Test: Chemical Equilibrium - 1 - Chemistry MCQ

As several moles of gaseous species are equal on both sides. Hence, for option (D), equilibrium is not affected by an increase in pressure.

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

There are 3 main factors that affect the equilibrium of a reaction:

• Concentration of reactants or products
• Temperature
• Pressure (Affects only gaseous reactions)

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.

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

K_p is a characteristic constant for a given reaction and changes only with temperature. 

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}

Since reaction is endothermic forward reaction is favoured by increase in temperature. K_p = [_pO₂]^1/2.

Thus, addition of NaNO₂ or NaNO₃ does not cause any change in K_p.

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.