Lakshya Education MCQs

Question: For a hypothetical reaction

4A + 5B 4P+6Q,

The equilibrium constant  Kc has units
Options:
A.molL−1
B.mol−1L
C.(molL−1)−2
D.unit less   
Answer: Option A
: A


Kc=[P]4[Q]6/([A]4[B]5)=(molL1)10/(molL1)9/=molL1

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Question 1. Equimolar concentrations of H2 and I2 are heated to equilibrium in a 2litre flask. At equilibrium, the forward and backward rate constants are found to be equal. What percentage of initial concentration of H2has reacted at equilibrium?
  1.    33.33%
  2.    75%
  3.    50%
  4.    15%
Answer: Option C
: C

Solution:Correct option is c).

H2(g)+l2(g)2HI(g)Initialmoles110Atequilibrium1x1x2xMolarconcentration(mol/L)(1x)/2(1x)/22x/2=x

K=x2/(1x)×(1x)2×2=4x2/(1x)24x2
K=Kf/Kb=1

4x2=1x=12,i.e.,50%
Question 2. For the reaction 2HI (g)  H2(g) + I2 (g), the degree of dissociation α of HI (g) is related to the equilibrium constant (Kp) by the expression
  1.    1+2√Kp2
  2.    √1+2Kp√2
  3.    √2Kp√2+2Kp
  4.    2√Kp1+2√Kp
Answer: Option D
: D



2HI(g)H2(g)+l2(g)Atequilibrium1αα/2α/2
Total moles at equilibrium =1
PHI=(1α)×P,PH2=α/2×PPl2=α/2×P
Kp=PH2×Pl2/(PHI)2

Substituting above values we get,

Kp=a2/[4×(1α2)]
solving,α/1α=2(Kp)12
α=2Kp1+2Kp
Question 3. The equilibrium constant for the reaction 2SO2 + O2  2 SO3 is 900 atm1 at 800K. A mixture containing SO3 and O2 having initial pressures of 1atm and 2atm is heated to equilibrate. The partial pressure of O2at equilibrium will be
  1.    0.97atm
  2.    0.012atm
  3.    2.01atm
  4.    0.024atm
Answer: Option C
: C

Correct option is c). Considering the backward reaction we have,
2SO32SO2+O2KP=1/900atmInitialpressure1atm02atmPressureatequilibrium1xx2+x/2 Let partial pressure of SO2=p1and partial pressure of O2=p2and partial pressure of SO3=p3
K<p=(p1)2×(p2)/(p3)2=x2(2+x/2)/(1x)2=1/900
AsKpfor this reaction is very small, x<<1.
Taking 2+x/2 as x and (1x) as x and solving for x, we get
X=0.0236
Hence, at equilibrium, partial pressure of SO3=1X=0.9764atm,
Partial pressure of SO2=x=0.0236atm
And Partial pressure of O2=2+x/2=2.0118atm
Question 4. The values of Kp1and KP2 for the reactions

X  Y + Z and A 2B Are in the ratio of 9:1. If the degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ration
  1.    3:1
  2.    1:9
  3.    36:1
  4.    1:1
Answer: Option C
: C


Suppose total pressure at equilibrium for reactions 1 and 2 areP1andP2respectively

XY+ZInitialmoles100Atequilibrium1xxx

Totalmoles=1+α
Partial pressure of X=px=(1x)XP11+x
Py<=(x)XP11+x=Pz
(KP1=PYPZ/Px=x2P1/(1x2)x2P1

A2BhlineInitialmoles10Atequilibrium1x2x

Totalmoles=1+α

PA=(1x)XP21+xPB=(2x)XP21+x

KP2=P2B/PA=4x2P2/(1x2)4x2P2
KP1/KP2=x2P1/4x2P2=9/1
P1/P2=36/1
Question 5. If concentration of OHions in the reaction  Fe(OH)3(s)  Fe3++(aq)+ 3 OH (aq) is decreased by 1/4 times, then equilibrium concentration of Fe3+will increase by
  1.    8 times
  2.    16 times
  3.    64 times
  4.    4 times.
Answer: Option C
: C


K=[Fe3+][OH]3 If concentration of OHion is decreased to 14 th to keep K constant
Concentration of Fe3+ions has to be increased by 64 times.
Question 6. At 30C,Kp for the dissociation reaction:

SO2Cl2 (g) SO2 (g) + Cl2 (g) is 2.9x 102~atm. If the total pressure is 1 atm, the degree of dissociation of SO2Cl2 is (assume 1α2 =1)
  1.    85%
  2.    12%
  3.    17%
  4.    35%
Answer: Option C
: C

Correct option is c).
SO2Cl2(g)SO2(g)+Cl2(g)1001ααα
Total moles = 1+α
Pso2Cl2=1α/1+α,Pso2=α/1+α,Pcl2=α/1+α
Kp=(α/1+α)2/(1α/1+α)=α2/1α2α2
α=Kp=(2.9×102=0.17,i.e.,17%

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