Equilibrium Test

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Equilibrium Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

In a given system, water water and ice are in equilibrium. If pressure is applied to the above system then

A
More of ice is formed

B
Amount of ice and water will remain same

C
More ice is melted

D
Either (a) or (c)

Solution

Ice $$\rightleftharpoons $$ Water
More vol $$\quad $$ Less vol
On applying pressure, equilibrium shifts to the side in which volume is less.

hence C is correct answer
Question 2
1 / -0

In the given reaction $$2\ X(g) +Y(g)\rightleftharpoons 2\ Z(g)+80\ kcal$$ which combination of pressure and temperature will give the highest yield of $$Z$$ at equilibrium?

A
$$1000\ atm$$ and $$200^oC$$

B
$$500\ atm$$ and $$500^oC$$

C
$$100\ atm$$ and $$200^oC$$

D
$$500\ atm$$ and $$100^oC$$

E
$$1000\ atm$$ and $$500^oC$$

Solution

As $$n_p < n_r$$ higher the pressure greater will be yield of $$Z$$. As reaction is exothermic in the forward direction lower the temperature, greater is the yield of $$Z$$ (As at low temperature the reaction is slow, usually optimum temperature is used).
Question 3
1 / -0

Assertion: $$K_p$$ is always greater than $$K_c$$.
Reason: The effect of pressure is greater on the rate of reaction than the effect of concentration.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

C
Assertion is correct but Reason is incorrect

D
Both Assertion and Reason are incorrect

E
If both Assertion and Reason are false.

Solution

$$K_p=K_c (RT)^{\Delta n}$$. If $$\Delta n=0, K_p=K_c$$. If $$\Delta n=+ve, K_p > K_c$$. If $$\Delta n=-ve, K_p < K_c$$. Both pressure and concentration affect the rate of reaction.
Question 4
1 / -0

The degree of dissociation of 0.1 M weak acid HA is 0.5%. If 2 ml of 1.0 M HA solution is diluted to 32 ml, the degree of dissociation of acid and $$H_{3}O^{+}$$ ion concentration in the resulting solution will be respectively:

A
$$0.02$$ and $$3.125 \times10^{-4}$$

B
$$1.25 \times10^{-3}$$ and $$0.02$$

C
$$0.632$$ and $$3.95 \times 10^{-4}$$

D
$$0.02$$ and $$8.0\times10^{-12}$$

Solution

Given,
$$\alpha =0.5 \%= 0.005$$ and $$C= 0.1\ M$$

The expression for the degree of dissociation is $$\alpha = \sqrt {\dfrac{K_a}{C}}$$.
$$0.005= \sqrt {\dfrac{K_a}{0.1}} $$

$$K_a=2.5 \times 10^{-6}$$.

When the solution is diluted, the molarity of the solution is given by the following expression.

$$M_1V_1=M_2V_2$$

or $$2 \times 1 = 32 \times M_2$$

Hence, $$M_2=\dfrac {1} {16} $$.

For this diluted solution, $$\alpha = \sqrt {\dfrac{K_a}{C}}=0.632\%$$.
The hydrogen ion concentration is $$[H^+]=c \alpha =\dfrac {1} {16} \times 0.00632 =3.955 \times 10^{-4}$$.
Question 5
1 / -0

Which of the following statements is correct?

A
The $$pH$$ of $$1.0 \times 10^{-8}M$$ solution of $$HCl$$ is $$8$$

B
The conjugate base of $$H_{2}PO_{4}^{-2}$$ is $$H PO_{4}^{-2}$$

C
Autoprotolysis constant of water increases with temperature

D
When a solution of weak monoprotic acid is titrated against a strong base, at half neutralization point $$pH = \frac{1}{2}p{Ka}$$

Solution

$$HCl$$ in water always creates an acidic solution. A solution with pH 8 is basic but the pH of the solution with $$10^{-8} $$ molarity $$HCl$$ must be less than 7.

Species formed after protonation of is called a conjugate base. The conjugate base of $$H_2PO_4^{-2}$$ is $$HPO_4^{-3}$$.

Autoprotolysis is the dissociation of water which increases with increase in temperature.

Half neutralization point occurs when half of the acid moles are consumed by strong base. Therefore, $$pH = 1/2 \left(pK_a - log (c/2) \right) $$.
Question 6
1 / -0

The first and second dissociation constants of an acid $$H_{2}A$$ are $$1.0\times10^{-5}$$ and $$ 5.0\times 10^{-10}$$ respectively. The overall dissociation constant of the acid will be:

A
$$5.0\times10^{-5}$$

B
$$5.0\times10^{15}$$

C
$$5.0\times10^{-15}$$

D
$$2.0\times10^{5}$$

Solution

For the dissociation of a dibasic acid, the overall dissociation constant is equal to the product of the first and second dissociation constants.
Therefore, $$K=K_1 \times K_2=1.0 \times 10^{-5} \times 5.0 \times 10^{-10}= 5.0 \times 10^{-15}$$
Hence, the overall dissociation constant is $$5.0\times 10^{-15}$$.
Question 7
1 / -0

The degree of dissociation of an electrolyte is :

A
directly proportional to its concentration

B
directly proportional to the square of its concentration

C
inversely proportional to its concentration

D
inversely proportional to the square root of its concentration

Solution

For a weak electrolyte, the dissociation constant can be written as:

$$K_d=\dfrac{\alpha^2}{1-\alpha} C$$, where $$\alpha$$ (degree of dissociation) $$<< 1$$ and C is the concentration of electrolyte.

$$\therefore \dfrac{K_d}{C} =\dfrac{\alpha^2}{1} $$.

$$\therefore \alpha=\sqrt{\dfrac{K_d}{C}}$$.

$$\therefore \alpha \propto \dfrac{1}{\sqrt{C}}$$.

