QBManager

Chemical Equilibrium

229088 At a fixed concentration, the molar conductance of aqueous sodium hydroxide at $298 K$ is found to be $10 {ohm}^{- 1} {cm}^{2}, {mol}^{- 1}$ . At infinite dilution nd 298 $K$ , its molar conductance is found to be $242 {ohm}^{-}$ $1, {cm}^{2}, {mol}^{- 1}$ . Find the degree of ionization of sodium hydroxide at the same concentration and temperature.

1

2.06 %

2

4.13 %

3

0.04 %

4

10 %

Explanation:

Degree of dissociation $(α) =$
$= \frac{molar conductivity at conc. (Λ_{m}^{c})}{molar conductivity at infinite dilution (Λ_{m}^{\infty})}$
$\begin{aligned} (α) & = \frac{10 Ω^{- 1} {cm}^{2} {mol}^{- 1}}{242 Ω^{- 1} {cm}^{2} {mol}^{- 1}} \\ = 0.0413 \\ = 4.13 % \end{aligned}$

Shift-I

Chemical Equilibrium

229089 At $298 K$ , the ratio of dissociated water to that of undissociated water is [Assume pure water]

1

1 \times 10^{- 7}

2

1.8 \times 10^{- 9}

3

1 \times 10^{- 3}

4 1,000

Explanation:

Assuming pure water $(c) = 55.55$
$\begin{array}{cccc} H_{2} O & ⇌ & H^{+} & + & {OH}^{-} \\ Initially & c & 0 & 0 \\ At eq. & (c - α) & α & 0.01 \\ (55.55 - 10^{- 7}) & (10^{- 7}) & (10^{- 7}) \end{array}$
$\begin{aligned} Undissociated water & = 55.55 - 10^{- 7} (∵ 10^{- 7} \approx 0) \\ = 55.55 \end{aligned}$
Dissociated water $= 10^{- 7}$
$∴$ The ratio of dissociated water to that of undissociated
$\begin{aligned} water = \frac{Dissociated water}{Undissociated water} \\ = \frac{10^{- 7}}{55.55} = 1.8 \times 10^{- 9} \end{aligned}$

TS-EAMCET 09.08.2021

Chemical Equilibrium

229090 The successive equilibrium constant for the stepwise dissociation of a tribasic acid are $K_{1}$ , $K_{2}$ , and $K_{3}$ , respectively. The equilibrium constant for the overall dissociation is

1

(K_{1} + K_{2} + K_{3})

2

\sqrt[3]{(K_{1} + K_{2} + K_{3})}

3

{(K_{1} \times K_{2} \times K_{3})}^{3}

4

K_{1} \times K_{2} \times K_{3}

Explanation:

The equilibrium constant at first dissociation constant-
Since, $H_{3} A ⇌ H_{2} A^{-} + H^{+}$
$K_{1} = \frac{[H_{2} A^{-}] [H^{+}]}{[H_{3} A]}$
$H_{2} A^{-} ⇌ {HA}^{2 -} + H^{+}$
$K_{2} = \frac{[{HA}^{2 -}] [H^{+}]}{[H_{3} A^{-}]}$
$\underset{―}{{HA}^{2 -}^{2 -} E^{K} e A^{3 -} + H^{+}}$
$K_{3} = \frac{[A^{3 -}] [H^{+}]}{[{HA}^{2 -}]}$
Thus, $H_{3} A ⇌ K^{K} A A^{3 -} + 3 H^{+}$
$K_{a} = \frac{[A^{- 3}] {[H^{+}]}^{3}}{[H_{3} A]}$
Thus, equilibrium constant for the over all dissociation.
$K_{a} = K_{1} \times K_{2} \times K_{3}$

TS-EAMCET (Engg.)

