Chemical Equilibrium Questions — JEE Chemistry

Q1

At temperature T, compound

AB_{2(g)}

dissociates as

AB_{2(g)} \rightleftharpoons AB_{(g)} + \dfrac{1}{2} B_{2(g)}

having degree of dissociation

x

(small compared to unity). The correct expression for

x

in terms of

K_p

and

p

is:

A

\sqrt{K_p}

B

\sqrt[3]{\dfrac{2 K_{\mathrm{p}}^2}{\mathrm{p}}}

C

\sqrt[3]{\dfrac{2 K_p}{p}}

D

\sqrt[4]{\dfrac{2 K_p}{p}}

Correct Answer

Option B

Solution

$A B_2(g) \rightleftharpoons A B_{(g)}+\dfrac{1}{2} B_2({g})$ Degree of dissociation $\rightarrow x$ (small compared to unity) Equilibrium constant interms of pressure $\rightarrow k_p$ Partial pressure of each gas $\rightarrow P_{A B_2}, P_{A B}$ and $P_{B_2}$ Total pressure $\rightarrow P$ ICE Table: $\begin{aligned} & \text{ partial pressure } \\ & \text{ at equilibrium }\end{aligned}=\dfrac{\text{ moles at equilibrium }}{\begin{array}{l}\text{ total moles at } \\ \text{ equilibrium }\end{array}} \times$ total pressure Partial pressure of $A B_2(g)$

P_{A B_2}=\left(\frac{1-x}{1+\frac{x}{2}}\right) P

Partial pressure of $A B(g)$

P_{A B}=\left(\frac{x}{1+\frac{x}{2}}\right) P

Partial pressure of $B_2(g)$

P_{B_2}=\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}\right)^P

$K_p$ for the equation $A_2(g) \rightleftharpoons A B_{(g)}+\dfrac{1}{2} B_2(g)$ can be written as $k_p=\dfrac{P_{A B} P_{B_2}^{1 / 2}}{P_{A B_2}}$ $\begin{aligned} & =\dfrac{\dfrac{x}{1+\dfrac{x}{2}} P 1 \times\left(\dfrac{\dfrac{x}{2}}{1+\dfrac{x}{2}} P\right)^{1 / 2}}{\dfrac{1-x}{1+\dfrac{x}{2}} P} \\ & =\dfrac{x \times\left(\dfrac{x}{1+\dfrac{x}{2}} P\right)^{1 / 2}}{1-x}\end{aligned}$

\begin{aligned} &\text{ given, } x \lll 1\\ &\begin{aligned} & \text{ So, } 1-x \approx 1 \\ & K_p=\frac{x \times\left(\frac{\frac{x}{2}}{1+\frac{x}{2}} p\right)^{1 / 2}}{1} \end{aligned} \end{aligned}

$=x \times \dfrac{\left(\dfrac{x}{2}\right)^{1 / 2}}{\left(\dfrac{2+x}{2}\right)^{1 / 2}} p^{1 / 2}$ $=\dfrac{x \times \dfrac{x^{1 / 2}}{2^{1 / 2}} \times p^{1 / 2}}{\dfrac{(2+x)^{1 / 2}}{2^{1 / 2}}}$ $=\dfrac{x^{3 / 2} \times p^{1 / 2}}{2^{1 / 2}} \times \dfrac{2^{1 / 2}}{(2+x)^{1 / 2}}$

\begin{gathered} x \lll 1 \\ 2+x \approx 2 \end{gathered}

\begin{aligned} & =\frac{x^{3 / 2} \times p^{1 / 2}}{2^{1 / 2}} \\ & =\left(\frac{x^3 p}{2}\right)^{1 / 2} \end{aligned}

\begin{aligned} &k_p=\left(\frac{x^3 p}{2}\right)^{1 / 2}\\ &\text{ Square on both sides, } \end{aligned}

\begin{aligned} k_p^2 & =\frac{x^3 p}{2} \\ x^3 p & =2 k_p^2 \\ x^3 & =\frac{2 k_p^2}{p} \\ x & =\sqrt[3]{\frac{2 k_p^2}{p}} \end{aligned}

Correct answer option (2) $\sqrt[3]{\dfrac{2 K_p^2}{p}}$

Q2

For the following three reactions a, b and c, equilibrium constants are given: a. CO (g) + H2O (g)

\leftrightharpoons

CO2(g) + H2 (g) ; K1 b. CH4 (g) + H2O (g)

\leftrightharpoons

CO(g) + 3H2 (g) ; K2 c. CH4 (g) + 2H2O (g)

\leftrightharpoons

CO2(g) + 4H2 (g) ; K3

A

{K_1}\sqrt {{K_2}} = {K_3}

B K2K3 = K1

C K3 = K1K2

D K3.

