Chemical Equilibrium — Page 10 | NEET Chemistry

Q91

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

Q92

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

Q93

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.