Chemical Equilibrium(12th Grade > Chemistry ) Questions and Answers
Question 1. Ammonia under a pressure of 15 atm at 27∘C is heated to
347∘C in a closed vessel in the presence of a catalyst. Under the conditions, NH3 is partially decomposed according to the equation,
2NH3 ⇋ N2 + 3H2 .The vessel is such that the volume remains effectively constant whereas pressure increases to
50 atm. Calculate the percentage of NH3 actually decomposed.
347∘C in a closed vessel in the presence of a catalyst. Under the conditions, NH3 is partially decomposed according to the equation,
2NH3 ⇋ N2 + 3H2 .The vessel is such that the volume remains effectively constant whereas pressure increases to
50 atm. Calculate the percentage of NH3 actually decomposed.
Explanation:-
Answer: Option B. -> 61.3%:
B
2NH3⇋N2+3H2
Initialmolea00
Mole at equilibrium (a−2x)x3x
Initial pressure of NH3 of a mole = 15 atm 27∘C
The pressure of 'a' mole of NH3=patmat347∘C
∴15300=p620
∴ p=31 atm
At constant volume and at 347∘C,mole α pressure
∴a+2xa=5031
∴x=1962
∴%ofNH3 decomposed = 2xa×100
=2×19a62×a×100=61.33%
Explanation:-
Answer: Option C. -> 2.67 × 104:
C
Partial pressure of I atoms
(pl)=40×105/100Pa=0.4×105Pa
Partial pressure of l2(p2)=60×105/100Pa=0.6×105Pa
Kp=(p1)2/p2=(0.4×105)2/(0.6×105)=2.67×104
Explanation:-
Answer: Option C. -> 0.05:
C
A+2B⇋2CInitialmoles232Molarconc.2/2=1mol/L3/2=1.5mol/L2/2=1mol/LAtEqui.Conc1+0.25=1.25mol/L1.5+0.5=2mol/L0.5mol/L
K=(0.5)2/(1.25×(2)2)=0.05 Note that concentration of C has decreased, i.e., to attain equilibrium, reaction has shifted backwards. Decrease in conc. of C will be 0.5mol/L and hence increase in concentration of B will be 0.5mol/L and that of A will be 0.25mol/L
Explanation:-
Answer: Option D. -> A is false but R is true.:
D
Equilibrium is possible only in a closed vessel. Hence A is false.
Boiling occurs at a constant temperature at a given pressure. Hence R is true.
Explanation:-
Answer: Option B. -> High temperature will favor forward reaction.:
B
a) Increasing pressure favors the reaction in the direction in which no. of moles decreases. As number of moles is decreasing in forward direction, hence increasing pressure will favor the production of SO3
b) Increasing temperature favors endothermic reaction. Here forward reaction is exothermic; hence high temperature will not favor forward reaction.
c) k[SO3]Hence it will decrease on reducing conc. of SO3
d) Increasing the volume reduces the pressure. Hence it will favor backward reaction in this case as number of moles is increasing in backward direction.
Explanation:-
Answer: Option A. -> increasing the temperature:
A
Since backward reaction is endothermic, increase in temperature favors backward reaction
Question 9. An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH4HS decomposition at this temperature is
Explanation:-
Answer: Option D. -> 0.11:
D
NH4HS(s)⇋NH3(g)+H2S(g)
Initially,a0.5atm0atequilibrium......0.5+xx
Total Pressure at eqilibrium =0.5+x+x=0.84
⇒x=0.17
Kp=PNH3(g)×PH2S(g)=(0.5+x)x=(0.5+0.17)×0.17=0.1139atm2
Explanation:-
Answer: Option C. -> Introducing an inert gas at constant pressure:
C
a) Introducing an inert gas at constant volume does not change the partial pressure of gases. Hence it does not affect the equilibrium.
b) Introducing chlorine gas at constant volume will favor the backward reaction.
c) Introducing inert gas at constant pressure means, one has to increase the volume of the container given temperature is also not changing.
PVmustincrease=nincreasingRT
Increasing container volume will favor the forward reaction. Henceoption (c) is correct.
d) Decreasing the container volume will favor backward reaction.