Chemical Equilibrium: Heterogeneous Equilibrium - Free MCQ Practice Test

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

B is correct.

At chemical equilibrium the rates of the forward and reverse reactions are equal, so the concentrations of reactants and products remain constant with time when external conditions (temperature, pressure, concentrations) are held fixed.

Equilibrium is dynamic: microscopic forward and reverse processes continue to occur, but there is no net change in concentrations because the rates are equal.

Changing the temperature alters the equilibrium constant K and typically shifts the equilibrium position; therefore concentrations generally change if temperature is changed.

For heterogeneous equilibria, the activities of pure solids and pure liquids are effectively constant and usually do not appear explicitly in the expression for K, but the gaseous or dissolved species present at equilibrium still have concentrations that remain constant with time.

Hence the best and most general statement is that the concentrations of all substances present remain constant with time once equilibrium is established under the given conditions.

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

The equilibrium pressure of water vapour for the dehydration of Na₂SO₄·10H₂O can be determined from the equilibrium constant K_p.

• The reaction is: Na₂SO₄·10H₂O (s) ⇌ Na₂SO₄ (s) + 10 H₂O (g).
• Since K_p = 2.56 x 10^-20, it represents the pressure of water vapour to the power of 10 at equilibrium.
• Let P be the pressure of water vapour. Then, P¹⁰ = K_p.
• Thus, P = (2.56 x 10^-20)^1/10.
• Calculating gives P ≈ 1.1 x 10^-2 atm.

Therefore, the equilibrium pressure of water vapour is approximately 1.1 x 10^-2 atm.

Option A is correct.

log₁₀Kp = 8 - 6400/T.

Convert temperature: T = 527 °C = 527 + 273 = 800 K.

Substitute T = 800 K into the formula: log₁₀Kp = 8 - 6400/800 = 8 - 8 = 0.

Therefore Kp = 10⁰ = 1.

For an equilibrium that involves only pure solids and CO₂(g), the activities of solids are unity, so Kp = p_CO2 (with pressure in atm). Hence p_CO2 = 1 atm, matching option A.

The correct answer is option C
CaCO₃ ​→ CaO + CO_2​
K_p​ = k₁ ​= Pco₂​​
total pressure of container P
k_1​ = p
NH₄​HS → NH₃ ​+ H2​S
PNH_3​​ = PH_2​S ​= P0​
P_0​ + P₀​ = p (total pressure)
P0 ​= p/₂
k_2​ = kp ​= [P_NH3​​][P_H2​s​] p2₄​
NH₂​CoNH₂ ​→ 2NH₃ ​+ CO_2​
P_NH3​​ = 2P₀​        P_CO2​ ​= P0​
2P_0​ + P₀ ​= P

C(s) + CO₂(g) <=========> 2CO(g)
K_p = pCO²/pCO²
GIven K_p = 63 and p_CO = 10p_CO2
Putting the value of pCO in above equation,
63 = 100(_pCO²)²/p_CO2
Or p_CO2 = 0.63
p_CO = 6.3
Therefore, total pressure = 6.3+0.63 = 6.93 atm

 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

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.

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%

Correct answer: B

Solids do not appear in the expression for the equilibrium pressure quotient; only gaseous species are included.

Qp = pCO₂ / pCO

Qp = 0.80 / 1.40 = 0.571

Kp = 0.265

Compare the two: Qp (0.571) > Kp (0.265). When Qp > Kp, the reaction shifts in the backward (reverse) direction to reach equilibrium, which will decrease pCO₂ and increase pCO.

Hence the reaction is displaced to the backward side.

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

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 =2.105×1.9
= 3.99 = 4
(partial pressure of co) =2

Option A is correct; K_p = 4p³/27.

Let the total pressure of the gases be p.

The decomposition produces 2 moles of NH₃ and 1 mole of CO₂ for each formula unit, so the total number of moles of gas formed is 3.

Therefore the mole fraction (and hence partial pressure fraction) of NH₃ is 2/3 and of CO₂ is 1/3.

So, P_NH3 = (2/3) p and P_CO2 = (1/3) p.

For the heterogeneous equilibrium, the solid is omitted and K_p = (P_NH3)²·(P_CO2).

Substituting the partial pressures gives K_p = [(2/3) p]²·[(1/3) p] = 4/27 · p³.

Using p = 0.116 atm, K_p = (4/27)·(0.116)³ ≈ 2.31×10^-4 atm³.

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

Hence, option(A) is correct.