K_2^3

=

K_1^2

Correct Answer

Option C

Solution

Reaction

(c)

can be obtained by adding reactions

(a)

and

(b)

therefore

{K_3} = {K_1}.{K_2}

Hence

(c)

is the correct answer.

Q3

For the reaction CO (g) + (1/2) O2 (g)

\leftrightharpoons

CO2 (g), Kp/Kc is :

A RT

B (RT)-1

C (RT)-1/2

D (RT)1/2

Correct Answer

Option C

Solution

{K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};

\Delta n = 1 - \left( {1 + {1 \over 2}} \right)

= 1 - {3 \over 2} = - {1 \over 2}.

$\therefore$

\,\,\,\,{{{K_p}} \over {{K_c}}} = {\left( {RT} \right)^{ - 1/2}}

Q4

In which one of the following equilibria, Kp

\ne

KC ?

A 2NO(g) ⇋ N2(g) + O2(g)

B 2C(s) + O2(g) ⇋ 2CO(g)

C 2HI(g) ⇋ H2(g) + I2(g)

D NO2(g) + SO2(g) ⇋ NO(g) + SO3(g)

Correct Answer

Option B

Solution

We know, Kp = KC(RT)

\Delta

ng Kp = KC when

\Delta

ng = 0 Kp $\ne$ KC when

\Delta

ng $\ne$ 0 and T $\ne$ 12 K In this reaction, 2C(s) + O2(g) ⇋ 2CO(g)

\Delta

ng = 1, so Kp $\ne$ KC

Q5

The equilibrium constant for the reaction

\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\dfrac{1}{2} \mathrm{O}_2(\mathrm{~g})

is

\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}

. The value of

\mathrm{K}_{\mathrm{c}}

for the reaction given below is

2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})

is :

A 49

B 416

C 41.6

D 4.9

Correct Answer

Option B

Solution

The reaction

2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3

can be formed from the given reaction by reverting it and multiplying coefficients by 2. Thus,

\begin{aligned} \mathrm{K}_c^{\prime} & =\mathrm{K}_{\mathrm{c}}^{-2}=\frac{1}{\mathrm{~K}_{\mathrm{c}}^2}=\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\ & =416 \end{aligned}

Q6

For the given hypothetical reactions, the equilibrium constants are as follows :

\begin{aligned} & \mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0 \\ & \mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0 \\ & \mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0 \end{aligned}

The equilibrium constant for the reaction

\mathrm{X} \rightleftharpoons \mathrm{W}

is

A 12.0

B 8.0

C 6.0

D 7.0

Correct Answer

Option B

Solution

To find the equilibrium constant for the overall reaction

\mathrm{X} \rightleftharpoons \mathrm{W}

, we need to combine the equilibrium constants for the individual reactions given.

Let's analyze this step-by-step: The given reactions and their equilibrium constants are:

\begin{aligned} & \mathrm{X} \rightleftharpoons \mathrm{Y} \quad K_1 = 1.0 \\ & \mathrm{Y} \rightleftharpoons \mathrm{Z} \quad K_2 = 2.0 \\ & \mathrm{Z} \rightleftharpoons \mathrm{W} \quad K_3 = 4.0 \end{aligned}

The equilibrium constant for the overall reaction

\mathrm{X} \rightleftharpoons \mathrm{W}

is the product of the equilibrium constants for the individual steps.

This is because the equilibrium constant for a composite reaction is the product of the equilibrium constants for the sequential reactions that lead to the composite reaction.

So, we have:

K_{\text{overall}} = K_1 \cdot K_2 \cdot K_3

Now, substituting the given values:

K_{\text{overall}} = 1.0 \cdot 2.0 \cdot 4.0 = 8.0

Therefore, the equilibrium constant for the reaction

\mathrm{X} \rightleftharpoons \mathrm{W}

is 8.0, which corresponds to Option B.

