Chemical Equilibrium — Page 6 | JEE Chemistry

Q51

For the given reaction, choose the correct expression of

\mathrm{K}_{\mathrm{C}}

from the following :-

\mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons(\mathrm{FeSCN})_{(\mathrm{aq})}^{2+}

A

\mathrm{K}_{\mathrm{C}}=\dfrac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}{\left[\mathrm{FeSCN}^{2+}\right]}

B

\mathrm{K}_{\mathrm{C}}=\dfrac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}

C

\mathrm{K}_{\mathrm{C}}=\dfrac{\left[\mathrm{FeSCN}^{2+}\right]^2}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}

D

\mathrm{K}_{\mathrm{C}}=\dfrac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]^2\left[\mathrm{SCN}^{-}\right]^2}

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{K}_{\mathrm{C}}=\frac{\text{ Products ion conc. }}{\text{ Reactants ion conc. }} \\ & \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]} \end{aligned}

Q52

Consider the equilibrium

\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g})

If the pressure applied over the system increases by two fold at constant temperature then (A) Concentration of reactants and products increases. (B) Equilibrium will shift in forward direction. (C) Equilibrium constant increases since concentration of products increases. (D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. Choose the correct answer from the options given below :

A (A) and (B) only

B (A), (B) and (D) only

C (B) and (C) only

D (A), (B) and (C) only

Correct Answer

Option A

Solution

Given equilibrium

\mathrm{\mathop {C{O_{(g)}} + 3{H_2}(g)}\limits_{(n = 4)}} \rightleftharpoons \mathrm{\mathop {C{H_4}_{(g)} + {H_2}O(g)}\limits_{(n = 2)}}

Temperature $\to$ Constant Pressure $\to$ Increases by two fold.

It is based on Le Chatelier's principle.

The principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

When pressure increases in a chemical equilibrium, the equilibrium shifts towards the side of the reaction with fewer moles of gas molecules.

According to the principle, the reaction will favor the side that provides fewer gas molecules to counteract the increased pressure.

In this equilibrium, fewer gas molecules are on the produce side and on increase in pressure, equilibrium shifts towards products - forwarded reaction occurs.

As a result, concentration of the products increases.

A) Concentration of reactants and products increases.

$-$ Correct With increase in pressure, at constant temperature, the concentration of both reactants and products will increase because the same amount of gas is now in a smaller volume.

(B) Equilibrium will shift in forward direction Correct Increase in pressure causes product formation as the equilibrium shift is in the forward direction.

(C) Equilibrium constant increases since concentration of products increases.

Not correct.

The expression for equilibrium constant (K $_c$ ) for the given equilibrium can be written as

{K_c} = {{[C{H_4}][{H_2}O]} \over {[CO]{{[{H_2}]}^3}}}

Equilibrium constant is not changed with increase in pressure and increase in concentration of reactants or products.

So, the equilibrium constant value is does not affected by change in concentration.

If the change increase in concentration of product takes place, the equilibrium position will be shifted to counteract the change.

So, equilibrium const value does not increase.

Statement is incorrect.

(D) Equilibrium constant remains unchanged as concentration of reactants and products remain same.

State is incorrect.

In the reaction, increase in pressure causes changes in the concentration of rectants and products but the equilibrium constant will not change.

Statements (A) and (B) are correct.

Q53

For a concentrated solution of a weak electrolyte (

\mathrm{K}_{\text{eq }}=

equilibrium constant)

\mathrm{A}_{2} \mathrm{B}_{3}

of concentration '

c

', the degree of dissociation '

\alpha

' is :

A

\left(\dfrac{K_{e q}}{25 c^{2}}\right)^{\dfrac{1}{5}}

B

\left(\dfrac{K_{e q}}{108 c^{4}}\right)^{\dfrac{1}{5}}

C

\left(\dfrac{K_{e q}}{5 c^{4}}\right)^{\dfrac{1}{5}}

D

\left(\dfrac{K_{e q}}{6 c^{5}}\right)^{\dfrac{1}{5}}

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{A}_2 \mathrm{~B}_3(\mathrm{aq} .) \rightleftharpoons 2 \mathrm{~A}_{\text{(aq.) }}^{3+}+3 \mathrm{~B}_{\text{(aq) }}^{2-} \\\\ & \mathrm{c}(1-\alpha) \quad\quad\quad 2 \mathrm{c} \alpha\quad\quad\quad 3 \mathrm{c}\alpha\\\\ & \mathrm{K}_{\mathrm{eq}}=\frac{\left[\mathrm{A}^{3+}\right]^2\left[\mathrm{~B}^{2-}\right]^3}{\left[\mathrm{~A}_2 \mathrm{~B}_3\right]}=\frac{4 \mathrm{c}^2 \alpha^2 \times 27 \mathrm{c}^3 \alpha^3}{\mathrm{c}(1-\alpha)} \\\\ & \mathrm{K}_{\mathrm{eq}}=\frac{108 \mathrm{c}^5 \alpha^5}{\mathrm{c}} \alpha=\left(\frac{\mathrm{K}_{\mathrm{eq}}}{108 \mathrm{c}^4}\right)^{\frac{1}{5}} \end{aligned}