Chemical Equilibrium

229091 Match the following
| List-I (Acid)| List-II ( $K_{a}$ ) (ionization constant) |
|----|----|
|

1 A- I,B- III,C - IV,D- V

2 A- V,B- II,C - III,D- IV

3 A- II,B- III,C - IV,D- V

4 A- III,B- IV,C - II,D- V

Explanation:

Ionization constant is a constant that depends upon the equilibrium between ions and molecules that are not ionized in a solution or liquid.
$\begin{array}{llc} Acid & K_{a} (Ionization constant) \\ (A) & HCN & 4.9 \times 10^{- 10} \\ (B) & H_{2} C_{2} O_{4} & 5.6 \times 10^{- 2} \\ (C) & H_{2} S & 8.9 \times 10^{- 8} \\ (D) & Niacin & 1.5 \times 10^{- 5} \end{array}$

Shift-I

Chemical Equilibrium

229092 If in the reaction, $N_{2} O_{4} ⇌ 2 {NO}_{2}; α$ is the degree of dissociation of $N_{2} O_{4}$ , then total number of moles at equilibrium is :

1

(1 - α)

2

(1 + α)

3

(1 - α)^{2}

4

(1 + α)^{2}

Explanation:

$\begin{aligned} N_{2} O_{4} & ⇌ & 2 {NO}_{2} \\ t = 0 & 1 & 0 \\ t = teq & 1 - α & 2 x \\ ∴ Total moles = 1 - α + 2 α \\ = (1 + α) \end{aligned}$

Manipal-2020

Chemical Equilibrium

229088 At a fixed concentration, the molar conductance of aqueous sodium hydroxide at $298 K$ is found to be $10 {ohm}^{- 1} {cm}^{2}, {mol}^{- 1}$ . At infinite dilution nd 298 $K$ , its molar conductance is found to be $242 {ohm}^{-}$ $1, {cm}^{2}, {mol}^{- 1}$ . Find the degree of ionization of sodium hydroxide at the same concentration and temperature.

1

2.06 %

2

4.13 %

3

0.04 %

4

10 %

Explanation:

Degree of dissociation $(α) =$
$= \frac{molar conductivity at conc. (Λ_{m}^{c})}{molar conductivity at infinite dilution (Λ_{m}^{\infty})}$
$\begin{aligned} (α) & = \frac{10 Ω^{- 1} {cm}^{2} {mol}^{- 1}}{242 Ω^{- 1} {cm}^{2} {mol}^{- 1}} \\ = 0.0413 \\ = 4.13 % \end{aligned}$

Shift-I

Chemical Equilibrium

229089 At $298 K$ , the ratio of dissociated water to that of undissociated water is [Assume pure water]

1

1 \times 10^{- 7}

2

1.8 \times 10^{- 9}

3

1 \times 10^{- 3}

4 1,000

Explanation:

Assuming pure water $(c) = 55.55$
$\begin{array}{cccc} H_{2} O & ⇌ & H^{+} & + & {OH}^{-} \\ Initially & c & 0 & 0 \\ At eq. & (c - α) & α & 0.01 \\ (55.55 - 10^{- 7}) & (10^{- 7}) & (10^{- 7}) \end{array}$
$\begin{aligned} Undissociated water & = 55.55 - 10^{- 7} (∵ 10^{- 7} \approx 0) \\ = 55.55 \end{aligned}$
Dissociated water $= 10^{- 7}$
$∴$ The ratio of dissociated water to that of undissociated
$\begin{aligned} water = \frac{Dissociated water}{Undissociated water} \\ = \frac{10^{- 7}}{55.55} = 1.8 \times 10^{- 9} \end{aligned}$

TS-EAMCET 09.08.2021

Chemical Equilibrium

229090 The successive equilibrium constant for the stepwise dissociation of a tribasic acid are $K_{1}$ , $K_{2}$ , and $K_{3}$ , respectively. The equilibrium constant for the overall dissociation is

1

(K_{1} + K_{2} + K_{3})

2

\sqrt[3]{(K_{1} + K_{2} + K_{3})}

3

{(K_{1} \times K_{2} \times K_{3})}^{3}

4

K_{1} \times K_{2} \times K_{3}

Explanation:

The equilibrium constant at first dissociation constant-
Since, $H_{3} A ⇌ H_{2} A^{-} + H^{+}$
$K_{1} = \frac{[H_{2} A^{-}] [H^{+}]}{[H_{3} A]}$
$H_{2} A^{-} ⇌ {HA}^{2 -} + H^{+}$
$K_{2} = \frac{[{HA}^{2 -}] [H^{+}]}{[H_{3} A^{-}]}$
$\underset{―}{{HA}^{2 -}^{2 -} E^{K} e A^{3 -} + H^{+}}$
$K_{3} = \frac{[A^{3 -}] [H^{+}]}{[{HA}^{2 -}]}$
Thus, $H_{3} A ⇌ K^{K} A A^{3 -} + 3 H^{+}$
$K_{a} = \frac{[A^{- 3}] {[H^{+}]}^{3}}{[H_{3} A]}$
Thus, equilibrium constant for the over all dissociation.
$K_{a} = K_{1} \times K_{2} \times K_{3}$

TS-EAMCET (Engg.)

