Test: Heterogeneous Equilibrium(Chemical Equilibrium)


22 Questions MCQ Test Chemistry Class 11 | Test: Heterogeneous Equilibrium(Chemical Equilibrium)


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This mock test of Test: Heterogeneous Equilibrium(Chemical Equilibrium) for Class 11 helps you for every Class 11 entrance exam. This contains 22 Multiple Choice Questions for Class 11 Test: Heterogeneous Equilibrium(Chemical Equilibrium) (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Heterogeneous Equilibrium(Chemical Equilibrium) quiz give you a good mix of easy questions and tough questions. Class 11 students definitely take this Test: Heterogeneous Equilibrium(Chemical Equilibrium) exercise for a better result in the exam. You can find other Test: Heterogeneous Equilibrium(Chemical Equilibrium) extra questions, long questions & short questions for Class 11 on EduRev as well by searching above.
QUESTION: 1

Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q. Vapour pressure of NH4HS (s)is 20 mm at 25°C, for

NH4HS (s)⇌ NH3(g) + NH3(g) + H2S(g)
Total pressure when NH4HS (s) dissociates at 25°Cin a vessel which already contains H2S (g)at a pressure of 15 mm, is

Solution:

The correct answer is Option A.
   NH4HS   ------->  NH3 + H2S
Let P be the pressure at eq. of NH3 and H2S.
Therefore, Kp = P2
= (20 / 2)2
= 100 mm
= 100
Also, Kp = (15 + P) (P)
100 = 15 P + P2
P2 + 15 P – 100 = 0
P = 5
Total pressure = 15 + 2(P)
= 15 + 2(5)
= 25 mm

QUESTION: 2

Once the equilibrium is reached under given condition:

Solution:

In a chemical reaction, chemical equilibrium is the state in which the forward reaction rate and the reverse reaction rate are equal. The result of this equilibrium is that the concentrations of the reactants and the products do not change. However, just because concentrations aren’t changing does not mean that all chemical reaction has ceased. Just the opposite is true; chemical equilibrium is a dynamic state in which reactants are being converted into products at all times, but at the exact rate that products are being converted back into reactants. The result of such a situation is analogous to a bridge between two cities, where the rate of cars going over the bridge in each direction is exactly equal. The result is that the net number of cars on either side of the bridge does not change.

QUESTION: 3

Ca(HCO3)2 is strongly heated and after equilibrium is attained, temperature changed to 25° C.

Ca(HCO3)2(s)⇌CaO(s) + 2CO2 (g) + H2O(g)
Kp = 36 (pressure taken in atm)
Thus, pressure set up due to CO2 is

Solution:

The reaction is as follow:-
Ca(HCO3)2(s)⇌CaO(s) + 2CO2 (g) + H2O(g)
At 25° C H2O goes in liquid state
Kp = (PCaO)1×(PCO2)2
(PCa(HCO3)2)
Since, Ca(HCO3)2, CaO and H2O are not in gaseous state, so their partial pressure is taken 1.
Putting all values, we have
36 = (PCO2)2 
Or PCO2 = 6 atm

QUESTION: 4

Na2SO4.10H2O dehydrates according to the equation,

Na2SO4· 10H2O (s)⇌ Na2sO4(s) + 10 H2O(g)     

; Kp = 2.56 x 10-20 

What is the pressure of water vapor at equilibrium with a mixture of Na2SO4· 10H2O and Na2sO4

Solution:
QUESTION: 5

Ca(HCO3)2 decomposes as,

Ca (HCO3)2(s) ⇌ CaCO3(s) + H2O(g) + CO2(g)
Equilibrium pressure is found to be 0.12 bar. What is pco2 if the reaction mixture also contains H2O(g)at 0.20 bar?

Solution:

QUESTION: 6

For the following equilibrium,

Solution:
QUESTION: 7

Assume following equilibria when total pressure set up in each are equal to 1 atm, and equilibrium constant (Kp) as K1; K2 and K3


Thus,

Solution:

The correct answer is option C
CaCO3 ​→ CaO + CO2​
Kp​ = k1 ​= Pco2​​
total pressure of container P
k1​ = p
NH4​HS → NH3 ​+ H2​S
PNH3​​ = PH2​S ​= P0​
P0​ + P0​ = p (total pressure)
P0 ​= p/2
k2​ = kp ​= [PNH3​​][PH2​s​] p24
NH2​CoNH2 ​→ 2NH3 ​+ CO2​
PNH3​​ = 2P0​        PCO2​ ​= P0​
2P0​ + P0 ​= P

