Chemistry Assignment: Analyzing Equilibrium and Redox Reactions

This chemistry assignment comprehensively addresses dynamic equilibrium and redox reactions. It begins by defining dynamic equilibrium and its requirements, then calculates equilibrium constants (Kc) for various reactions. The assignment further explores the solubility product (Ksp) and the effect of concentration changes on equilibrium. Finally, it delves into redox reactions, determining oxidation states, identifying reducing agents, and balancing redox equations. The document concludes with detailed references. Desklib provides a platform to access this and many other solved assignments to aid students in their studies.

Chemistry Assignment 1
CHEMISTRY ASSIGNMENT
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Chemistry Assignment 2
Solution
Question 1
Dynamic equilibrium is a state in which the rate of forward reaction is the same as that of
backward reaction (Toh, 2013). From the equation provided, it implies that the rate at which
Sulphur (IV) Oxide reacts with Oxygen to form Sulphur (VI) Oxide is equal and opposite to the
rate at which Sulphur (VI) Oxide decomposes to form Sulphur (IV) Oxide and Oxygen. It
implies that reactions experiencing dynamic equilibrium are reversible. However, for dynamic
equilibrium to be possible, the physical conditions affecting the reaction e.g. temperature must
remain constant to avoid shifting the equilibrium either way. At dynamic equilibrium, the
quantities of the reactants (Oxygen to form Sulphur (VI) Oxide) and the products formed (form
Sulphur (IV) Oxide) remain more or less the same but are not equal (Verma, et al., 2016).
Question 2
I a) 2NH3 (g)  N2 (g) + 3H2 (g)
b) Kc (T )= [ N2 ]1
[ H2 ]3
[ N H3 ]2
c) 1 mol/dm3 = 1 Molar (M)
Kc ( T ) = [ M ]1 [ M ]3
[ M ]
2 =M 2= [ mol d m−3 ] 2
=mol2 d m−6
II a) N2O4 (g)  2NO2 (g)
b) Kc (T )= [ NO2 ]2
[ N2 O4 ]1

Chemistry Assignment 3
c) 1 mol/dm3 = 1 Molar (M) Kc ( T ) = [ M ]2
[ M ]
1 =M =mol d m−3
III a) CaCO3 (s)  CaO (s) + CO2 (g)
b) Kc (T )= [ CaO ]1
[ CO2 ]1
[ CaCO3 ]1
c) 1 mol/dm3 = 1 Molar (M) Kc (T )= [ M ]1 [ M ]1
[ M ]1 =M =mol d m−3
Question 3
a) To calculate the equilibrium constant Kc ( T ) , one needs to know the concentration of the
reactants and products and not necessarily the volume. Since the volume of the vessel
remains constant, we can calculate the value of equilibrium constant without necessarily
knowing the volume.
b) Kc (T )= [ CH 3 COOC H2 C H3 ]1
[ H2 O ]1
[ C H3 COOH ]1
[ C H3 C H2 OH ]1
c) Mole ratio is 1:1:1:1. Since 0.82 Mole of ethylethanoate was formed, a similar mole
water was formed as mole ratio is 1:1. The amount of ethanol that has not reacted is 1.18
mole (2-0.82) while that of ethanol is 0.18 (1-0.82). The Equilibrium constant will
therefore be
Kc (T )= [ 0.82 ]1 [ 0.82 ]1
[ 1.18 ]1
[ 0.8 ]1 =3.17 ,no units .
Question 4

Chemistry Assignment 4
a) i)
PbCl2 (s)  Pb2+(aq) + 2Cl-(aq)
ii)
Dynamic equilibrium. The reaction is reversible i.e. the association of the lead and chloride irons
will result in the formation of lead chloride in solid state (Evans & Lewis, 2018).
iii)
Ksp =¿ ¿
The units will be mol3 dm-9
iv)
Mole ratio of Chloride ions to lead ions is 2:1. Therefore, concentration of lead ions will be
0.0317
2 =0.01585 mol /dm3
Since mole ratio between the solid lead and lead ions is 1:1 hence have similar concentration.
Ksp = [ 0.01585 ]1 [ 0.0317 ] 2=1. 59× 10−5 mol3 dm−9
b) i)
Increasing the concentration of chloride ions will shift the reaction to the left. The rate of reverse
reaction will be faster.
ii)