The degree of dissociation is inversely proportional to the square root of concentration.
Question 8
1 / -0

For the reaction $$A(g) \leftrightharpoons B(g) + C(g), K_{p}$$ at $$400^{\circ} C$$ is $$1.5 \times 10^{-4}$$ and $$K_{p}$$ at $$600^{\circ}C$$ is $$6 \times 10^{-3}$$. Then, which of the following statements is incorrect?

A
The reaction is an exothermic one.

B
An increase in temperature increases the formation of $$B$$.

C
An increase in pressure increase the formation of $$A$$.

D
A decrease in temperature and increase in pressure shifts the equilibrium towards left.

Solution

As the temperature increases from $$400^oC$$ to $$600^oC$$, the value of the equilibrium constant increases from $$1.5 \times 10^{-4}$$ to $$6 \times 10^{-3}$$. This indicates that the reaction is endothermic so that with an increase in temperature, the equilibrium shifts to right, heat is absorbed which nullifies the effect of the increased temperature.

In an endothermic reaction, an increase in pressure shifts the equilibrium towards less mole of gas. So, in the given reaction when pressure is increased the formation of A increases.
Question 9
1 / -0

The volume of the reaction vessel containing an equilibrium mixture in the reaction is increased, when equilibrium is re-established:
$$SO_{2}Cl_{2}(g) \leftrightharpoons SO_{2}(g) + Cl_{2}(g)$$

A
the amount of $$SO_{2}$$ will have decreased

B
the amount of $$SO_{2}Cl_{2}$$ will have increased

C
the amount of $$Cl_{2}$$ will have increased

D
the amount of $$Cl_{2}$$ will have remained unchanged

Solution

The equilibrium reaction is $$SO_{2}Cl_{2}(g) \leftrightharpoons SO_{2}(g) + Cl_{2}(g)$$.

The forward reaction proceeds with an increase in the number of moles from 1 to 2. When the volume of the vessel is increased, the pressure decreases. The equilibrium will shift in the forward direction so as to nullify the effect of decreased pressure.

Hence, the amount of chlorine will have increased.
Question 10
1 / -0

Blood $$pH$$ is controlled by the concentrations of $$H_{2}CO_{3}$$ and $$HCO_{3}{^{-}}$$. In the presence of $$NaHCO_{3}$$, $$pH$$ of blood:

A
increases

B
decreases

C
does not change

D
first decreases and then increases

Solution

Acidic buffer solution is prepared by mixing a weak acid and its salt with a strong base.
Thus, a mixture of $$H_{2}CO_{3}$$ and $$HCO_{3}^{-} $$ acts as an acidic buffer solution.
The expression for the pH of the acidic buffer solution is as given below:
$$pH=pK_a+log \frac {[salt]} {[acid]}$$
In the presence of $$NaHCO_{3}$$, the concentration of salt increases. Hence, the term $$log \frac {[salt]} {[acid]}$$ increases. This increases the $$pH$$ of blood.

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Equilibrium Test - 44

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Equilibrium Test - 44

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Question 1

More of ice is formed

Amount of ice and water will remain same

More ice is melted

Either (a) or (c)

Solution

Question 2

$$1000\ atm$$ and $$200^oC$$

$$500\ atm$$ and $$500^oC$$

$$100\ atm$$ and $$200^oC$$

$$500\ atm$$ and $$100^oC$$

$$1000\ atm$$ and $$500^oC$$

Solution

Question 3

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Both Assertion and Reason are incorrect

If both Assertion and Reason are false.

Solution

Question 4

$$0.02$$ and $$3.125 \times10^{-4}$$

$$1.25 \times10^{-3}$$ and $$0.02$$

$$0.632$$ and $$3.95 \times 10^{-4}$$

$$0.02$$ and $$8.0\times10^{-12}$$

Solution

Question 5

The $$pH$$ of $$1.0 \times 10^{-8}M$$ solution of $$HCl$$ is $$8$$

The conjugate base of $$H_{2}PO_{4}^{-2}$$ is $$H PO_{4}^{-2}$$

Autoprotolysis constant of water increases with temperature

When a solution of weak monoprotic acid is titrated against a strong base, at half neutralization point $$pH = \frac{1}{2}p{Ka}$$

Solution

Question 6

$$5.0\times10^{-5}$$

$$5.0\times10^{15}$$

$$5.0\times10^{-15}$$

$$2.0\times10^{5}$$

Solution

Question 7

directly proportional to its concentration

directly proportional to the square of its concentration

inversely proportional to its concentration

inversely proportional to the square root of its concentration

Solution

Question 8

The reaction is an exothermic one.

An increase in temperature increases the formation of $$B$$.

An increase in pressure increase the formation of $$A$$.

A decrease in temperature and increase in pressure shifts the equilibrium towards left.

Solution

Question 9

the amount of $$SO_{2}$$ will have decreased

the amount of $$SO_{2}Cl_{2}$$ will have increased

the amount of $$Cl_{2}$$ will have increased

the amount of $$Cl_{2}$$ will have remained unchanged

Solution

Question 10

increases

decreases

does not change

first decreases and then increases

Solution

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