Equilibrium Test

Result

Equilibrium Test - 45

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The equilibrium constant in reversible reaction at a given temperature-

A
Depends on the initial concentration of the reactants.

B
Depends on the concentration of the products at equilibrium.

C
Does not depend on the initial concentrations.

D
It is not characteristic of the reaction.

Solution

Explanation:

Consider the reaction,
A + B $$ \rightleftharpoons $$ C + D.
For the above given reversible reaction, the reaction constant depends on equilibrium concentration of product and reactant as follows:
$$ K_c $$= $$\dfrac {[C][D]} {[A][B]} $$.
As the above reaction shows that the rate constant of reaction depends upon the ratio equilibrium concentration of the product and reactants and not on the Initial concentration.
The equilibrium constant is affected by a change in temperature only and not on the concentration.

Correct Option: $$C$$.
Question 2
1 / -0

$$PCl_5$$ dissociates as $$PCl_{5}(g)\rightleftharpoons PCl_{3} (g) + Cl_{2}(g)$$. If $$PCl_{5}$$ is 10% dissociated at 1 atm, then the percentage of dissociation at 4 atm is:

A
2.5%

B
4%

C
5%

D
10%

Solution

$$PCl_{5}(g)\rightleftharpoons PCl_{3} (g) + Cl_{2}(g)$$
$$P_{1} $$
$$P_{1}(1-\alpha) \quad \quad P_{1}\alpha \quad \quad \quad P_{1}\alpha$$
$$0.9P_{1} \quad \quad \quad 0.1P_{1} \quad \quad \quad 0.1P_{1}$$

Now,
$$K_{P}=\dfrac{P_{PCl_3}\times P_{Cl_2}}{P_{PCl_5}}=\displaystyle\frac{(0.1P_{1})^{2}}{0.9P_{1}}=\displaystyle\frac{P_{1}}{90}$$

Again,
$$1.1P_{1}=1 atm$$ -- Total pressure
$$\Rightarrow P_{1}=\displaystyle\frac{1}{1.1}$$

So, $$K_{P}=\displaystyle\frac{1}{99}$$

For new condition,
$$PCl_{5}(g)\rightleftharpoons PCl_{3} (g) + Cl_{2}(g)$$
$$P_{2} $$
$$P_{2}(1-\alpha) \quad \quad P_{2}\alpha \quad \quad P_{2}\alpha$$

$$\Rightarrow P_{2}(1+\alpha)=4$$ -- Total pressure

$$K_{P}=\displaystyle\frac{P_{2}\alpha ^{2}}{1-\alpha} = \displaystyle\frac{1}{99}$$

Note: $$1-\alpha^2 \approx 1$$ since $$\alpha^2<<1$$

$$\alpha ^{2}= \displaystyle\frac{1}{396}\Rightarrow \alpha = 0.05$$
Question 3
1 / -0

A reversible chemical reaction having two reactants in equilibrium. If the concentration of the reactants are doubled, then the equilibrium constant will______.

A
Also be doubled.

B
Be halved.

C
Become one-fourth.

D
Remain the same.

Solution

When the concentrations of the reactants are changed, the concentrations of the products also change in such a way that the overall value of the equilibrium constant remains same.
Thus the equilibrium constant is the characteristic of a given reaction and is unaffected by the changes in the concentrations of the reactants and products.
Question 4
1 / -0

The equilibrium constant for the reaction $$H_{2}+I_{2} \rightleftharpoons 2HI$$ is 50. If the volume of the container is reduced to half of its original value, the value of the equilibrium constant will be:

A
25

B
50

C
75

D
100

Solution

At equilibrium the concentration of $$H_{2} $$ $$I_{2}$$ and HI is ($$1-x), (1-x), (2x)$$ respectively.

$$K_{c} = 50$$

$$= \dfrac{\left [ HI \right ]^{2}}{\left [ H_{2} \right ]I_{2}}$$

$$K_{c}= \dfrac{4x^{2}}{(1-x^{2})}= 50$$

$$K_{c}$$ value does not depend on initial concentration. When volume of reaction vessel is reduced to half, concentration is doubled. $$K_{c}$$ remains same.

It can also be seen in another way that since $$\Delta n $$ of reaction is zero there would be no impact of a change in concentration.
Question 5
1 / -0

Increase in pressure shifts the equilibrium of the reaction

$$N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)$$

A
to produce more $$N_{2}$$

B
to produce more $$NH_{3}$$

C
to reduce the temperature of the reaction

D
to produce more $$H_{2}$$

Solution
Question 6
1 / -0

The following reaction is at equilibrium in a given cylinder of constant volume.
$$N_2 (g) + 3 H_2 (g) \Leftrightarrow 2NH_3 (g)$$
Addition of argon gas to the cylinder will

A
reduce the formation of $$NH_3$$ with no change in temperature of the equilibrium system.

B
Increase the formation of $$NH_3$$ with increase in temperature.

C
reduce the formation of $$NH_3$$ with increase in temperature.

D
increase the formation of $$NH_3$$ with no change in temperature of the equilibrium system.
Question 7
1 / -0

When $$NaNO_3$$ is heated in a closed vessel, oxygen is liberated and $$NaNO_2$$ is left behind. At equilibrium:

A
addition of $$NaNO_3$$ favours reverse reaction

B
addition of $$NaNO_3$$ favours forward reaction

C
increasing temperature favours forward reaction

D
increasing pressure favours the forward reaction

Solution

$$NaNO_3$$ is a solid and its concentration is unity. Thus the addition of $$NaNO_3$$ does not affect the forward reaction or backward reaction.

