Direction (Q. Nos. 110) 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 NH_{4}HS (s)is 20 mm at 25°C, for
NH_{4}HS (s)⇌ NH_{3}(g) + NH_{3}(g) + H_{2}S(g)
Total pressure when NH_{4}HS (s) dissociates at 25°Cin a vessel which already contains H_{2}S (g)at a pressure of 15 mm, is
The correct answer is Option A.
NH_{4}HS > NH_{3} + H_{2}S
Let P be the pressure at eq. of NH_{3} and H_{2}S.
Therefore, K_{p} = P^{2}
= (20 / 2)^{2}
= 100 mm
= 100
Also, K_{p} = (15 + P) (P)
100 = 15 P + P2
P^{2} + 15 P – 100 = 0
P = 5
Total pressure = 15 + 2(P)
= 15 + 2(5)
= 25 mm
Once the equilibrium is reached under given condition:
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.
Ca(HCO_{3})_{2} is strongly heated and after equilibrium is attained, temperature changed to 25° C.
Ca(HCO_{3})_{2}(s)⇌CaO(s) + 2CO_{2} (g) + H_{2}O(g)
K_{p} = 36 (pressure taken in atm)
Thus, pressure set up due to CO_{2} is
The reaction is as follow:
Ca(HCO_{3})_{2}(s)⇌CaO(s) + 2CO_{2} (g) + H_{2}O(g)
At 25° C H_{2}O goes in liquid state
Kp = (PCaO)1×(PCO_{2})_{2}
(PCa(HCO_{3})_{2})
Since, Ca(HCO_{3})_{2}, CaO and H_{2}O are not in gaseous state, so their partial pressure is taken 1.
Putting all values, we have
36 = (PCO_{2})_{2}
Or PCO_{2} = 6 atm
Na_{2}SO_{4}.10H_{2}O dehydrates according to the equation,
Na_{2}SO_{4}· 10H_{2}O (s)⇌ Na_{2}sO_{4}(s) + 10 H_{2}O(g)
; K_{p} = 2.56 x 10^{20}
What is the pressure of water vapor at equilibrium with a mixture of Na_{2}SO_{4}· 10H_{2}O and Na_{2}sO_{4}
Ca(HCO_{3})_{2} decomposes as,
Ca (HCO_{3})_{2}(s) ⇌ CaCO_{3}(s) + H_{2}O(g) + CO_{2}(g)
Equilibrium pressure is found to be 0.12 bar. What is pco_{2} if the reaction mixture also contains H_{2}O(g)at 0.20 bar?
For the following equilibrium,
Assume following equilibria when total pressure set up in each are equal to 1 atm, and equilibrium constant (K_{p}) as K_{1}; K_{2} and K_{3}
Thus,
The correct answer is option C
CaCO_{3} → CaO + CO_{2}
K_{p} = k_{1} = Pco_{2}
total pressure of container P
k_{1} = p
NH_{4}HS → NH_{3} + H2S
PNH_{3} = PH_{2}S = P0
P_{0} + P_{0} = p (total pressure)
P0 = p/_{2}
k_{2} = kp = [P_{NH3}][P_{H2s}] p2_{4}
NH_{2}CoNH_{2} → 2NH_{3} + CO_{2}
P_{NH3} = 2P_{0} P_{CO2} = P0
2P_{0 }+ P_{0} = P
For the equilibrium,
at 1000 K. If at equilibrium p_{CO} = 10_{} then total pressure at equilibrium is
C(s) + CO_{2}(g) <=========> 2CO(g)
K_{p} = pCO^{2}/pCO^{2}
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
Ammonium carbamate dissociates as,
In a closed vessel containing ammonium carbamate in equilibrium with its vapour, ammonia is added such that partial pressure of NH_{3} now equals the original total pressure. Thus, ratio of the total pressure to the original pressure is
Kc forthe decomposition of NH_{4}HS(s) is 1.8x 10^{4} at 25°C.
If the system already contains [NH_{3}] = 0.020 M, then when equilibrium is reached, molar concentration are
NH4HS (s) ⇋ NH3 (g) + H2S (g)
Initial 1  
At eqm 1x x+0.02 x
Kc = [NH_{3}][H_{2}S] (Since NH4HS is solid, we ignore it.)
1.8×104 = (x+0.02)(x)
x2+0.02x1.8×104 = 0
Applying quadratic formula; x = 0.02+√{(0.02)24×1.8×104}
= 0.0330.020/2 = 0.0065
Therefore, concn of NH_{3} at equilibrium = x+0.020 = 0.0265
concn of H2S at equilibrium = x = 0.0065
So, option b is correct
Direction (Q. Nos. 1114) 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 NH_{4}CI is taken in a closed vessel of 1 L capacity and heated to 275° C.
Q. Partial pressure of NH_{3}(g) or HCI (g) at equilibrium is
The reaction is as follow:
NH_{4}Cl(s) ⇋ NH_{3}(g) + HCl(g)
Kp = (pNH_{3})(pHCl)
1×10^{2} = p_{2} (SInce reactant dissociates into same ratio, so the partial pressure will be same for both)
100×10^{4} = p_{2}
Or p = 10×10^{2} = 0.10
So, the partial pressure of NH_{3} and HCl are 0.10 atm.
Passage I
Solid ammonium chloride is in equilibrium with ammonia and hydrogen chloride gases
0.980 g of solid NH_{4}CI is taken in a closed vessel of 1 L capacity and heated to 275° C.
Q. Percentage decomposition of the original sample is
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 = 1.00×102 at 275° C
Kp = kc(RT)2
1.00×102 = kc(0.0821×548)2
Or kc = 4.94×106
NH4Cl(s) ⇌ NH3(g) + HCl(g)
Initial 1  
At eqm 1x x x
Kc = x2
x = √(4.94×106)
= 2.22×103
Therefore, NH4Cl dissociated at eqm = 2.22×103 × 53.5 = 0.118
%age decomposition = 0.118/0.980×100 = 12.13%
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
CO_{2}(g) = 0.80 atm
Q. Under the given partial pressure, reaction is
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.
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
CO_{2}(g) = 0.80 atm
Q. When equilibrium is attained,
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 0.80p atm
Kp = pCO2/pCO = 0.80p/1.4+p
0.265 = 0.80p/1.4+p
Or p = 0.339 atm
Therefore, partial pressure at eqm, pCO2 = 0.800.339 = 0.461 atm
And pCO = 1.4+0.339 = 1.739 atm
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 O_{2} are left in 1 L of a gas supply after equilibrium has been reached?
At 1000 K, pressure of CO_{2} in equilibrium with CaCO_{3} 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 CaCO_{3} 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)?
K= (partial pressure of co^{2}/(partial pressure of co^{2})
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
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
Solid ammonium carbonate (NH_{2}CO_{2}NH_{4}) 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 K_{p}.
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 dm^{3} flask, what will be the partial pressure of H_{2}S gas at equilibrium?
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: (0.2α) (1α) 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 = (0.2α) + (1α) + 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
Graphite is added to a vessel that contains CO_{2}(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.
For the equilibrium,
K_{p} = 0.166 at 1000 K. Exactly 10.0 g of CaCO_{3} is placed in a 10.0 L flask at 1000 K. After equilibrium is reached, what mass of CaCO_{3} remains?
Equilibrium constant for the reaction PCL_{5}⇋PCL_{3}+CL_{2} is 0.0205 at 230°C and 1 atmospheric pressure if at equilibrium concentration of PCL_{5} is 0.235 moles liter−1liter1and that of CL_{2}= 0.028 moleslit−1lit1 then conc. of PCL_{3} at equilibrium is
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 