Chemical Equilibrium

229091 Match the following
| List-I (Acid)| List-II ( $K_{a}$ ) (ionization constant) |
|----|----|
|

1 A- I,B- III,C - IV,D- V

2 A- V,B- II,C - III,D- IV

3 A- II,B- III,C - IV,D- V

4 A- III,B- IV,C - II,D- V

Explanation:

Ionization constant is a constant that depends upon the equilibrium between ions and molecules that are not ionized in a solution or liquid.
$\begin{array}{llc} Acid & K_{a} (Ionization constant) \\ (A) & HCN & 4.9 \times 10^{- 10} \\ (B) & H_{2} C_{2} O_{4} & 5.6 \times 10^{- 2} \\ (C) & H_{2} S & 8.9 \times 10^{- 8} \\ (D) & Niacin & 1.5 \times 10^{- 5} \end{array}$

Shift-I

Chemical Equilibrium

229092 If in the reaction, $N_{2} O_{4} ⇌ 2 {NO}_{2}; α$ is the degree of dissociation of $N_{2} O_{4}$ , then total number of moles at equilibrium is :

1

(1 - α)

2

(1 + α)

3

(1 - α)^{2}

4

(1 + α)^{2}

Explanation:

$\begin{aligned} N_{2} O_{4} & ⇌ & 2 {NO}_{2} \\ t = 0 & 1 & 0 \\ t = teq & 1 - α & 2 x \\ ∴ Total moles = 1 - α + 2 α \\ = (1 + α) \end{aligned}$

Manipal-2020

Chemical Equilibrium

229088 At a fixed concentration, the molar conductance of aqueous sodium hydroxide at $298 K$ is found to be $10 {ohm}^{- 1} {cm}^{2}, {mol}^{- 1}$ . At infinite dilution nd 298 $K$ , its molar conductance is found to be $242 {ohm}^{-}$ $1, {cm}^{2}, {mol}^{- 1}$ . Find the degree of ionization of sodium hydroxide at the same concentration and temperature.

1

2.06 %

2

4.13 %

3

0.04 %

4

10 %

Explanation:

Degree of dissociation $(α) =$
$= \frac{molar conductivity at conc. (Λ_{m}^{c})}{molar conductivity at infinite dilution (Λ_{m}^{\infty})}$
$\begin{aligned} (α) & = \frac{10 Ω^{- 1} {cm}^{2} {mol}^{- 1}}{242 Ω^{- 1} {cm}^{2} {mol}^{- 1}} \\ = 0.0413 \\ = 4.13 % \end{aligned}$

Shift-I

Chemical Equilibrium

229089 At $298 K$ , the ratio of dissociated water to that of undissociated water is [Assume pure water]

1

1 \times 10^{- 7}

2

1.8 \times 10^{- 9}

3

1 \times 10^{- 3}

4 1,000

Explanation:

Assuming pure water $(c) = 55.55$
$\begin{array}{cccc} H_{2} O & ⇌ & H^{+} & + & {OH}^{-} \\ Initially & c & 0 & 0 \\ At eq. & (c - α) & α & 0.01 \\ (55.55 - 10^{- 7}) & (10^{- 7}) & (10^{- 7}) \end{array}$
$\begin{aligned} Undissociated water & = 55.55 - 10^{- 7} (∵ 10^{- 7} \approx 0) \\ = 55.55 \end{aligned}$
Dissociated water $= 10^{- 7}$
$∴$ The ratio of dissociated water to that of undissociated
$\begin{aligned} water = \frac{Dissociated water}{Undissociated water} \\ = \frac{10^{- 7}}{55.55} = 1.8 \times 10^{- 9} \end{aligned}$

TS-EAMCET 09.08.2021

Chemical Equilibrium

229090 The successive equilibrium constant for the stepwise dissociation of a tribasic acid are $K_{1}$ , $K_{2}$ , and $K_{3}$ , respectively. The equilibrium constant for the overall dissociation is

1

(K_{1} + K_{2} + K_{3})

2

\sqrt[3]{(K_{1} + K_{2} + K_{3})}

3

{(K_{1} \times K_{2} \times K_{3})}^{3}

4

K_{1} \times K_{2} \times K_{3}

Explanation:

The equilibrium constant at first dissociation constant-
Since, $H_{3} A ⇌ H_{2} A^{-} + H^{+}$
$K_{1} = \frac{[H_{2} A^{-}] [H^{+}]}{[H_{3} A]}$
$H_{2} A^{-} ⇌ {HA}^{2 -} + H^{+}$
$K_{2} = \frac{[{HA}^{2 -}] [H^{+}]}{[H_{3} A^{-}]}$
$\underset{―}{{HA}^{2 -}^{2 -} E^{K} e A^{3 -} + H^{+}}$
$K_{3} = \frac{[A^{3 -}] [H^{+}]}{[{HA}^{2 -}]}$
Thus, $H_{3} A ⇌ K^{K} A A^{3 -} + 3 H^{+}$
$K_{a} = \frac{[A^{- 3}] {[H^{+}]}^{3}}{[H_{3} A]}$
Thus, equilibrium constant for the over all dissociation.
$K_{a} = K_{1} \times K_{2} \times K_{3}$

TS-EAMCET (Engg.)

Chemical Equilibrium

229091 Match the following
| List-I (Acid)| List-II ( $K_{a}$ ) (ionization constant) |
|----|----|
|

1 A- I,B- III,C - IV,D- V

2 A- V,B- II,C - III,D- IV

3 A- II,B- III,C - IV,D- V

4 A- III,B- IV,C - II,D- V

Explanation:

Ionization constant is a constant that depends upon the equilibrium between ions and molecules that are not ionized in a solution or liquid.
$\begin{array}{llc} Acid & K_{a} (Ionization constant) \\ (A) & HCN & 4.9 \times 10^{- 10} \\ (B) & H_{2} C_{2} O_{4} & 5.6 \times 10^{- 2} \\ (C) & H_{2} S & 8.9 \times 10^{- 8} \\ (D) & Niacin & 1.5 \times 10^{- 5} \end{array}$

Shift-I

Chemical Equilibrium

229092 If in the reaction, $N_{2} O_{4} ⇌ 2 {NO}_{2}; α$ is the degree of dissociation of $N_{2} O_{4}$ , then total number of moles at equilibrium is :

1

(1 - α)

2

(1 + α)

3

(1 - α)^{2}

4

(1 + α)^{2}

Explanation:

$\begin{aligned} N_{2} O_{4} & ⇌ & 2 {NO}_{2} \\ t = 0 & 1 & 0 \\ t = teq & 1 - α & 2 x \\ ∴ Total moles = 1 - α + 2 α \\ = (1 + α) \end{aligned}$

Manipal-2020

Chemical Equilibrium

229088 At a fixed concentration, the molar conductance of aqueous sodium hydroxide at $298 K$ is found to be $10 {ohm}^{- 1} {cm}^{2}, {mol}^{- 1}$ . At infinite dilution nd 298 $K$ , its molar conductance is found to be $242 {ohm}^{-}$ $1, {cm}^{2}, {mol}^{- 1}$ . Find the degree of ionization of sodium hydroxide at the same concentration and temperature.

1

2.06 %

2

4.13 %

3

0.04 %

4

10 %

Explanation:

Degree of dissociation $(α) =$
$= \frac{molar conductivity at conc. (Λ_{m}^{c})}{molar conductivity at infinite dilution (Λ_{m}^{\infty})}$
$\begin{aligned} (α) & = \frac{10 Ω^{- 1} {cm}^{2} {mol}^{- 1}}{242 Ω^{- 1} {cm}^{2} {mol}^{- 1}} \\ = 0.0413 \\ = 4.13 % \end{aligned}$

Shift-I

Chemical Equilibrium

229089 At $298 K$ , the ratio of dissociated water to that of undissociated water is [Assume pure water]

1

1 \times 10^{- 7}

2

1.8 \times 10^{- 9}

3

1 \times 10^{- 3}

4 1,000

Explanation:

Assuming pure water $(c) = 55.55$
$\begin{array}{cccc} H_{2} O & ⇌ & H^{+} & + & {OH}^{-} \\ Initially & c & 0 & 0 \\ At eq. & (c - α) & α & 0.01 \\ (55.55 - 10^{- 7}) & (10^{- 7}) & (10^{- 7}) \end{array}$
$\begin{aligned} Undissociated water & = 55.55 - 10^{- 7} (∵ 10^{- 7} \approx 0) \\ = 55.55 \end{aligned}$
Dissociated water $= 10^{- 7}$
$∴$ The ratio of dissociated water to that of undissociated
$\begin{aligned} water = \frac{Dissociated water}{Undissociated water} \\ = \frac{10^{- 7}}{55.55} = 1.8 \times 10^{- 9} \end{aligned}$

