Test: Chemical Equilibrium - 1
20 Questions MCQ Test Physical Chemistry | Test: Chemical Equilibrium - 1
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For the pair of reactions given below,
i) 
(ii) 1/2N2(g) + 3/2H2(g) ⇔ NH3(g)
If at a particular temperature, Kp1 and Kp2 are the equilibrium constants for reaction (i) and (ii) respectively then,
Among the following, the equilibrium which is not affected by an increase in pressure is:
As several moles of gaseous species are equal on both sides. Hence, for option (D), equilibrium is not affected by an increase in pressure.
Under the equilibrium condition for the reaction, 
the total pressure is 12 atm. If 50% of CO2 reacts then value of Kp is:
the total pressure is 12 atm. If 50% of CO2 reacts then value of Kp is:For the reaction, 
the equilibrium constant Kp changes with:
Pure ammonia is placed in a vessel at a temperature where its dissociation constant (α) is appreciable. At equilibrium,
One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is:
N2O4 → 2NO2
moles of unreacted N2O4 = 1 (1 - 0.2) = 0.8
moles of NO2 = 2 * 0.2 = 0.4
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (+)
total moles n2 = 0.8 + 0.4 = 1.2
P1/(T1 n1) = P2/(T2 * n2)
1/(300 * 1) = P2/(600 * 1.2)
P2 = 2.4 atm
For the reaction:
At a given temperature, the equilibrium amount of CO2(g) can be increased by.
For the chemical reaction,
The amount of X3 Y at equilibrium is affected by
There are 3 main factors that affect the equilibrium of a reaction:
- Concentration of reactants or products
- Temperature
- Pressure (Affects only gaseous reactions)
pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product (Ksp) of Ba(OH)2 is
To find out Ksp we need to know solubility of OH- which we can calculate in this case by using the value of pH.
pH+pOH=14
pOH=14-12 = 2
[OH-]=102
The following equilibrium is established when Barium hyfroxide is dissolved.
= s + 2s
it'sthe 2s which is equal to 10-2
so s = 5 * 10-3
the solubility product (Ksp) of Barium Hydroxide is 4s3
so putting 102 to 4s3 we get 5 * 10-7.
When two reactants, A and B are mixed to give products, C and D, the reaction quotient, (Q) at the initial stages of the reaction:
Given reaction is A+B⇌C+D
The reaction quotient Q is written by multiplying concentrations of products and dividing by concentration products of reactants
Q = [C][D]/[A][B]
At initial stages of reaction, formation of products C and D takes place; i.e, the concentration od C & D increases at the expenses of A & B concentration, which decreases.
∴Q increases with respect to time
At constant temperature, the equilibrium (Kp) for the decomposition reaction
is expressed by
where, p = pressure,
X = extent of decomposition. Which one of the following statement is true:
Kp is a characteristic constant for a given reaction and changes only with temperature.
Consider the following equilibrium in a closed container
At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements hold true regarding the equilibrium constant (Kp) and degree of dissociation (α)?
Which is correct statement if N2 is added at equilibrium condition?
N2 + 3H2 ⇌ 2NH3
The relationship between the equilibrium constant K1 for the reaction.
CO(g) +1/2O2 
(g) CO2(g)
And the equilibrium constant K2 for the reaction:
For the reaction H2 + I2 ⇋ 2HI the equilibrium constant (K) is
Correct Answer :- d
Explanation : aA(g) + bB(g) ----> cC(g) + dD(g)
K = {[C]c[D]d}/{[A]a[B]b}
H2 + I2 ⇋ 2HI
K = {[H2][I2]}/{[2HI]2}
For the gas phase reaction.
Carried out in a vessel, the equilibrium concentration of C2H4 can be increased by
(I) Increasing the temperature
(II) Decreasing the pressure
(III) Removing some H2
(IV) Adding some C2H6
When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium.
Since reaction is endothermic forward reaction is favoured by increase in temperature. Kp = [pO2]1/2.
Thus, addition of NaNO2 or NaNO3 does not cause any change in Kp.
The equilibrium
is attained at 250C in a closed container and an inert gas, helium is introduced. Which of the following statements are correct?
For the reaction, 
The forward reaction at constant temperature is favoured by
(I) Introducing an inert gas at constant volume
(II) Introducing chlorine gas at constant volume
(III) Introducing an inert gas at constant pressure
(IV) Increasing the volume of the container
(V) Introducing PCl5 at constant volume
According to Le chatelier's principle, if a system in equilibrium is disturbed by changes in determining factors, such as temperature, pressure, and concentration of components then the system will tend to shift its equilibrium position in such a way so as to counteract the effect of the disturbance.
At constant volume, there will be increase in total pressure of the system but the partial pressure of both the reactants and products remain the same. Hence, there will be no effect on equilibrium.
At constant pressure, there will be increase in total volume. Therefore, there will be decrease in number of moles per unit volume of each reactant and product. Hence, the equilibrium will shift towards the side where number of moles are increased.
And if there is an increase in volume of the container, it will directly decrease number of moles per unit volume of each reactant and product, thus, shifting equilibrium towards the side where number of moles are increased.
For the reaction, 
the equilibrium constant Kp changes with:
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