Q54

At a certain temperature in a

5

L

vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction, CO + Cl2

\rightleftharpoons

COCl2 At equilibrium, if one mole of CO is present then equilibrium constant (Kc) for the reaction is :

A 2

B 2.5

C 3

D 4

Correct Answer

Option B

Solution

CO + Cl2 $\rightleftharpoons$ COCl2 Initially number of moles 2 3 0 At equilibrium number of moles 1 2 1 The equilibrium constant, Kc =

{{\left[ {COC{l_2}} \right]} \over {\left[ {CO} \right]\left[ {C{l_2}} \right]}}

=

{{\left( {{1 \over 5}} \right)} \over {\left( {{1 \over 5}} \right) \times \left( {{2 \over 5}} \right)}}

=

{5 \over 2}

= 2.5

Q55

The INCORRECT match in the following is :

A

\Delta

Go = 0, K = 1

B

\Delta

Go < 0, K < 1

C

\Delta

Go > 0, K < 1

D

\Delta

Go < 0, K > 1

Correct Answer

Option B

Solution

We know,

\Delta {G^o} = - RT\ln K

Case 1 : If

\Delta

Go < 0 $\Rightarrow$

- RT\ln K

< 0 $\Rightarrow$

\ln K

> 0 $\Rightarrow$ K > 1 Case 2 : If

\Delta

Go > 0 $\Rightarrow$

- RT\ln K

> 0 $\Rightarrow$

\ln K

< 0 $\Rightarrow$ K < 1 Case 2 : If

\Delta

Go = 0 $\Rightarrow$

- RT\ln K

= 0 $\Rightarrow$

\ln K

= 0 $\Rightarrow$ K = 1

Q56

Given below are two statements : Statement I : On passing

\mathrm{HCl}_{(\mathrm{g})}

through a saturated solution of

\mathrm{BaCl}_2

, at room temperature white turbidity appears. Statement II : When

\mathrm{HCl}

gas is passed through a saturated solution of

\mathrm{NaCl}

, sodium chloride is precipitated due to common ion effect. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are correct

B Statement I is incorrect but Statement II is correct

C Both Statement I and Statement II are incorrect

D Statement I is correct but Statement II is incorrect

Correct Answer

Option D

Solution

To determine the correctness of the statements, we'll analyze each one individually by applying principles of solubility, common ion effect, and chemical equilibria.

Statement I: On passing $\mathrm{HCl}_{(g)}$ through a saturated solution of $\mathrm{BaCl}_2$ at room temperature, white turbidity appears.

Analysis: Dissolution of HCl Gas: When $\mathrm{HCl}_{(g)}$ is bubbled through water, it dissolves and dissociates completely: $\mathrm{HCl}_{(aq)} \rightarrow \mathrm{H}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}$ This increases the concentration of $\mathrm{Cl}^-$ ions in the solution.

Effect on BaCl₂ Solubility: The solubility equilibrium of $\mathrm{BaCl}_2$ in water is: $\mathrm{BaCl}_2 \leftrightarrow \mathrm{Ba}^{2+}_{(aq)} + 2\mathrm{Cl}^-_{(aq)}$ Adding more $\mathrm{Cl}^-$ shifts the equilibrium to the left (Le Chatelier's Principle), causing $\mathrm{BaCl}_2$ to precipitate.

The precipitation of $\mathrm{BaCl}_2$ manifests as a white turbidity.

Conclusion: Statement I is correct.

Statement II: When $\mathrm{HCl}$ gas is passed through a saturated solution of $\mathrm{NaCl}$ , sodium chloride is precipitated due to common ion effect.

Analysis: Dissolution of HCl Gas: Similar to before, $\mathrm{HCl}$ increases $\mathrm{Cl}^-$ ion concentration.

Effect on NaCl Solubility: The solubility equilibrium of $\mathrm{NaCl}$ is: $\mathrm{NaCl} \leftrightarrow \mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}$ However, $\mathrm{NaCl}$ is highly soluble in water, and its solubility is not significantly affected by the common ion effect from $\mathrm{Cl}^-$ .

The solubility product ( $K_{sp}$ ) of $\mathrm{NaCl}$ is large, and the addition of $\mathrm{Cl}^-$ ions does not cause $\mathrm{NaCl}$ to precipitate under normal conditions.

No precipitation occurs; the solution remains clear.

Conclusion: Statement II is incorrect.

Final Answer: Statement I is correct but Statement II is incorrect.