Q7

The ratio

\dfrac{K_P}{K_C}

for the reaction :

\mathrm{CO}_{(\mathrm{g})}+\dfrac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}

is :

A 1

B

(\mathrm{RT})^{1 / 2}

C RT

D

\mathrm{ \dfrac{1}{\sqrt{R T}}}

Correct Answer

Option D

Solution

To solve this problem, we need to understand the relationship between the equilibrium constant in terms of pressure,

K_P

, and the equilibrium constant in terms of concentration,

K_C

. The relationship between these two constants for a general reaction is given by:

K_P = K_C (RT)^{\Delta n}

where:

\Delta n

is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants),

R

is the ideal gas constant, and

T

is the temperature in Kelvin. For the given reaction:

\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}

we need to calculate

\Delta n

. On the reactants side, we have: 1 mole of

\mathrm{CO}

(g) 0.5 moles of

\mathrm{O}_2

(g) So, the total number of moles of reactants is:

1 + \frac{1}{2} = \frac{3}{2} = 1.5

On the products side, we have: 1 mole of

\mathrm{CO}_2

(g) The total number of moles of products is:

1

Therefore,

\Delta n

is:

\Delta n = \text{moles of products} - \text{moles of reactants} = 1 - 1.5 = -0.5

Now, we can use the relationship between

K_P

and

K_C

:

K_P = K_C (RT)^{\Delta n}

Substituting

\Delta n = -0.5

, we get:

K_P = K_C (RT)^{-0.5}

or equivalently:

\frac{K_P}{K_C} = (RT)^{-0.5}

Therefore, the correct option is: Option D:

\frac{1}{\sqrt{R T}}

Q8

An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules :

A leaves the solution increases

B leaves the vapour increases

C leaves the vapour decreases

D leaves the solution decreases

Correct Answer

Option D

Solution

With addition of solute in solvent, surface area for vapourisation decreases causes lowering in vapour pressure.

Q9

A vessel at 1000 K contains

\mathrm{CO}_2

with a pressure of 0.5 atm . Some of

\mathrm{CO}_2

is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm , then Kp is :

A 0.18 atm

B 0.3 atm

C 3 atm

D 1.8 atm

Correct Answer

Option D

Solution

The reaction is:

\mathrm{CO}_2(g) + \mathrm{C}(s) \rightleftharpoons 2\mathrm{CO}(g)

Initially, the pressure of $\mathrm{CO}_2$ is 0.5 atm, and the pressure of $\mathrm{CO}$ is 0 atm.

Let 'x' be the change in pressure of $\mathrm{CO}_2$ at equilibrium.

Since the stoichiometric coefficient of CO is twice that of $\mathrm{CO}_2$ , the pressure of CO formed at equilibrium is 2x.

At equilibrium: Pressure of $\mathrm{CO}_2 = 0.5 - x$ Pressure of $\mathrm{CO} = 2x$ The total pressure at equilibrium is given as 0.8 atm.

Therefore:

(0.5 - x) + 2x = 0.8

0.5 + x = 0.8

x = 0.8 - 0.5 = 0.3

So, at equilibrium: Pressure of $\mathrm{CO}_2 = 0.5 - 0.3 = 0.2 \text{ atm}$ Pressure of $\mathrm{CO} = 2(0.3) = 0.6 \text{ atm}$ The equilibrium constant $K_p$ is given by:

K_p = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}}

K_p = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \text{ atm}

Therefore, the value of $K_p$ is 1.8 atm.

Q10

Consider the following chemical equilibrium of the gas phase reaction at a constant temperature :

\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})

If

p

being the total pressure,

K_p

is the pressure equilibrium constant and

\alpha

is the degree of dissociation, then which of the following is true at equilibrium?

A If

K_p

value is extremely high compared to

p, \alpha

becomes much less than unity

B When p increases

\alpha

increases

C If p value is extremely high compared to

\mathrm{K}_{\mathrm{p}}, \alpha \approx 1

D When

p

increases

\alpha

decreases

Correct Answer

Option D

Solution

a moles of $A(g)$ taken initially and at time Now moles fraction of $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are

\begin{aligned} & \mathrm{X}_{\mathrm{A}}=\frac{\mathrm{a}-\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{1-\alpha}{1+\alpha} \\ & \mathrm{X}_{\mathrm{B}}=\frac{\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{\alpha}{1+\alpha} \\ & \mathrm{X}_{\mathrm{C}}=\frac{\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{\alpha}{1+\alpha} \end{aligned}

Now if P is total pressure then partial pressure of $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are

\begin{aligned} & \mathrm{P}_{\mathrm{A}}=\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P} \\ & \mathrm{P}_{\mathrm{B}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\ & \mathrm{P}_{\mathrm{C}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \end{aligned}

\begin{aligned} & \mathrm{K}_{\mathrm{P}}=\frac{\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}}{\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P}} \\ & \mathrm{~K}_{\mathrm{P}}=\frac{\alpha^2 \mathrm{P}}{1-\alpha^2} \end{aligned}

As $K_P$ is only function of temperature. So as $\mathrm{P} \uparrow \quad \alpha \downarrow$