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Two flasks A and B of equal volume containing 1 mole and 2 mole of O_{3} respectively, are heated to the same temperature. When the reaction 2O_{3} → 3O_{2} practically stops, then both the flasks shall have
For Reaction
= will be same for both the containers
A 10L container at 300K contains CO_{2} gas at pressure of O.2 atm and an excess solid CaO (neglect the volume of solid CaO). The volume of container is now decreased by moving the movable piston fitted in the container. What will be the maximum volume of container when pressure of CO_{2} attains its maximum value given that
CaCO_{3} (s) → CaO(s) + CO_{2}(g) K_{ph} = 0.800 atm
K_{p} = 0.800 atm = = maximum pressure of CO2 in the container to calculate maximum volume of container the = 0.8 atm and none of = 0.8 should get converted into CaCO_{3} (s)
so V (0.800 atm) = (10 L) (o.2 atm)
so v = 2.5 L
An equilibrium mixture at 700 K of 0.50 M N_{2}, 3.00 M H_{2} and 2.00 M NH_{3} is present in a container. Now if this equilibrium is disturbed by adding N_{2} so that its concentration becomes 1.50 M just after addition then which of the following graphs represents the above situation more appropriately –
Let reaction be
N_{2}+3H_{2 }→ 2NH_{3}
Now after distrubing equilibrium then N_{2} concentration will be > 0.5 M and there will be some time taken for attainment of equilibrium hence graphs (B) and (C) can be discarded
from calculations it can be proved that the amount by which N_{2} should reaction should be less than 0.2 . Hence the result.
At 1400 K, K_{c}= 2.5x10 ^{+3} for the reaction CH_{4}(g) + 2H_{2}S(g) CS_{2}(g) + 4H_{2}(g) . A 10.0L reaction vessel at 1400 K contains 2.00 mole of CH_{4}, 3.0 mol of CS_{2}, 3.0 mole of H_{2} and 4.0 mole of H_{2}S. Then
The graph which will be representing all the equilibrium concentrations for the reaction
N_{2}O_{4} → 2NO_{2} (g) will be :
(the concentrations of N_{2}O_{4} (g) and of NO (g) for which the following eaction will be at equilibrium will lie
All the equilibrium will statisfy the equation
the required parabolic graph
In a basic aqueous solution chloromethane undergoes a substitution reaction in which Cl is replaced by OH^{–} as.
CH_{3}Cl (aq) + OH^{} → CH_{3}OH(aq) + Cl(aq)
The equilibrium constant of above reaction C, =1 x 10'6. If a solution is prepared by mixing equal volumes of 0.1 M CH_{3}CI and 0.2 M NaOH (100% dissociated) then [OH^{}] concentration at equilibrium in mixture will be:
CH_{3}Cl (aq) + OH^{} → CH_{3}OH(aq) + Cl^{}(aq)
As equilibrium constant is very large so reaction will be almost 100% complete.
For the chemical equilibrium,
CaCO_{3 }(S) → CaO(s) + CO_{2} (g)
ΔHf° can be determined from which one of the following plots ?
CaCO_{3 }(S) → CaO(s) + CO_{2} (g)
Graph (a) represents (i) and its slope will be used to determine the heat of the reaction.
At a certain temperature the following equilibrium is established, CO(g) + N0_{2}(g) → CO(g) + NO(g) One mole of each of the four gas is mixed in one litre container and the reaction is allowed to reach equilibrium state. When excess of baryta water (Ba(OH)_{2}) is added to the equilibrium mixture, the weight of white ppt (BaCO_{3}) obtained is 236.4 gm. The equilibrium constant C_{c} of the reaction is (Ba = 137)
CO(g) + N0_{2}(g) → CO(g) + NO(g)
or 1+x = 1.2
x= 0.2
At temperature T, the compound AB_{2} (g) dissociates according to the reaction, 2AB_{2} (g) 2 AB(g) + B_{2}(g).With a degree of dissociation x, which is small compared with unity. Deduce the expression for x in terms of the equilibrium constant, K_{p} and the total pressure, P.
