Chapter 7 Equilibrium Solutions
Question - 21 : - Equilibrium constant, Kc for the reaction
at500 K is 0.061.
At a particular time, the analysis showsthat composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction atequilibrium? If not in which direction does the reaction tend to proceed toreach equilibrium?
Answer - 21 : -
Thegiven reaction is:
Now,we know that,
Since
,the reaction is not at equilibrium.
Since 
,the reaction will proceed in the forward direction to reach equilibrium.
Question - 22 : - Brominemonochloride, BrCl decomposes into bromine and chlorine and reaches theequilibrium:
for which Kc= 32 at 500 K. Ifinitially pure BrCl is present at a concentration of 3.3 × 10–3 molL–1, what is its molarconcentration in the mixture at equilibrium?
Answer - 22 : -
Let the amount of bromine and chlorineformed at equilibrium be x. The given reaction is:
Now,we can write,
Therefore,at equilibrium,
Question - 23 : - At 1127 K and 1 atm pressure, a gaseousmixture of CO and CO2 in equilibrium with solid carbon has90.55% CO by mass
Calculate Kc for this reaction atthe above temperature.
Answer - 23 : -
Letthe total mass of the gaseous mixture be 100 g.
Massof CO = 90.55 g
And, mass of CO2 = (100 – 90.55) =9.45 g
Now,number of moles of CO, 
Number of moles of CO2, 
Partialpressure of CO,
Partial pressure of CO2,
Forthe given reaction,
Δn = 2 – 1 = 1
Weknow that,
Calculate a) ΔG°and b) theequilibrium constant for the formation of NO2 from NO and O2 at 298 K
where ΔfG° (NO2) = 52.0 kJ/mol
ΔfG° (NO) = 87.0 kJ/mol
ΔfG° (O2) = 0 kJ/mol
Answer - 24 : -
(a) For the given reaction,
ΔG° = ΔG°( Products) – ΔG°(Reactants)
ΔG° = 52.0 – {87.0 + 0}
= – 35.0 kJ mol–1
(b) We know that,
ΔG° = RT log Kc
ΔG° = 2.303 RT log Kc
Hence,the equilibrium constant for the given reaction Kc is 1.36 × 106
Question - 25 : - Doesthe number of moles of reaction products increase, decrease or remain same wheneach of the following equilibria is subjected to a decrease in pressure byincreasing the volume?
(a) 
(b) 
(c) 
Answer - 25 : -
(a) The number of moles ofreaction products will increase. According to Le Chatelier’s principle, ifpressure is decreased, then the equilibrium shifts in the direction in whichthe number of moles of gases is more. In the given reaction, the number ofmoles of gaseous products is more than that of gaseous reactants. Thus, thereaction will proceed in the forward direction. As a result, the number ofmoles of reaction products will increase.
(b) The number of moles ofreaction products will decrease.
(c) The number of moles of reaction productsremains the same.
Question - 26 : - Whichof the following reactions will get affected by increasing the pressure?
Also,mention whether change will cause the reaction to go into forward or backwarddirection.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Answer - 26 : -
Thereactions given in (i), (iii), (iv), (v), and (vi) will get affected byincreasing the pressure.
Thereaction given in (iv) will proceed in the forward direction because the numberof moles of gaseous reactants is more than that of gaseous products.
Thereactions given in (i), (iii), (v), and (vi) will shift in the backwarddirection because the number of moles of gaseous reactants is less than that ofgaseous products.
Question - 27 : - The equilibrium constant for the followingreaction is 1.6 ×105 at 1024 K.
Findthe equilibrium pressure of all gases if 10.0 bar of HBr is introduced into asealed container at 1024 K.
Answer - 27 : -
Given,
for the reaction i.e., 
Therefore, for the reaction
theequilibrium constant will be,
Now, let p be the pressureof both H2 and Br2 at equilibrium.
Now,we can write,
Therefore,at equilibrium,
Question - 28 : - Dihydrogengas is obtained from natural gas by partial oxidation with steam as perfollowing endothermic reaction:
(a) Write as expression for Kp for the above reaction.
(b) How will the values of Kp and composition ofequilibrium mixture be affected by
(i)Increasing the pressure
(ii)Increasing the temperature
(iii)Using a catalyst?
Answer - 28 : -
(a) For the given reaction,
(b) (i) According to LeChatelier’s principle, the equilibrium will shift in the backward direction.
(ii)According to Le Chatelier’s principle, as the reaction is endothermic, theequilibrium will shift in the forward direction.
(iii)The equilibrium of the reaction is not affected by the presence of a catalyst.A catalyst only increases the rate of a reaction. Thus, equilibrium will beattained quickly.
Question - 29 : - Describethe effect of:
a) Addition of H2
b) Addition of CH3OH
c)Removal of CO
d) Removal of CH3OH
onthe equilibrium of the reaction:
Answer - 29 : -
(a) According to LeChatelier’s principle, on addition of H2, the equilibrium of thegiven reaction will shift in the forward direction.
(b) On addition of CH3OH, the equilibrium willshift in the backward direction.
(c) On removing CO, theequilibrium will shift in the backward direction.
(d) On removing CH3OH, the equilibrium willshift in the forward direction.
Question - 30 : - At 473 K, equilibrium constant Kc for decomposition ofphosphorus pentachloride, PCl5 is 8.3 ×10-3. If decomposition isdepicted as,
ΔrH° = 124.0 kJmol–1
a) Write an expression for Kc for the reaction.
b) What is the value of Kc for the reversereaction at the same temperature?
c)What would be the effect on Kc if(i) more PCl5 isadded (ii) pressure is increased? (iii) The temperature is increased?
Answer - 30 : -
(a) 
(b) Value of Kc for the reversereaction at the same temperature is:
(c) (i) Kc would remain the samebecause in this case, the temperature remains the same.
(ii) Kc is constant at constanttemperature. Thus, in this case, Kc would not change.
(iii)In an endothermic reaction, the value of Kc increases with an increasein temperature. Since the given reaction in an endothermic reaction, the valueof Kc will increase if the temperature isincreased.