Test: Chemical Equilibrium: Heterogeneous Equilibrium - NEET MCQ

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

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.

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 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%

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.

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

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