Chemistry Assignment 5
Solid deposits will be seen. The shifting of the reaction to the left means that more of solid lead
chloride will be formed to counteract the increase in the concentration of chloride ions.
iii)
Ksp =1.59× 10−5 mol3 dm−9
Ksp =¿ ¿
P b2+ ¿=0.00637mol dm−3 ¿
Question 5
For partial pressure, the equilibrium constant will be given by
K p = [ C2 H4 Cl2 ] 1
[ C2 H4 ] 1
[ Cl2 ]
1
From the formula above, we can see that increasing the pressure of the reactants, more of the
product will be formed as Kp remains unchanged. There are two molecules producing one (Reno,
2015). Increasing pressure will result increase in forward reaction as only one molecule is
formed hence a reduction in pressure. The system will try to reduce the added pressure by
forming more of the product.
The reaction is exothermic, meaning that heat is evolved. Reducing the heat would induce the
system to increase the forward reaction which evolves heat, with an aim of re-establishing the
equilibrium (Aryangat, 2016). By reducing heat, one can get more of the product.
Question 6
A. a)
1st Cr is +3, 1st Al is 0, 2nd Al is +3, 2nd Cr is 0.
b)
Al is being oxidized.
c)
Al is the reducing agent.
d)
Not disproportionate
B. a)

Chemistry Assignment 6
1st Fe is +3, 1st C is 0, 2nd Fe is 0 and 2nd C is +4.
b)
Carbon atom is being oxidized.
c)
Carbon is the reducing agent.
d)
Not disproportionate.
C. a)
1st Bromine is -7, 2nd Br is -1, 1st H is +1, 2nd Br is 0 and 3rd H is +1.
b)
Bromine ions are being oxidized.
c)
Bromine ion is the reducing agent.
d)
Not disproportionate
D. a)
1st Sulphur is +2, 2nd Sulphur is +4 and 3rd Sulphur is 0.
b)
Sulphur is being oxidized from +2 in S2O32- to +4 in HSO3- .
c)
Some of the Sulphur atoms in S2O32- act as reducing agent and they are themselves oxidized.
d)
This is a disproportionate reaction. Sulphur is simultaneously oxidized and reduced. Sulphur is
being oxidized from +2 in S2O32- to +4 in HSO3- and reduced to 0 in S.
Question 7
a) i)
Fe2+(aq) → Fe3+(aq) + e-
ii)

Chemistry Assignment 7
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7 H2O(aq)
b)
Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) - → 2Cr3+(aq) + 6Fe3+(aq) + 7 H2O(aq) (Wolfe, 2017).
References
Aryangat, A., 2016. The MCAT Chemistry Book. 7th ed. Los Angeles: Nova Press.
Evans, W. & Lewis, R., 2018. Chemistry. 5th ed. London: Palgrave.
Reno, S., 2015. Chemistry: LebreTexts. [Online]
Available at:
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/
Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/
Chemical_Equilibria/Calculating_an_Equilibrium_Constant%2C_Kp
%2C_with_Partial_Pressures
[Accessed 24 March 2019].
Toh, C. S., 2013. A-Level Study Guide Chemistry (Higher 1). 1st ed. Singapore: Step-by-Step
International.
Verma, N. K., Chandigarh & Verma, M., 2016. Comprehensive Science and Technology
Chemistry. Ist ed. New Delhi: Laxmi Publications (P) Ltd..
Wolfe, E., 2017. Medium: 7 Simple Steps to Balancing Redox Reactions. [Online]
Available at: https://medium.com/countdown-education/7-simple-steps-to-balancing-redox-
reactions-dcf1b843ed1a
[Accessed 24 March 2019].