Equilibrium constant for the given reaction is C = 10^{20} at temperature 300 K
A(s) +2B(aq) 2C (s) + D(aq.)K = 10^{20}
The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is
10lt box contain O_{3} and O_{2} at equilibrium at 2000 K. The ΔG^{0} = –534.52 kJ at 8 atm equilibrium
pressure.The following equilibrium is present in the container
2o_{3}(g) 3O_{2} (g). the partial pressure of O_{3} will be ( In 10 = 2.3, R = 8.3 J mol^{1} K^{1)}
ΔG^{0} = Rt In K = 2.3 x 2000 x 8.3 log K
Solid ammonium carbamate dissociates to give ammonia and carbon dioxid as follows:
NH_{2}COONH_{4}(s) 2NH_{5}(g) + CO_{2}(g)
At equilibrium, ammonia is added such that partial pressures of NH_{5} now equals the original total pressure. Calculate the ratio of the total pressures now to the original total pressure.
The reactions, PCl_{5 }(g) PCl_{5}(g) + Cl_{2}(g) and Cl_{2}(g) CO_{2}(g) + Cl_{2} (g) are simultaneously in equilibrium in an equilibrium box at constant volume. Afew moles of CO(g) are later introduced into the vessel. After some time, the new equilibrium concentration of
If CO is added 2nd equilibrium will proceede in the backward dkection and concentration of Cl_{2} will decrease. This Cl_{2} will be furtherformed by the decomposition of PCI_{5}.
In the Haber process for the industrial manufacture of ammonia involving the reaction, atm pressure in the presence of a catalyst a temperature of about 500°C is used. This is considered as optimum temperature for the process because
Formation of ammonia is an exothermix process therefore it is favorable at lower temperature. But at lower temperature rate of the reaction becomes slow.
For the equilibrium of the reaction ,k_{P} for the reaction at total pressure of P is:
What is the minimum mass of CaC0_{5} (s), below which it decomposes completely, required to establish equilibrium in a 6.50 litre container for the reaction : [ K_{C }= mole/litre]
CaCO_{3} (s) CaO(s) + CO_{2}(g)
K_{c} = [Cl_{2}] = 0.05 mole/itre
so moles of CO_{2}= 6.50 x 0.05 moles = 0.3250 moles
CaCl_{3} CaO + Cl_{2}
1 mole of CO_{2} =1 mole of CaCO_{3}
0.3250 moles of CO_{2} = 0.3250 moles of CaCO_{3}= 0.3250 x 100 gm of CaCO_{3} = 32.5 gm of CaCO_{3 }
The value of k_{p} for the reaction at 27°C
is '1 atom'. At equilibrium in a closed container partial pressure of BrCI gas is 0.1 atm and at this temperature the vapour pressure of Br_{2}(l) is also 0.1 atm. Then what will be minimum moles of Br_{2}(l) to be added to 1 mole of C1_{2} , initially, to get above equilibrium situation :
5 mol PCI_{3}(g) and one mole N_{2} gas is placed in a closed vessel. At equilibrium PCI_{5}(g) decomposes 20% and total pressure in to the container is found to be 1 atm. The k_{P} for equilibrium
Degree of association can be defined as the number of moles of a particular substance associated per mole of the substance taken.
For example : If out of 10 mole of N_{2} , 3 mol of N_{2} combine with H_{2} to form NHL. then degree of association of N_{2} = 0.3.Consider the equilibrium situation : N_{2}(g) + 3H_{2}(g) 2NH_{3}(g)
Initially N_{2} & H_{2} were mixed in 1 : 3 molar ratio and after long time the mean molar mass of the mixture was found to be 34/3g. The degree of association of N_{2} is
The correct answer is Option A
N2(g) + 2H2(g) ⇌ 2NH3(g)
t=0 1 3 0
t=t_{eq} 1  x 3  3x 2x
Total moles at equilibrium = 4 − 2x
Mean molar mass =
Value of mean molar mass is given. Find ′x′.
Attainment of the equilibrium A(g) 2C(g) +B(g) gave the gave the followeing graph. Find the correct option. (% dissociation = fraction dissociated x 100)
The correct answer is Option D.
Kc = ([C]^{2} * [B]) / [A]
= (6^{2} * 4) / 8
= 18 M^{2}
And, % dissociation of A = ([10  8] / 10)*100
= 20 %
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