TS-EAMCET 09.08.2021

Chemical Equilibrium

229090 The successive equilibrium constant for the stepwise dissociation of a tribasic acid are $K_{1}$ , $K_{2}$ , and $K_{3}$ , respectively. The equilibrium constant for the overall dissociation is

1

(K_{1} + K_{2} + K_{3})

2

\sqrt[3]{(K_{1} + K_{2} + K_{3})}

3

{(K_{1} \times K_{2} \times K_{3})}^{3}

4

K_{1} \times K_{2} \times K_{3}

Explanation:

The equilibrium constant at first dissociation constant-
Since, $H_{3} A ⇌ H_{2} A^{-} + H^{+}$
$K_{1} = \frac{[H_{2} A^{-}] [H^{+}]}{[H_{3} A]}$
$H_{2} A^{-} ⇌ {HA}^{2 -} + H^{+}$
$K_{2} = \frac{[{HA}^{2 -}] [H^{+}]}{[H_{3} A^{-}]}$
$\underset{―}{{HA}^{2 -}^{2 -} E^{K} e A^{3 -} + H^{+}}$
$K_{3} = \frac{[A^{3 -}] [H^{+}]}{[{HA}^{2 -}]}$
Thus, $H_{3} A ⇌ K^{K} A A^{3 -} + 3 H^{+}$
$K_{a} = \frac{[A^{- 3}] {[H^{+}]}^{3}}{[H_{3} A]}$
Thus, equilibrium constant for the over all dissociation.
$K_{a} = K_{1} \times K_{2} \times K_{3}$

TS-EAMCET (Engg.)

Chemical Equilibrium

229091 Match the following
| List-I (Acid)| List-II ( $K_{a}$ ) (ionization constant) |
|----|----|
|

1 A- I,B- III,C - IV,D- V

2 A- V,B- II,C - III,D- IV

3 A- II,B- III,C - IV,D- V

4 A- III,B- IV,C - II,D- V

Explanation:

Ionization constant is a constant that depends upon the equilibrium between ions and molecules that are not ionized in a solution or liquid.
$\begin{array}{llc} Acid & K_{a} (Ionization constant) \\ (A) & HCN & 4.9 \times 10^{- 10} \\ (B) & H_{2} C_{2} O_{4} & 5.6 \times 10^{- 2} \\ (C) & H_{2} S & 8.9 \times 10^{- 8} \\ (D) & Niacin & 1.5 \times 10^{- 5} \end{array}$

Shift-I

Chemical Equilibrium

229092 If in the reaction, $N_{2} O_{4} ⇌ 2 {NO}_{2}; α$ is the degree of dissociation of $N_{2} O_{4}$ , then total number of moles at equilibrium is :

1

(1 - α)

2

(1 + α)

3

(1 - α)^{2}

4

(1 + α)^{2}

Explanation:

$\begin{aligned} N_{2} O_{4} & ⇌ & 2 {NO}_{2} \\ t = 0 & 1 & 0 \\ t = teq & 1 - α & 2 x \\ ∴ Total moles = 1 - α + 2 α \\ = (1 + α) \end{aligned}$

Manipal-2020

Chemical Equilibrium

229088 At a fixed concentration, the molar conductance of aqueous sodium hydroxide at $298 K$ is found to be $10 {ohm}^{- 1} {cm}^{2}, {mol}^{- 1}$ . At infinite dilution nd 298 $K$ , its molar conductance is found to be $242 {ohm}^{-}$ $1, {cm}^{2}, {mol}^{- 1}$ . Find the degree of ionization of sodium hydroxide at the same concentration and temperature.

