Nonionic equilibrium
(General questions)
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Q: 1 2 3
4 5
6 7 8 9
10 11

1

For each of the following
reactions write an algebraic expression for the equilibrium constants
K_{p} and K_{c} and give the relationship
between K_{p} and K_{c}.
(1.)
CH_{4}(g) + Cl_{2}(g)
CH_{3}Cl(g) + HCl(g)
(2.)
2NO(g) + O_{2}(g) 2NO_{2}(g)
(3.)
4HCl(g) + O_{2}(g) 2Cl_{2}(g)
+ 2H_{2}O(g)
(4.)
2Pb(NO_{3})_{2}(s)
2PbO(s) + 4NO_{2}(g) + O_{2}(g)
(5.)
(NH_{4})_{2}CO_{3}(s)
NH_{3}(g) + NH_{4}HCO_{3}(s)
(6.)
NH_{4}HCO_{3}(s)
NH_{3}(g) + H_{2}O(g) + CO_{2}(g)
(7.)
CuO(s) + SO_{3}(g) CuSO_{4}(s)


2 
Give an expression for K_{c} for each of the following
reactions.
(1.)
2SO_{2}(g) + O_{2}(g)
2SO_{3}(g)
(2.)
SO_{2}(g) + ½O_{2}(g)
SO_{3}(g)
(3.)
2SO_{3}(g) 2SO_{2}(g)
+ O_{2}(g)
(4.)
CaCO_{3}(s) CaO(s) +
CO_{2}(g)


3 
When some nitrogen oxide, oxygen and nitrogen dioxide gases are
injected into an evacuated container at a certain temperature the
equilibrium 2NO(g) + O_{2}(g)
2NO_{2}(g) is established, for which DH
= 5.66 kJ mol^{1}.
Predict how the following
disturbances will affect the concentration of nitrogen dioxide gas
in the vessel at equilibrium. (In (1.)
to (4.)
the volume of the container is fixed.) Write "increase", "decrease"
or "no change".
(1.)
Increasing the temperature
(2.)
Removing nitrogen(II) oxide gas from the vessel
(3.)
Adding a catalyst
(4.)
Adding oxygen gas to the vessel
(5.)
Compressing the vessel to a smaller volume


4 
At a suitable temperature the following gas equilibrium is established
in a container of fixed volume:
N_{2}(g) + O_{2}(g)
2NO(g)
DH
for the forward reaction is +180 kJ mol^{1}.
(1.)
Find an expression for the equilibrium constant for the reaction.
(2.)
State the effect upon the equilibrium of each of the following ("L
to R", "R to L", or "no effect"):
(a)
Increased temperature
(b)
Decreased pressure
(c)
Increased concentration of oxygen
(d)
Increased concentration of nitrogen(II) oxide
(e)
Presence of a catalyst
(f)
Addition of argon at constant volume


5 
For the equilibrium
H_{2}(g) + I_{2}(g)
2HI(g), DH
= 13 kJ mol^{1} at 715K.
At 750 K, is the equilibrium
constant "greater", "unchanged" or "smaller"?


6 
When hydrogen iodide (1.000 mole) in a sealed 1.00 litre flask
is maintained at a temperature of 225 °C,
it decomposes until hydrogen (0.182 mole) is formed. What is the
value of Kc for the reaction:
2HI(g)
H_{2}(g) + I_{2}(g) at 225 °C ?


7 
For the equilibrium 2SO_{2}(g) + O_{2}(g)
2SO_{3}(g) K_{c }= 800 M^{1 }at
525 °C. Calculate the following
(at 525 °C):
(1.)
K_{c} and K_{p} for the equilibrium
SO_{2}(g) + ½O_{2}(g)
SO_{3}(g)
(2.)
K_{p} for the equilibrium SO_{3}(g)
SO_{2}(g) + ½O_{2}(g)


8 
In a sealed container, the reaction:
CO(g) + Cl_{2}(g)
COCl_{2}(g)
was allowed to come to
equilibrium at a certain temperature. The concentrations of carbon
monoxide and carbonyl chloride gases in the mixture were respectively
0.30 M and 0.80 M. If the concentration equilibrium constant is
13.3, find the concentration of chlorine gas in the mixture.


9 
At 294 K, Kc for
the dissociation of dinitrogen tetroxide,
N_{2}O_{4}(g)
2NO_{2}(g)
is 0.48. Calculate the concentration of nitrogen dioxide in equilibrium
with 0.36 M dinitrogen tetroxide at 294 K.


