Consider the reaction $$\mathrm{X}_2 \mathrm{Y}(\mathrm{~g}) \rightleftharpoons \mathrm{X}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Y}_2(\mathrm{~g})$$ The equation representing correct relationship between

To determine the relationship between the degree of dissociation ($x$) of $\mathrm{X}_2 \mathrm{Y}(\mathrm{g})$ and its equilibrium constant $K_p$, let's analyze the reaction: $ \mathrm{X}_2 \mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{X}_2(\mathrm{g}) + \frac{1}{2} \mathrm{Y}_2(\mathrm{g}) $ Initial and Change in Moles Initially, we start with 1 mole of $\mathrm{X}_2 \mathrm{Y}$. At equilibrium: $\mathrm{X}_2 \mathrm{Y}$ is $(1-x)$ moles. $\mathrm{X}_2$ is $x$ moles. $\frac{1}{2} \mathrm{Y

Chemical Equilibrium — Page 9 | NEET Chemistry

Q81

The following reaction occurs in the Blast furnance where iron ore is reduced to iron metal

\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightleftharpoons \mathrm{Fe}_{(\mathrm{l})}+3 \mathrm{CO}_{2(g)}

Using the Le-chatelier's principle, predict which one of the following will not disturb the equilibrium.

A Addition of

\mathrm{CO}_2

B Removal of

\mathrm{CO}

C Addition of

\mathrm{Fe}_2 \mathrm{O}_3

D Removal of

\mathrm{CO}_2

Correct Answer

Option C

Solution

For the reaction :

\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}+3 \mathrm{CO}_{(\mathrm{g})} \rightleftharpoons \mathrm{Fe}_{(\mathrm{l})}+3 \mathrm{CO}_{2(\mathrm{~g})}

Addition or removal of

\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{s})}

and/or

\mathrm{Fe}_{\text{(l) }}

will not affect the equilibrium quotient and the equilibrium.

Q82

The gas phase reaction 2NO2(g)

\to

N2O4(g) is an exothermic reaction. The decomposition of N2O4, in equilibrium mixture of NO2(g) and N2O4(g), can be increased by :

A lowering the temperature.

B increasing the pressure.

C addition of an inert gas at constant volume.

D addition of an inert gas at constant pressure.

Correct Answer

Option D

Solution

2NO2 (g)

\overset{\,}\longrightarrow

N2O4 (g);

\Delta

H = $-$ ve At equilibrium, N2O4 $\rightleftharpoons$ 2NO2 ;

\Delta

H = + ve On adding the inert gas at constant pressure, the number of moles per unit volume of various reactants and products will decrease.

Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

Therefore, in above case, reaction will move towards forward direction which will lead to the decomposition of dinitrogen tetraoxide (N2O4).

Q83

In the following system,

\mathrm{PCl}_5(\mathrm{~g}) \leftrightharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})

at equilibrium, upon addition of xenon gas at constant T \& p , the concentration of

A

\mathrm{PCl}_5, \mathrm{PCl}_3 \& \mathrm{Cl}_2

remain constant

B

\mathrm{PCl}_3

will increase

C

\mathrm{Cl}_2

will decrease

D

\mathrm{PCl}_5

will increase

Correct Answer

Option B

Solution

When xenon gas, an inert gas, is added to the equilibrium system $\mathrm{PCl}_5(\mathrm{g}) \leftrightharpoons \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$ at a constant temperature and pressure, it affects the equilibrium according to Le Chatelier's principle.

The addition of an inert gas at constant pressure effectively increases the volume of the system, reducing the concentration of the gases involved.

Since the system is at constant pressure, adding an inert gas increases the total volume, which favors the side of the reaction with more moles of gas.

In this reaction, there are two moles of gas on the right side ( $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$ ) compared to one mole on the left ( $\mathrm{PCl}_5$ ).

Thus, the equilibrium will shift toward the right to increase the number of moles and counteract the change, decreasing the concentration of $\mathrm{PCl}_5$ and increasing the concentrations of $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$ .

However, due to the increase in overall volume, the individual concentrations of all species $\left([\mathrm{PCl}_5], [\mathrm{PCl}_3], [\mathrm{Cl}_2]\right)$ will decrease as a result of more space being available for the same amount of gas particles.

Q84

For the following reactions, equilibrium constants are given : S(s) + O2(g) ⇋ SO2(g); K1 = 1052 2S(s) + 3O2(g) ⇋ 2SO3(g); K2 = 10129 The equilibrium constant for the reaction, 2SO2(g) + O2(g) ⇋ 2SO3(g) is :

A 10181

B 1025

C 1077

D 10154

Correct Answer

Option B

Solution

S(s) + O2(g) ⇋ SO2(g); K1 = 1052 By reversing the equation, we get SO2(g); ⇋ S(s) + O2(g) ;

{1 \over {{K_1}}} = {1 \over {{{10}^{52}}}}

..............(

1) 2S(s) + 3O2(g) ⇋ 2SO3(g); K2 = 10129 ....................(

2) Performing (2) - 2 $\times$ (1), we get 2SO2(g) + O2(g) ⇋ 2SO3(g) $\therefore$ Equilibrium constant of this reaction is =

{{{K_2}} \over {K_1^2}}

=

{{{{10}^{129}}} \over {{{\left( {{{10}^{52}}} \right)}^2}}}

= 1025

Q85

The exothermic formation of ClF3 is represented by the equation: Cl2 (g) + 3F2 (g)

\leftrightharpoons

2ClF3 (g);

\Delta H

= -329 kJ Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3?

