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A comprehensive set of exercises and solutions for chem 104 module 1, covering key concepts in chemical kinetics and equilibrium. It includes problems on reaction rates, rate laws, equilibrium constants, and the effect of various factors on equilibrium. Designed to help students understand and apply these concepts through practical examples and calculations.
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Module 1:
Question 1
In the reaction of gaseous N 2
to yield NO 2
gas and O 2
gas as shown
below, the following data table is obtained:
2 (g)
2 (g)
Data Table #
2 N 2
O 5 (g)
Time (sec) [N 2 O 5 ]^ [O 2 ]
0 0.300 M 0
300 0.272 M 0.014 M
600 0.224 M 0.038 M
900 0.204 M 0.048 M
1200 0.186 M 0.057 M
1800 0.156 M 0.072 M
2400 0.134 M 0.083 M
3000 0.120 M 0.090 M
Question 2
The following rate data was obtained for the hypothetical reaction: A + B
Experiment
[A] [B] rat
e 1 0.50 0.50 2.
n [B]
m (filling in the correct
exponents).
Your
Answer: rate
= k [A]
x [B]
y
rate 1 / rate 2 = k [0.50]
x [0.50]
y / k
x [0.50]
y 2.0 / 8.0 = [0.50]
x /
x
x
x = 2
rate 2 / rate 3 = k [1.00]
x [0.50]
y / k
x [1.00]
y 8.0 / 64.0 = [0.50]
y /
y
y y = 3
rate = k [A]
2.0 = k [0.50]
k = 64
Question 3
ln [A] - ln [A] 0
= - k t 0.693 = k t 1/
An ancient sample of paper was found to contain 19.8 %
C content as
compared to a present-day sample. The t 1/
for
C is 5720 yrs. Show the
calculation of the decay constant (k) and the age of the paper.
(g)
2 (g)
4 (g)
(g)
Your Answer:
0.309 mole of H 2
O formed = 0.309 mole of CH 4
formed
0.309 mole of H 2
O formed = 0.800 - 0.309 = 0.491 mole CO
0.309 mole of H 2
O formed = 3 x 0.309 mole H 2
reacted = 2.40 - (3 x 0.309) =
1.473 mole H 2
[CO] = 0.491 mole / 8.00 L = 6.1375 x 10
[H2] = 1.473 mole / 8.00 L = 18.4125 x 10
[CH4] = 0.309 mole / 8.00 L = 3.8625 x 10
[H2O] = 0.309 mole / 8.00 L = 3.8625 x 10
c
= [3.8625 x 10
] [3.8625 x 10
] / [6.1375 x 10
] [18.4125 x 10
c
Question 6
Explain the terms substrate and active site in regard to an enzyme.
Your Answer:
Enzymes are large protein molecules that act as catalysts and increases the
rate of a reaction. The catalysts act only on one type of substance to cause
one type of reaction and this is called a substrate. Active sites are enzymes
that are spherical in shape together ith a group of atoms on a surface of
protein. These active sites bind with the substrate causing it to undergo a
reaction. The substrate and active site forms a complex that creates a new
pathway with lower activation energy and thus speed up a reaction.
Question 7
The reaction below has the indicated equilibrium constant. Is the
equilibrium mixture made up of predominately reactants, predominately
products or significant amounts of both products and reactants. Be sure to
explain your answer.
= [0.0380], CO = [0.0620] and H 2
= [0.186] with
c
If the concentration of CO at equilibrium is increased to [0.200], how and
for what reason will the equilibrium shift? Be sure to calculate the value of
the reaction quotient, Q, and use this to confirm your answer.
Your
Answer:
c
(g)
2 (g)
4 (g)
(g)
c
c
c
is greater than K c
and thus the equilibrium will shift to the left towards
the direction of the reactants.
c
= 3.62 when H 2
= [0.0380], CO = [0.0620] and H 2
When CO = [0.200], Q = [0.0380] [0.0380] = 1.
The reaction must shift briefly in the forward direction to decrease the
[CO] to come back to equilibrium.This is in agreement with Q c
c
which also predicts the reaction will proceed to the right.
new CO concentration is 0.
Question 9
The equilibrium reaction below has the K c
= 3.93. If the volume of the
system at equilibrium is decreased from 6.00 liters to 2.00 liters, how and
for what reason will the equilibrium shift? Be sure to calculate the value of
the reaction quotient, Q, and use this to confirm your answer.
(g)
2 (g)
4 (g)
(g)
Your Answer:
Initial volume =
6.00 L Final volume
= 2.00 L Volume is
reduced by 3.
Pressure is inversely proportional to volume, thus pressure is tripled and
concentration of gases is also tripled.
Final concentration = Initial concentration x 3
Your Answer:
Temperature is decreased in the reaction at equilibrium that has a - ∆H
thus reaction shifts in the forward direction. The reaction shift in the
direction that produces some heatIn this reaction, the concentration of the
product increases and the concentration of the reactants decreases and thus
increases the value of K c
4
4
4
Module 2:
Question 1
Identify each of the compounds below as ACID, BASE or SALT on the basis
of their formula and explain your answer.
(1) Cr(OH) 3
(2) HAsO 4
(3) CoC
Your
Answer:
Question 2
For the Brønsted-Lowry acid base reactions shown below, list the stronger
acid, stronger base, weaker acid and weaker base in the answer blanks
provided:
Stronger acid:
Stronger base:
Weaker acid:
Weaker base:
Your Answer:
Stronger acid:
Stronger
base: NH 3
Weaker
4
HClO 4
or HBrO 4
or
Stronger acid
is: Explanation:
Formula of conjugate base of the stronger acid:
Your Answer:
Stronger acid is: HClO 4
HBrO 4
Cl is more electronegative than Br so HClO 4
is stronger acid than HBrO 4
and so is Br to I, Br is more electronegative than I so HBrO 4
is stronger
than HIO 4
Stronger acid is: HClO 4
Explanation: Cl has the higher electronegativity (of Cl, Br or I) which
makes the H-O bond of HClO 4
most polar and most likely to form H
Formula of conjugate base of stronger acid: conjugate base of HClO 4
is ClO
Question 4
Show the calculation of the [H
] and pH of a 0.00350 M solution of the
strong acid H 2
. Your Answer:
] = 2 x [H 2
] = 2 x
pH = - log [H
] = - log
( 0.007) pH = 2.
Question 5
In the titration of 15.0 mL of H 2
of unknown concentration, the
phenolphthalein indicator present in the colorless solution turns pink when
26.4 mL of 0.130 M NaOH is added. Show the calculation of the molarity of
the H 2
2
a
= 1.8 x 10
a
1.8 x 10
= (x) (x) /
(0.645-x) x
= 1.8 x 10
x = 3.41x
2 2 3 2
4
4
pH = - log [H
] = - log
(3.41x
) pH = 2.
% ionization = ([C 2
]/ [HC H O ]) x 100
% ionization = (3.41x
/ 0.645) x 100
% ionization = 0.53%
Question 7
Predict and explain whether a solution of LiF is acidic, basic or neutral.
Your Answer:
Basic since Li hydrolyzes to form strong base [LiOH].
LiF: Basic since F
hydrolyzes to form a weak acid (HF) and OH
Question 8
Show calculation of the pH of a buffer prepared by mixing 0.100 M
Cl and 0.0750 M NH 3
(liq)
a
= 1.8 x 10
Your Answer:
(liq)