QUESTION: 8

For the equilibrium,

at 1000 K. If at equilibrium pCO = 10 then total pressure at equilibrium is 

Solution:

C(s) + CO2(g) <=========> 2CO(g)
Kp = pCO2/pCO2
GIven Kp = 63 and pCO = 10pCO2
Putting the value of pCO in above equation,
63 = 100(pCO2)2/pCO2
Or pCO2 = 0.63
pCO = 6.3
Therefore, total pressure = 6.3+0.63 = 6.93 atm

QUESTION: 9

Ammonium carbamate dissociates as,

In a closed vessel containing ammonium carbamate in equilibrium with its vapour, ammonia is added such that partial pressure of NH3 now equals the original total pressure. Thus, ratio of the total pressure to the original pressure is

Solution:

QUESTION: 10

Kc forthe decomposition of NH4HS(s) is 1.8x 10-4 at 25°C.

If the system already contains [NH3] = 0.020 M, then when equilibrium is reached, molar concentration are

Solution:

 NH4HS (s)  ⇋ NH3 (g) + H2S (g)
Initial    1                   -               -
At eqm     1-x                 x+0.02     x
Kc = [NH3][H2S]   (Since NH4HS is solid, we ignore it.)
1.8×10-4    = (x+0.02)(x)
x2+0.02x-1.8×10-4 = 0
Applying quadratic formula; x = -0.02+√{(0.02)2-4×1.8×10-4}
= 0.033-0.020/2 = 0.0065
Therefore, concn of NH3 at equilibrium = x+0.020 = 0.0265
concn of H2S at equilibrium = x = 0.0065
So, option b is correct

QUESTION: 11

Direction (Q. Nos. 11-14) This section contains  a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d)

Passage I

Solid ammonium chloride is in equilibrium with ammonia and hydrogen chloride gases

0.980 g of solid NH4CI is taken in a closed vessel of 1 L capacity and heated to 275° C.

Q. Partial pressure of NH3(g) or HCI (g) at equilibrium is

Solution:

The reaction is as follow:-
NH4Cl(s)  ⇋  NH3(g) + HCl(g)
Kp = (pNH3)(pHCl)
1×10-2 = p2   (SInce reactant dissociates into same ratio, so the partial pressure will be same for both)
100×10-4 = p2
Or p = 10×10-2 = 0.10
So,  the partial pressure of NH3 and HCl are 0.10 atm.

QUESTION: 12

Passage I

Solid ammonium chloride is in equilibrium with ammonia and hydrogen chloride gases

0.980 g of solid NH4CI is taken in a closed vessel of 1 L capacity and heated to 275° C.

Q. Percentage decomposition of the original sample is

Solution:

The state of HCl is given wrong. It will be in gaseous state.
So, the reaction be like;-
NH4Cl(s)  ⇌  NH3(g) + HCl(g)        kp = 1.00×10-2 at 275° C
Kp = kc(RT)2
1.00×10-2 = kc(0.0821×548)2
Or kc = 4.94×10-6
                          NH4Cl(s)  ⇌  NH3(g) + HCl(g)
Initial  1                     -             -
At eqm 1-x                  x            x 
Kc = x2
x = √(4.94×10-6)
=  2.22×10-3
Therefore, NH4Cl dissociated at eqm = 2.22×10-3 × 53.5 = 0.118
%age decomposition = 0.118/0.980×100 = 12.13%

QUESTION: 13

Passage lI

One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and carbon dioxide.

Initial partial pressure
CO(g) = 1.40 atm
CO2(g) = 0.80 atm

Q. Under the given partial pressure, reaction is

Solution:

We know that with increase in pressure on one side, reaction shifts to that side which has less no. of moles(of gaseous species). However the no. of moles are same on both sides. So the REACTION WILL REMAIN IN EQUILIBRIUM.

QUESTION: 14

Passage lI

One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and carbon dioxide.

Initial partial pressure
CO(g) = 1.40 atm
CO2(g) = 0.80 atm

Q. When equilibrium is attained,

Solution:

Initially                1.4 atm                0.80 atm
Qp = pCO2/pCO 0.8/1.4 = 0.571
SInce Qp>Kp ; reaction proceeds in the backward direction. So pressure of CO2 decreases and that of CO increases.
At eqm.               1.4+p atm            0.80-p atm
Kp = pCO2/pCO = 0.80-p/1.4+p
0.265 = 0.80-p/1.4+p
Or p = 0.339 atm
Therefore, partial pressure at eqm, pCO2 = 0.80-0.339 = 0.461 atm
And pCO = 1.4+0.339 = 1.739 atm

*Answer can only contain numeric values
QUESTION: 15

Direction (Q. Nos. 15 and 16) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive)

Hot copper tunnings can be used as an “oxygen getter" for inert gas supplies by slowly passing the gas over the copper tunnings at 650 K.

