Test: Homogeneous Equilibria & Heterogeneous Equilibria (July 14) - NEET MCQ

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

Correct answer is A.

For the equilibrium reaction:
A+2B ⇌ 2C+D
volume of flask = 1L
Initial moles of A = 2 mol
initial concentration of A=[A]i = 2 M
initial mole of B = 1 mol 
[B]i = 1 M
[A]_eq = 2-x, [B]_eq = 1-2x, [C]_eq = x, [D]_eq = 3x
Given [D]_eq = 1 * 1L
= 1 M
Thus x = 1M
[A]_eq = 1, [B]_eq = -1, [C]_eq = 1, [D] = 3
Kc = {([D]eq)³ * ([C]eq)}/{[A]eq * ([B]eq)²} 
= Kc = {(3)^3*1}/{1*(-1)²}
= 27/1
= 27

Squaring on both sides 



⇒
⇒
⇒
=> 10 bar approximately

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 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%