A Increasing the temperature

B Removing Cl2

C Increasing the volume of the container

D Adding F2

Correct Answer

Option D

Solution

The reaction given is an exothermic reaction thus accordingly to Lechatalier's principle lowering of temperature, addition of

{F_2}

and or

C{l_2}

favour the for ward direction and hence the production of

Cl{F_3}.

Q86

An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm. Pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH4HS decomposition at this temperature is :

A 0.30

B 0.11

C 0.17

D 0.18

Correct Answer

Option B

Solution

\mathop {N{H_4}HS\left( s \right)\,\rightleftharpoons\,}\limits_{\begin{array}{ll}{start} \\ {At\,\,equib.} \end{array}} \,\,\mathop {N{H_3}\left( g \right)}\limits_{\begin{array}{ll}{0.5\,\,atm} \\ {0.5 + x\,\,atm} \end{array}} \,\, + \mathop {{H_2}S\left( g \right)}\limits_{\begin{array}{ll}{0\,\,atm} \\ {x\,\,atm} \end{array}}

Then

0.5 + x + x = 2x + 0.5 = 0.84\,\,

(Given)

\Rightarrow x = 0.17\,\,atm.

{P_{N{H_3}}} = 0.5 + 0.17 = 0.67\,atm;

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

{P_H}_{_2S} = 0.17\,atm

K = {P_{N{H_3}}} \times {P_{{H_2}S}}

= 0.67 \times 0.17\,at{m^2}

= 0.1139 = 0.11

Q87

Consider the reaction

\mathrm{X}_2 \mathrm{Y}(\mathrm{~g}) \rightleftharpoons \mathrm{X}_2(\mathrm{~g})+\dfrac{1}{2} \mathrm{Y}_2(\mathrm{~g})

The equation representing correct relationship between the degree of dissociation (x) of

\mathrm{X}_2 \mathrm{Y}(\mathrm{g})

with its equilibrium constant Kp is __________. Assume

x

to be very very small.

A

x=\sqrt[3]{\dfrac{\mathrm{Kp}}{\mathrm{p}}}

B

x=\sqrt[3]{\dfrac{\mathrm{Kp}}{2 \mathrm{p}}}

C

x=\sqrt[3]{\dfrac{2 \mathrm{Kp}^2}{\mathrm{p}}}

D

x=\sqrt[3]{\dfrac{2 \mathrm{Kp}}{\mathrm{p}}}

Correct Answer

Option C

Solution

To determine the relationship between the degree of dissociation ( $x$ ) of $\mathrm{X}_2 \mathrm{Y}(\mathrm{g})$ and its equilibrium constant $K_p$ , let's analyze the reaction: $\mathrm{X}_2 \mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{X}_2(\mathrm{g}) + \dfrac{1}{2} \mathrm{Y}_2(\mathrm{g})$ Initial and Change in Moles Initially, we start with 1 mole of $\mathrm{X}_2 \mathrm{Y}$ .

At equilibrium: $\mathrm{X}_2 \mathrm{Y}$ is $(1-x)$ moles. $\mathrm{X}_2$ is $x$ moles. $\dfrac{1}{2} \mathrm{Y}_2$ is $\dfrac{x}{2}$ moles.

Partial Pressures The total pressure $\mathrm{P}_\text{total}$ is given by: $\mathrm{P}_{\mathrm{X}_2 \mathrm{Y}} = \dfrac{1-x}{1+\dfrac{x}{2}} \times \mathrm{P}$ $\mathrm{P}_{\mathrm{X}_2} = \dfrac{x}{1+\dfrac{x}{2}} \times \mathrm{P}$ $\mathrm{P}_{\mathrm{Y}_2} = \dfrac{x/2}{1+\dfrac{x}{2}} \times \mathrm{P}$ Equilibrium Constant Expression The equilibrium constant $K_p$ can be written in terms of partial pressures: $K_p = \dfrac{\left(\dfrac{x}{1+\dfrac{x}{2}} \cdot \mathrm{P}\right) \cdot \left(\dfrac{\dfrac{x}{2}}{1+\dfrac{x}{2}} \cdot \mathrm{P}\right)^{1/2}}{\dfrac{1-x}{1+\dfrac{x}{2}} \cdot \mathrm{P}}$ Simplifying the expression, considering $x$ to be very small, results in: $K_p = \left(\dfrac{x}{1-x}\right)\left(\dfrac{x}{2\left(1+\dfrac{x}{2}\right)}\right)^{1/2} \cdot \mathrm{P}^{\dfrac{1}{2}}$ For small $x$ , $(1-x) \approx 1$ , thus: $K_p = \dfrac{x^{3/2}}{\dfrac{1}{3}} \cdot \mathrm{P}^{\dfrac{1}{2}}$ $x^{3/2} = \dfrac{K_p \cdot 2^{1/2}}{\mathrm{P}^{1/2}}$ Finally, by cubing both sides: $x^3 = \dfrac{2 \cdot K_p^2}{\mathrm{P}}$ $x = \left(\dfrac{2 \cdot K_p^2}{\mathrm{P}}\right)^{1/3}$ Thus, the degree of dissociation $x$ relates to $K_p$ by the equation: $x = \left(\dfrac{2 \cdot K_p^2}{\mathrm{P}}\right)^{1/3}$

Q88

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}

Q89

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.

Q90

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}