10 
At 600 °C, K_{c}
for the reaction 2SO_{2}(g) + O_{2}(g)
2SO_{3}(g) is 4.5. What amount of sulfur trioxide was placed
in a 1.0 litre reaction vessel if, at equilibrium at 600 °C,
the mixture arising from the above reaction was found to contain
oxygen gas (0.200 mole)?


11 
O_{2}(g) + 2NO(g)
2NO_{2}(g) (K_{c1})
2NO(g) + Cl_{2}(g)
2NOCl(g) (K_{c2})
NO_{2}(g) + ½Cl_{2}(g)
NOCl(g) + ½O_{2} (K_{c3})
Express K_{c3}
in terms of K_{c1} and K_{c2}.



Nonionic equilibrium (General answers)


1

K_{p}
is the equilibrium constant defined in terms of partial pressures,
and is related to K_{c} by the relationship K_{p}
= K_{c}(RT)^{Dn
gas}, where
R = 0.0821 L atm K^{1} mol^{1}, T
is temperature in K, and Dn gas is the change in no. of moles
of gas over the course of reaction (Dn
gas = (no. mol. product gases  no. mol recatant gases)).
(1.)
CH_{4}(g) + Cl_{2}(g)
CH_{3}Cl + HCl
K_{p} = pCH_{3}Cl . pHCl / pCH_{4}
. pCl_{2}
K_{c} = [CH_{3}Cl] [HCl] / [CH_{4}]
[Cl_{2}]
K_{p} = K_{c}. In this example
K_{p} is equal to K_{c} because there
is no change in the no. moles gas from reactants to products.
(2.)
2NO(g) + O_{2}(g) 2NO_{2}(g)
K_{p} = p^{2}NO_{2} / p^{2}NO
. pO_{2}
K_{c} = [NO_{2}]^{2} / [NO]^{2}
[O_{2}]
K_{p} = K_{c}(RT)^{1}
(3.)
4HCl(g) + O_{2}(g) 2Cl_{2}(g)
+ 2H_{2}O(g)
K_{p} = p^{2}Cl_{2 }. p^{2}H_{2}O
/ p^{4}HCl . pO_{2}
K_{c} = [Cl_{2}]^{2} [H_{2}O]^{2}
/ [HCl]^{4} [O_{2}]
K_{p} = K_{c}(RT)^{1}
(4.)
2Pb(NO_{3})_{2}(s)
2PbO(s) + 4NO_{2}(g) + O_{2}(g)
Note that solids and pure liquids do not appear as terms in expressions
for K:
K_{p} = p^{4}NO_{2} . pO_{2}
K_{c} = [NO_{2}]^{4} [O_{2}]
K_{p} = K_{c}(RT)^{5}
(5.)
(NH_{4})_{2}CO_{3}(s)
NH_{3}(g) + NH_{4}HCO_{3}(s)
K_{p} = pNH_{3}
K_{c} = [NH_{3}]
K_{p} = K_{c}RT
(6.)
NH_{4}HCO_{3}(s)
NH_{3}(g) + H_{2}O(g) + CO_{2}(g)
K_{p} = pNH_{3} . pH_{2}O . pCO_{2}
K_{c} = [NH_{3}] [H_{2}O] [CO_{2}]
K_{p} = K_{c}(RT)^{3}
(7.)
CuO(s) + SO_{3}(g) CuSO_{4}(s)
K_{p} = 1 / pSO_{3} = p^{1}SO_{3}
K_{c} = [SO_{3}]^{1}
K_{p} = K_{c}(RT)^{1}


2 
Note that (1.)
and (2.)
are different equations describing the
same reaction, and that the expression for K_{c}
varies according to the way the equation is written. Also
note that solids and pure liquids do not appear as terms in expressions
for K.
(1.)
K_{c} = [SO_{3}]^{2} / [SO_{2}]^{2}
[O_{2}]
(2.)
K_{c} = [SO_{3}] / [SO_{2}] [O_{2}]^{½}
(3.)
K_{c} = [SO_{2}]^{2} [O_{2}]
/ [SO_{3}]^{2}
(4.)
K_{c} = [CO_{2}]