Thermal decomposition of $$NaNO_3$$ is an endothermic reaction. When the temperature of the reaction is increased, the heat is absorbed and the reaction proceeds in the forward direction so that the effect of increased temperature is nullified.

During thermal decomposition of $$NaNO_3$$, reactants contain 2 moles and products contain 3 moles. Thus when pressure is increased, the reaction shifts in the backward direction so that the number of moles decreases.
Question 8
1 / -0

The following equilibrium in a closed container $$N_2O_4(g)\rightleftharpoons 2NO_2(g)$$
At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant $$(K_p)$$ and degree of dissociation?

A
Neither $$K_p$$ and nor $$\alpha$$ changes.

B
Both $$K_p$$ and $$\alpha$$ changes.

C
$$ \alpha $$ changes

D
$$K_p$$ does not change

Solution

When the volume of the container is halved, the pressure is doubled. Due to this the equilibrium will shift in the backward direction which will decrease the number of moles and pressure. Hence the value of the equllibrium constant $$K_p$$ and the degre of dissociation $$\alpha$$ changes.
Question 9
1 / -0

A 10L container at 300 K contains CO$$_2$$ gas at a pressure of 0.2 atm and an excess solid CaO (neglect the volume of solid CaO). The volume of a container is now decreased by moving the movable piston fitted in the container. What will be the maximum volume of a container when a pressure of CO$$_2$$ attains its maximum value?
Given that, $$CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) ; \ K_p = 0.80\ atm$$

A
$$5 L$$

B
$$2.5 L$$

C
$$1 L$$

D
The information is insufficient.

Solution

$$K_p = 0.800 atm =P_{co_2} = $$ maximum pressure of CO$$_2$$ in the container.

To calculate maximum volume of container $$P_{co_2} = 0.8$$ atm and none of CO$$_2$$ should get converted into CaCO$$_3$$(s).

So, $$PV=constant$$

$$\therefore$$ $$V\times (0.800 atm) = (10 L)\times (0.2 atm)$$

$$\Rightarrow V= 2.5 L$$
Question 10
1 / -0

Which of the following statements is incorrect?

A
In equilibrium mixture of ice and water kept in a perfectly insulated flask, mass of ice and water does not change with time.

B
The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.

C
On addition of a catalyst, the equilibrium constant value is not affected.

D
Equilibrium constant for a reaction with negative $$\Delta H$$ value decreases as the temperature increases.

Solution

Iron (III) nitrate and potassium thiocyanate react to give deep red coloration.
The reaction is:
$$Fe^{3+}_{aq}+SCN^-\leftrightarrow [FeSCN]^{2+}_{aq}$$
yellow deep red
Now, to this solution, if oxalic acid $$(H_2C_2O_4)$$ is added, then it reacts with $$Fe^{3+}$$ ions to form the stable complex ion $$[Fe(C_2O_4)_3]^{3-}$$, thus decreasing the concentration of $$Fe^{3+}$$ ions. In accordance with the Le-Chatelier's principle, the decrease in the concentration of $$Fe^{3+}$$ ions, leads to the decrease in the concentration of $$[FeSCN]^{2+}$$ ions which is indicated by the fact that the colour of the solution becomes lighter.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Equilibrium Test - 45

Result

Equilibrium Test - 45

Score

-

Rank

-

SHARING IS CARING

Question 1

Depends on the initial concentration of the reactants.

Depends on the concentration of the products at equilibrium.

Does not depend on the initial concentrations.

It is not characteristic of the reaction.

Solution

Question 2

2.5%

4%

5%

10%

Solution

Question 3

Also be doubled.

Be halved.

Become one-fourth.

Remain the same.

Solution

Question 4

25

50

75

100

Solution

Question 5

to produce more $$N_{2}$$

to produce more $$NH_{3}$$

to reduce the temperature of the reaction

to produce more $$H_{2}$$

Solution

Question 6

reduce the formation of $$NH_3$$ with no change in temperature of the equilibrium system.

Increase the formation of $$NH_3$$ with increase in temperature.

reduce the formation of $$NH_3$$ with increase in temperature.

increase the formation of $$NH_3$$ with no change in temperature of the equilibrium system.

Question 7

addition of $$NaNO_3$$ favours reverse reaction

addition of $$NaNO_3$$ favours forward reaction

increasing temperature favours forward reaction

increasing pressure favours the forward reaction

Solution

Question 8

Neither $$K_p$$ and nor $$\alpha$$ changes.

Both $$K_p$$ and $$\alpha$$ changes.

$$ \alpha $$ changes

$$K_p$$ does not change

Solution

Question 9

$$5 L$$

$$2.5 L$$

$$1 L$$

The information is insufficient.

Solution

Question 10

In equilibrium mixture of ice and water kept in a perfectly insulated flask, mass of ice and water does not change with time.

The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.

On addition of a catalyst, the equilibrium constant value is not affected.

Equilibrium constant for a reaction with negative $$\Delta H$$ value decreases as the temperature increases.

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.