Chapter 7 Equilibrium Solutions
Question - 11 : - A sample of HI(g) is placed in flaskat a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. Whatis Kp for the given equilibrium?
Answer - 11 : -
Theinitial concentration of HI is 0.2 atm. At equilibrium, it has a partialpressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 – 0.04= 0.16. The given reaction is:
Therefore,
Hence,the value of Kp for the given equilibrium is 4.0.
Question - 12 : - A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vesselat 500 K. At this temperature, the equilibrium constant, Kc for the reaction 
Isthe reaction mixture at equilibrium? If not, what is the direction of the netreaction?
Answer - 12 : -
Thegiven reaction is:
Now, reaction quotient Qc is:
Since
,the reaction mixture is not at equilibrium.
Again, 
. Hence, the reaction will proceed in thereverse direction.
Question - 13 : - Theequilibrium constant expression for a gas reaction is,
Writethe balanced chemical equation corresponding to this expression.
Answer - 13 : -
Thebalanced chemical equation corresponding to the given expression can be writtenas:
Question - 14 : - One mole of H2O and one mole of CO aretaken in 10 L vessel and heated to
725K. At equilibrium 40% of water (by mass) reacts with CO according to theequation,Calculatethe equilibrium constant for the reaction.
Answer - 14 : -
Thegiven reaction is:
Therefore,the equilibrium constant for the reaction,
Question - 15 : - At700 K, equilibrium constant for the reaction
is 54.8. If 0.5 molL–1 of HI(g) is present atequilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that weinitially started with HI(g) and allowed it toreach equilibrium at 700 K?
Answer - 15 : - It is given that equilibrium constant
for the reaction
is54.8. Therefore,at equilibrium, the equilibrium constant 
forthe reaction
will be 
Let the concentrations of hydrogen andiodine at equilibrium be x molL–1
.
Hence, at equilibrium,
Whatis the equilibrium concentration of each of the substances in the equilibriumwhen the initial concentration of ICl was 0.78 M?
2 ICl(g) ⇌ I2(g) + Cl2(g) ; KC = 0.14
Answer - 16 : -
Thegiven reaction is:
2 ICl(g) ⇌ I2(g) + Cl2(g) Initialconc. 0.78 M 0 0 Atequilibrium (0.78 – 2x)M x M x M
Hence,at equilibrium,
[ICl]=[I2] = 0.167 M[ICl] = (0.78 -2×0.167)M =0.446 M
Question - 17 : - Kp = 0.04 atm at 899 Kfor the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed ina flask at 4.0 atm pressure and allowed to come to equilibrium?
Answer - 17 : -
Let p be the pressureexerted by ethene and hydrogen gas (each) at equilibrium.
Now,according to the reaction,
Wecan write,
Hence,at equilibrium,
Question - 18 : - Ethylacetate is formed by the reaction between ethanol and acetic acid and theequilibrium is represented as:
(i) Write the concentration ratio (reactionquotient), Qc, for this reaction (note: water is not in excess andis not a solvent in this reaction)
(ii)At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol,there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculatethe equilibrium constant.
(iii)Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining itat 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium beenreached?
Answer - 18 : - (i) Reactionquotient,
(ii) Let the volume of thereaction mixture be V. Also, here we will consider that water is asolvent and is present in excess.
Thegiven reaction is:
Therefore,equilibrium constant for the given reaction is:
(iii) Let the volume ofthe reaction mixture be V.
Therefore,the reaction quotient is,
Since
, equilibrium has not been reached.
A sample of pure PCl5 was introduced intoan evacuated vessel at 473 K. After equilibrium was attained, concentration ofPCl5 was found to be 0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are theconcentrations of PCl3 and Cl2 at equilibrium?
Answer - 19 : -
Let the concentrations of both PCl3 and Cl2 at equilibriumbe x molL–1. The given reaction is:
Nowwe can write the expression for equilibrium as:
Therefore,at equilibrium,
Question - 20 : - One of the reactions that takes place inproducing steel from iron ore is the reduction of iron (II) oxide by carbonmonoxide to give iron metal and CO2.
FeO(s) + CO (g) 
Fe (s) + CO2 (g); Kp= 0.265 at 1050 K.
Whatare the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressuresare: pCO =1.4 atm and 
= 0.80 atm?
Answer - 20 : -
Forthe given reaction,
Since
,the reaction will proceed in the backward direction.
Therefore, we can say that the pressure ofCO will increase while the pressure of CO2 will decrease.
Now, let the increase in pressure of CO =decrease in pressure of CO2 be p.
Then,we can write,
Therefore, equilibrium partial of 
And, equilibrium partial pressure of 