3 
2NO(g) + O_{2}(g)
2NO_{2}(g) (DH
= 5.66 kJ mol^{1})
(1.)
The negative DH value indicated
the reaction is exothermic. As a result, increasing the temperature
will favour the reactants, leading to a decrease in concentration
of NO_{2}.
(2.)
Upon removal of NO the system will reequilibrate in such a way
as to replace the NO just removed. This will lead to a decrease
in concentration of NO_{2}.
(3.)
Adding a catalyst will have no effect on the equilibrium concentration
of NO_{2}. Catalysts decrease the time taken to reach equilibrium,
but do not affect the equilibrium concentrations once equilibrium
is reached.
(4.)
Upon addition of O_{2} the system will reequilibrate in
such a way as to use the O_{2} just added. This will lead
to an increase in concentration of NO_{2}.
(5.)
Since there are fewer moles of gas on the product side (2 mol) than
the reactant side (3 mol), increasing the pressure by reducing the
volume will move the reaction towards the side which contains fewer
moles of gas. This will lead to an increase in concentration of
NO_{2} with increased applied pressure.


4

(1.)
K_{c} = [NO]^{2} / [N_{2}] [O_{2}]
(2.)
2NO(g) + O_{2}(g)
2NO_{2}(g) (DH = +180
kJ mol^{1})
(a)
Since the reaction is endothermic (positive DH
value), increasing the temperature will make the reaction move from
L ® R.
(b)
Since there are equal amounts of gas on each side of the equation,
changing the pressure will have no effect on the equilibrium.
(c)
Increasing the amount of oxygen will make the reaction move from
L ® R.
(d)
Increasing the concentration of NO will move the reaction
from R ® L.
(e)
The addition of a catalyst will not affect the direction of reaction
(ie the equilibrium concentrations will be the same as no catalyst),
it will only make the system equilibrate faster.
(f)
Addition of an inert gas (like argon) at constant volume has no
effect on the equilibrium position since the partial pressures
of the gaseous components remain the same.


5 
H_{2}(g) + I_{2}(g)
2HI(g) (DH
= 13 kJ mol^{1} at 715K)
The equilibrium constant
can be expressed in terms of forward and reverse reaction constants:
K = k_{forward} / k_{reverse}.
Since the forward reaction is exothermic, an increase in temperature
will favour the reverse reaction, so k_{forward}
will decrease, and k_{reverse} will increase. From
the above equation, this will lead to a smaller value of K.


6 
2HI(g)
H_{2}(g) + I_{2}(g)
K_{c}
= [H_{2}] [I_{2}] / [HI]^{2}.
From the above reaction
equation, 1 mole of HI reacting will produce 0.5 mole H_{2}
and 0.5 mole I_{2}. So if 0.182 mol H_{2} is formed
at equlibrium, 0.182 mol I_{2} must also be present.
Since 2 moles HI react
to give 1 mole H_{2}, if there are 0.182 mol H_{2}
present, then 2 x 0.182 mol = 0.364 mol HI must have reacted. Therefore
the number of moles HI remaining at equilibrium = no. mol HI reactant
 no. mol HI consumed = 1.000 mol HI  0.364 mol HI = 0.636 mol.
Substituting these values
into the above expression for K_{c} we obtain
K_{c}
= (0.182 mol L^{1}) x (0.182 mol L^{1}) / (0.636
mol L^{1})^{2} = 0.0819.
Note that in this case
K_{c} is unitless. Referring to the K_{c}
expression it can be seen that units of mol^{2} L^{2}
appear on both numerator and demoninator, hence cancelling and leaving
the resultant K_{c} value unitless.