1

2.06 %

2

4.13 %

3

0.04 %

4

10 %

Explanation:

Degree of dissociation $(α) =$
$= \frac{molar conductivity at conc. (Λ_{m}^{c})}{molar conductivity at infinite dilution (Λ_{m}^{\infty})}$
$\begin{aligned} (α) & = \frac{10 Ω^{- 1} {cm}^{2} {mol}^{- 1}}{242 Ω^{- 1} {cm}^{2} {mol}^{- 1}} \\ = 0.0413 \\ = 4.13 % \end{aligned}$

Shift-I

Chemical Equilibrium

229089 At $298 K$ , the ratio of dissociated water to that of undissociated water is [Assume pure water]

1

1 \times 10^{- 7}

2

1.8 \times 10^{- 9}

3

1 \times 10^{- 3}

4 1,000

Explanation:

Assuming pure water $(c) = 55.55$
$\begin{array}{cccc} H_{2} O & ⇌ & H^{+} & + & {OH}^{-} \\ Initially & c & 0 & 0 \\ At eq. & (c - α) & α & 0.01 \\ (55.55 - 10^{- 7}) & (10^{- 7}) & (10^{- 7}) \end{array}$
$\begin{aligned} Undissociated water & = 55.55 - 10^{- 7} (∵ 10^{- 7} \approx 0) \\ = 55.55 \end{aligned}$
Dissociated water $= 10^{- 7}$
$∴$ The ratio of dissociated water to that of undissociated
$\begin{aligned} water = \frac{Dissociated water}{Undissociated water} \\ = \frac{10^{- 7}}{55.55} = 1.8 \times 10^{- 9} \end{aligned}$

TS-EAMCET 09.08.2021

Chemical Equilibrium

229090 The successive equilibrium constant for the stepwise dissociation of a tribasic acid are $K_{1}$ , $K_{2}$ , and $K_{3}$ , respectively. The equilibrium constant for the overall dissociation is

1

(K_{1} + K_{2} + K_{3})

2

\sqrt[3]{(K_{1} + K_{2} + K_{3})}

3

{(K_{1} \times K_{2} \times K_{3})}^{3}

4

K_{1} \times K_{2} \times K_{3}

Explanation:

The equilibrium constant at first dissociation constant-
Since, $H_{3} A ⇌ H_{2} A^{-} + H^{+}$
$K_{1} = \frac{[H_{2} A^{-}] [H^{+}]}{[H_{3} A]}$
$H_{2} A^{-} ⇌ {HA}^{2 -} + H^{+}$
$K_{2} = \frac{[{HA}^{2 -}] [H^{+}]}{[H_{3} A^{-}]}$
$\underset{―}{{HA}^{2 -}^{2 -} E^{K} e A^{3 -} + H^{+}}$
$K_{3} = \frac{[A^{3 -}] [H^{+}]}{[{HA}^{2 -}]}$
Thus, $H_{3} A ⇌ K^{K} A A^{3 -} + 3 H^{+}$
$K_{a} = \frac{[A^{- 3}] {[H^{+}]}^{3}}{[H_{3} A]}$
Thus, equilibrium constant for the over all dissociation.
$K_{a} = K_{1} \times K_{2} \times K_{3}$

TS-EAMCET (Engg.)

Chemical Equilibrium

229091 Match the following
| List-I (Acid)| List-II ( $K_{a}$ ) (ionization constant) |
|----|----|
|

1 A- I,B- III,C - IV,D- V

2 A- V,B- II,C - III,D- IV

3 A- II,B- III,C - IV,D- V

4 A- III,B- IV,C - II,D- V

Explanation:

Ionization constant is a constant that depends upon the equilibrium between ions and molecules that are not ionized in a solution or liquid.
$\begin{array}{llc} Acid & K_{a} (Ionization constant) \\ (A) & HCN & 4.9 \times 10^{- 10} \\ (B) & H_{2} C_{2} O_{4} & 5.6 \times 10^{- 2} \\ (C) & H_{2} S & 8.9 \times 10^{- 8} \\ (D) & Niacin & 1.5 \times 10^{- 5} \end{array}$

Shift-I

Chemical Equilibrium

229092 If in the reaction, $N_{2} O_{4} ⇌ 2 {NO}_{2}; α$ is the degree of dissociation of $N_{2} O_{4}$ , then total number of moles at equilibrium is :

1

(1 - α)

2

(1 + α)

3

(1 - α)^{2}

4

(1 + α)^{2}

Explanation:

$\begin{aligned} N_{2} O_{4} & ⇌ & 2 {NO}_{2} \\ t = 0 & 1 & 0 \\ t = teq & 1 - α & 2 x \\ ∴ Total moles = 1 - α + 2 α \\ = (1 + α) \end{aligned}$

Manipal-2020

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