Chemistry Assignment 1

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