7 
2SO_{2}(g) +
O_{2}(g)
2SO_{3}(g) (K_{c }= 800 L mol^{1}
at 525 °C)
...Equation 1
(1.)
SO_{2}(g) + ½O_{2}(g)
SO_{3}(g) ...Equation 2
By considering the expressions of K_{c} for each
of the two equations, the relationship between them can be determined:
K_{c} (Equation 1) = [SO_{3}]^{2}
/ [SO_{2}]^{2} [O_{2}]
K_{c} (Equation 2) = [SO_{3}] / [SO_{2}]
[O_{2}]^{½}
It is evident that the expression for K_{c} (Equation
2) is the square root of the expression for K_{c}
(Equation 1). Therefore the numerical value of K_{c}
(Equation 2) = (800 L mol^{1})^{½} = 28.3 L^{½}
mol ^{½}.
This is the basis of the general rule for converting K_{c}
values between equations which differ only by a common factor: If
reaction equation with an equilibrium constant K_{c}
is multiplied through by a factor x, the resultant equilibrium
constant is (K_{c})^{x}. So in this
example, where Equation 2 is Equation 1 multiplied through by the
factor 0.5, the K_{c} value for Equation 2 is the
square root of K_{c} for Equation 1.
K_{p} = K_{c}(RT)^{Dn
gas}
In this case R = 0.0821 L atm K^{1} mol^{1},
T = (273 + 525) = 798 K, and Dn
gas = 0.5. This gives K_{p} = 28.3 L^{½}
mol ^{½} x (0.0821 L atm K^{1} mol^{1}
x 798 K) ^{½} = 3.49 atm^{ ½}. Note that the units
in this expression cancel to give K_{p} in atm. From
this a general rule can be established  if Dn
gas = x, then the units of K_{p} will be atm^{x}
(\ if Dn
gas = 0, K_{p} will be unitless).
(2.)
SO_{3}(g) SO_{2}(g)
+ ½O_{2}(g) ...Equation 3
This is the reverse of Equation 2. We can compare the expressions
of K_{p} for the two equations to determine the relationship
between them:
K_{p} (Equation 2) = pSO_{3} / pSO_{2}
p^{½}O_{2}
K_{p} (Equation 3) = pSO_{2} p^{½}O_{2}
/ pSO_{3}
This shows that the expression for K_{p} (Equation
3) is the inverse of the expression for K_{p} (Equation
2). The K_{p} value for Equation 3 will therefore
be the inverse of the K_{p} value of Equation 2.
ie K_{p} (Equation 3) = (K_{p} (Equation
2))^{1} = (3.49 atm ^{½} ) ^{1} = 0.289
atm^{½}. Note here that the units for K_{p}
are also inverted.


8 
CO(g) + Cl_{2}(g)
COCl_{2}(g)
[CO] = 0.30 mol L^{1}
[COCl_{2}] =
0.80 mol L^{1}
K_{c}
= 13.3 L mol^{1} = [COCl_{2}] / [CO] [Cl_{2}]
\[Cl_{2}]
= [COCl_{2}] / [CO] K_{c} = 0.80 mol L^{1}
/ 0.30 mol L^{1} x 13.3 L mol^{1} = 0.20 mol L^{1}


9 
N_{2}O_{4}
2NO_{2} (K_{c} = 0.48 mol L^{1}
at 294 K)
K_{c}
= [NO_{2}]^{2} / [N_{2}O_{4}]
\
[NO_{2}] = (K_{c} [N_{2}O_{4}])^{½}
= (0.48 mol L^{1} x 0.36 mol L^{1})^{½}
= 0.42 mol L^{1}.


10 
2SO_{2} + O_{2}
2SO_{3} K_{c} = 4.5 L mol^{1} at
600 °C
K_{c}
= [SO_{3}]^{2} / [SO_{2}]^{2} [O_{2}]
[O_{2}] = 0.200
mol L^{1}
\
[SO_{2}] = 0.400 mol L^{1}
[SO_{3}] = (K_{c}
[SO_{2}]^{2} [O_{2}])^{½} = (4.5
L mol^{1} x 0.400^{2} mol^{2} L^{2}
x 0.200 mol L^{1}) = (0.144 mol^{2} L^{2})^{½}
= 0.38 mol L^{1} at equilibrium.
\ [SO_{3}]
before equilibrium = amount of SO_{3} present at equilibrium
+ amount of SO_{3} converted into SO_{2} and O_{2}
= 0.38 mol L^{1} + 0.400 mol L^{1} = 0.78 mol
L^{1}. Volume of reaction vessel = 1L, \ 0.78 mol SO_{3}
was placed into the reaction vessel.


11 
K_{c1}
= [NO_{2}]^{2} / [NO]^{2} [O_{2}]
K_{c2}
= [NOCl]^{2} / [Cl_{2}] [NO]^{2}
K_{c3}
= [NOCl]^{2} [O_{2}]^{½} / [NO_{2}]
[Cl_{2}]^{½}
Rearranging K_{c1}
and K_{c2}:
K_{c1}^{½}
= [NO] [O_{2}]^{½} / [NO_{2}]
K_{c2}^{½}
= [NOCl] / [Cl_{2}]^{½} [NO]
K_{c1}^{½}
x K_{c2}^{½} = [NO] [O_{2}]^{½}
[NOCl] / [Cl_{2}]^{½} [NO] [NO_{2}] = [O_{2}]^{½}
[NOCl] / [Cl_{2}]^{½} [NO_{2}] = K_{c3}.


