
5
10
15
20
25
30
0.1000
0.0707
0.0500
0.0354
0.0250
0.0177
0.0125
[CO(NH2 )2 ] (M)
0.1
0.08
0.06
0.04
0.02
0
0
2
4
6
8
10
12 14 16 18
Time (hours)
20
22
24
26
28
30
Figure 1: Urea decomposition graph
Ahmed is trying to formulate the rate law for this reaction.
(a) Find the rate law for the reaction given the data above.
(b) Using the rate law and the student’s results, determine the value of the rate constant, k
Problem 5. The Contact process is used to create sulfuric acid for use in industrial applications.
The key step is the catalyzed oxidation of SO2 into SO3 , shown below:
2 SO2 (g) + O2 (g) ⇌ 2SO3 (g)
∆H = -198
kJ
mol
The reaction is carried out using Vanadium (V) Oxide. The following is the proposed mechanism:
1. SO2 (g) + V2 O5 (s) ⇌ SO3 (g) + V2 O4
∆H = H1
1
2. V2 O4 (s)+ O2 (g) ⇌ V2 O5 (s)
2
∆H = H2
(a) Write the Hess’s Law expression, relating the enthalpy changes of these reactions.
(b) This reaction is carried out at 500◦ C. Explain what increasing the temperature does to the
concentration of products at equilibrium .
Increasing the temperature of an exothermic reaction shifts the equilibrium to the reactants side, decreasing the products
concentration
(c) This reaction is carried out at a pressure of 1atm. Explain what decreasing the pressure would
have on the equilibrium concentration.
Decreasing pressure favors the side with more moles. There would be less products at equilibrium
(d) What is the order of this reaction in SO2 ? Assume that step 1 is the rate determining step, and
write the rate law expression.
It is first order in SO2. rate = -k[SO2]
(e) Explain the effect on the equilibrium yield that increasing the amounts of each substance would
be:
(I) SO2
increases the concentration of products
(II) V2 O5
nothing
Problem 6. Hydrogen peroxide decomposes spontaneously. The equation is shown below:
H2 O2 (aq) → H2 O(l) + 12 O2 (g)
Potential energy (kJ)
This reaction is exothermic. The energy profile is shown below.
Ea
2H2 O2 (aq)
∆H
2H2 O(l) + O2 (g)
Progress of reaction
The activation energy for the decomposition of hydrogen peroxide is Ea = 80
initially performed at T = 300K.
kJ
. This reaction is
mol
(a) Using the Arrhenius rate equation, k = Ae-Ea /(RT ) , the pre-exponential factor is A = 1.4 ×
J
, calculate the rate constant k for this reaction.
1011 s-1 and R = 8.314 mol·K
1.36*10^11
jus put in da numbers man
(b) M nO2 is added to catalyze the reaction. What effect does this change have on the activation
energy?
increases rate of reaction
The temperature of the reaction is increased to T = 400K and the activation energy when the
kJ
reaction is catalyzed by M nO2 is Ea = 25
.
mol
(c) What is the rate constant now and what does it say about the rate of the reaction?
1.38*10^11
rate increased
Problem 7. Sulfur tetrafluoride breaks down (hydrolysis) in water as follows:
SF4 (g) + H2 O(l) ⇌ SO2 (g) + 4 HF(g)
∆H = -52.91
kJ
mol
Raneem bubbled 68g of SF4 into a 1L container full of water.
(a) Calculate the initial molar concentration of SF4 .
(b) Khalid 1 and Khalid 2 found that the equilibrium constant with respect to molar concentrations of this reaction is 13.6 at a temperature of 631◦ C. This is a supercritical temperature
for water, but for simplicity’s sake, assume water remains a liquid, and write the equilibrium
constant expression for this reaction.
(c) Did you include H2 O(l) in your expression? Why or why not? What is the concentration of a
pure liquid?
(d) Raneem measures the concentration after a short while, and plugs in her numbers into the
expression you wrote in (b). She got a value of 6.8. Which way will the equilibrium shift?
(e) After leaving the container for a while, Raneem measured and found only 22g of SF4 . Calculate
the amount of SO2 and HF produced at this moment.
(f) According to the values you found in (e), will the equilibrium shift? If so, how? Justify your
answer.
(g) A blank ICE Table is drawn out for you. Fill it and calculate all the equilibrium concentrations.
Use the initial concentrations you calculated in part (e). Write the expression you would
use to solve this equation. You will get a 5th order polynomial. Do not attempt to solve it.
The expression is all we need.
SF4
SO2
HF
Initial (I)
Change (C)
Equilibrium (E)
(h) Yasser said that the equilibrium constant in terms of molar concentration Kc will be the same
as the equilibrium constant in terms of partial pressures Kp . Is he right? Calculate Kp and
explain why he is right, or wrong. Assume the x you put in the change row is equal to
0.0167 M.
(i) After telling her the equilibrium concentrations, Raneem wanted to increase the yield of SO2 .
What are three things she could do?
Note: the assumption that water is a liquid at 631C is not a right one to make. In real life, you
will need to include water as a gas in your equilibrium equation, but because water is supercritical
here, it behaves in a... weird way. Stay curious!
--- (: good job! you finished the worksheet! :) ---