Q. How many molecules of O2 are left in 1 L of a gas supply after equilibrium has been reached?


Solution:
*Answer can only contain numeric values
QUESTION: 16

At 1000 K, pressure of CO2 in equilibrium with CaCO3 and CaO is equal to 2.105 atm. The equilibrium constant for the reaction,

is 1.9 at the same temperature when pressure are in atm. Solid C, CaO, and CaCO3 are mixed and allowed to come to equilibrium at 1000 K in a closed vessel.

Q. What is the pressure of CO (g)at equilibrium (in atm)?


Solution:

K= (partial pressure of co2/(partial pressure of co2)
since k =1.9
So 1.9 = (partial pressure of co)2/2.105
(partial pressure of co)=2.105×1.9
= 3.99 = 4
(partial pressure of co) =2

QUESTION: 17

Direction (Q. Nos. 17) Choice the correct combination of elements and column I and coloumn II  are given as option (a), (b), (c) and (d), out of which ONE option is correct.

Q. Given at 298 K




Solution:
QUESTION: 18

Solid ammonium carbonate (NH2CO2NH4) dissociates completely into ammonia and carbon dioxide when it evaporates :

At 298 K, the total pressure of the gases in equilibrium with the solid is 0.116 atm. Derive the value of equilibrium constant Kp.

Solution:
QUESTION: 19

At 90° C , the following equilibrium is established :

If 0.20 mole of hydrogen and 1.0 mole of sulphur are heated to 90°C in a 1.0 dm3 flask, what will be the partial pressure of H2S gas at equilibrium?

Solution:

Initial moles of H₂ = 0.2
Initial moles of S = 1  
Kp = 6.8 * 10⁻²
Given equation:
H2(g) + S(s) ⇋ H2S(g)
Initial moles:          0.2        1
At equilibrium: (0.2-α)   (1-α)       α
Here, in the above equation we can see that hydrogen is the limiting reagent.  
∴ Kp = α/(0.2 – α)
⇒ 6.8 * 10⁻²  = α/(0.2 – α)
⇒ 1.36*10⁻² – (6.8*10⁻²)α = α
⇒ α + 0.068α = 1.36*10⁻²
⇒ α = 1.36*10⁻² / 1.068 = 1.273 * 10⁻² ← moles of H₂S
So, at equilibrium moles of H₂ = 0.2 – α = 0.2 – 1.273 * 10⁻² = 0.1873
Now, using the Ideal Gas equation,
PV = nRT ….. (i)
Where P = total pressure of the vessel
n = total no. of moles = (0.2-α) + (1-α) + α = 1.2 – α = 1.2 – 1.273*10⁻² = 1.1873
V = volume of vessel = 1 litre
R = Ideal gas constant = 0.082 L atm K⁻¹mol⁻¹
T = total temperature = 90℃ = 90+273 = 363 K
Substituting all the values in eq. (i), we get
P * 1 =  1.1873 * 0.082 * 363  
⇒ P = 35.34 atm
Thus,  
The partial pressure of H₂S at equilibrium
= (mole fraction of H₂S) * (total pressure)
= [1.273*10⁻² /  1.1873] * 35.34
= 0.3789 atm
≈ 0.38 atm

QUESTION: 20

Graphite is added to a vessel that contains CO2(g) at a pressure of 0.830 atm at a certain high temperature. The pressure rises due to a reaction that produces CO (g). The total pressure reaches an equilibrium value of 1.366 atm. Calculate the equilibrium constant of the following reaction.

Solution:

QUESTION: 21

For the equilibrium,

Kp = 0.166 at 1000 K. Exactly 10.0 g of CaCO3 is placed in a 10.0 L flask at 1000 K. After equilibrium is reached, what mass of CaCO3 remains?

Solution:

QUESTION: 22

Equilibrium constant for the reaction PCL5⇋PCL3+CL2 is 0.0205 at 230°C and 1 atmospheric pressure if at equilibrium concentration of PCL5 is 0.235 moles liter−1liter-1and that of CL2= 0.028 moleslit−1lit-1 then conc. of PCL3 at equilibrium is

Solution: