Chemical Equilibrium — Page 3 | NEET Chemistry

Q21

For the reaction, N 2(g) + O 2(g)

\rightleftharpoons

2NO (g), the equilibrium constant is K 1 . The equilibrium constant is K 2 for the reaction, 2NO (g) + O 2(g)

\rightleftharpoons

2NO 2(g) What is K for the reaction, NO 2(g)

\rightleftharpoons

{1 \over 2}

N 2(g) + O 2(g)

A

{1 \over {2{K_1}{K_2}}}

B

{1 \over {4{K_1}{K_2}}}

C

{\left[ {{1 \over {{K_1}{K_2}}}} \right]^{1/2}}

D

{1 \over {{K_1}{K_2}}}

Correct Answer

Option C

Solution

N 2(g) +

{1 \over 2}

O 2(g) $\rightleftharpoons$ NO (g),

{\left( {{K_1}} \right)^{{1 \over 2}}}

NO (g) +

{1 \over 2}

O 2(g) $\rightleftharpoons$ NO 2(g) ,

{\left( {{K_2}} \right)^{{1 \over 2}}}

---------------------------------------------------

{1 \over 2}

N 2(g) + O 2(g) $\rightleftharpoons$ NO 2(g)

{\left( {{K_1}{K_2}} \right)^{{1 \over 2}}}

So, for the reverse reaction which is the desired one the value of K will be reciprocal of this value. i.e. K =

{\left[ {{1 \over {{K_1}{K_2}}}} \right]^{1/2}}

Q22

The value of

\Delta

H for the reaction X 2(g) + 4Y 2(g)

\rightleftharpoons

2XY 4(g) is less than zero. Formation of XY 4(g) will be favoured at

A high temperature and high pressure

B low pressure and low temperature

C high temperature and low pressure

D high pressure and low temperature

Correct Answer

Option D

Solution

X 2(g) + 4Y 2(g) $\rightleftharpoons$ 2XY 4(g)

\Delta

n g = -ve and

\Delta

H = -ve As

\Delta

H < 0 i.e., the given reaction is exothermic.

According to Le-Chatelier principle, for exothermic reaction, forward reaction is favoured when temperature becomes low.

Also, there are 5 gaseous moles on reactant side and 2 gaseous moles on products side.

So, forward reaction is favoured when pressure of the reaction mixture becomes high.

The reason is that at high pressure reaction tends to more in direction where there is lessee number of gaseous moles.

Q23

The reaction, 2A (g) + B (g)

\rightleftharpoons

3C (g) + D (g) is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measuread and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression

A [(0.75) 3 (0.25)]

\div

[(1.00) 2 (1.00)]

B [(0.75) 3 (0.25)]

\div

[(0.50) 2 (0.75)]

C [(0.75) 3 (0.25)]

\div

[(0.50) 2 (0.25)]

D [(0.75) 3 (0.25)]

\div

[(0.75) 2 (0.25)]

Correct Answer

Option B

Solution

<table class=tg> <tbody><tr> <th class=tg-c3ow></th> <th class=tg-9wq8>2A(g)</th> <th class=tg-9wq8>+</th> <th class=tg-c3ow>B(g)</th> <th class=tg-baqh>⇌</th> <th class=tg-baqh>3C(g)</th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>D(g)</th> </tr> <tr> <td class=tg-c3ow>Initial mole</td> <td class=tg-9wq8>1</td> <td class=tg-9wq8></td> <td class=tg-c3ow>1</td> <td class=tg-baqh></td> <td class=tg-baqh>0</td> <td class=tg-c3ow></td> <td class=tg-c3ow>0</td> </tr> <tr> <td class=tg-baqh>At equilibrium</td> <td class=tg-baqh>1 - (2 $\times$ 0.25)<br>= 0.5</td> <td class=tg-baqh></td> <td class=tg-baqh>1 - 0.25<br>= 0.75</td> <td class=tg-baqh></td> <td class=tg-baqh>3 $\times$ 0.25<br>= 0.75</td> <td class=tg-baqh></td> <td class=tg-baqh>0.25</td> </tr> </tbody></table> <br><br>Equilibrium constant, K =

{{{{\left[ C \right]}^3}\left[ D \right]} \over {{{\left[ A \right]}^2}\left[ B \right]}}

<br><br> $\therefore$ K =

{{{{\left[ {0.75} \right]}^3}\left[ {0.25} \right]} \over {{{\left[ {0.5} \right]}^2}\left[ {0.75} \right]}}

Q24

In which of the following equilibrium K c and K p are not equal?

A 2NO (g)

\rightleftharpoons

N 2(g) + O 2(g)

B SO 2(g) + NO 2(g)

\rightleftharpoons

SO 3(g) + NO (g)

C H 2(g) + I 2(g)

\rightleftharpoons

2HI (g)

D 2C (s) + O 2(g)

\rightleftharpoons

2CO 2(g)

Correct Answer

Option D

Solution

As we know, K p = K c × (RT)

\Delta

n g So, for reaction having same number of gaseous moles on reactants and products side will have same value of Kc and Kp otherwise their values are different.

For reaction, 2NO (g) $\rightleftharpoons$ N 2(g) + O 2(g)

\Delta

n g = 2 – 2 = 0 $\Rightarrow$ K p = K c × (RT) 0 $\therefore$ K p = K c For reaction, SO 2(g) + NO 2(g) $\rightleftharpoons$ SO 3(g) + NO (g)

\Delta

n g = 2 – 2 = 0 $\Rightarrow$ K p = K c × (RT) 0 $\therefore$ K p = K c For reaction, H 2(g) + I 2(g) $\rightleftharpoons$ 2HI (g)

\Delta

n g = 2 – 2 = 0 $\Rightarrow$ K p = K c × (RT) 0 $\therefore$ K p = K c For reaction, H 2(g) + I 2(g) $\rightleftharpoons$ 2HI (g)

\Delta

n g = 2 – 2 = 0 $\Rightarrow$ K p = K c × (RT) 0 $\therefore$ K p = K c For reaction, 2C (s) + O 2(g) $\rightleftharpoons$ 2CO 2(g)

\Delta

n g = 2 – 3 = -1 $\Rightarrow$ K p = K c × (RT) -1 $\therefore$ K p $\ne$ K c

Q25

The dissociation constants for acetic acid and HCN at 25 o C are 1.5

\times

10

-

5 and 4.5

\times

10

-

10 respectively. The equilibrium constant for the equilibrium CN

-

+ CH 3 COOH

\rightleftharpoons

HCN + CH 3 COO

-

would be

A 3.0

\times

10

-

5

B 3.0

\times

10

-

4

C 3.0

\times

10 4

D 3.0

\times

10 5

Correct Answer

Option C

Solution

CH 3 COOH $\rightleftharpoons$ CH 3 COO – + H + , K 1 = 1.5 $\times$ 10 $-$ 5 $\Rightarrow$ K 1 =

{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {C{H_3}COOH} \right]}}

= 1.5 $\times$ 10 $-$ 5 HCN $\rightleftharpoons$ CN – + H + , K 2 = 4.5 × 10 –10 K 2 =

{{\left[ {C{N^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {HCN} \right]}}

= 4.5 × 10 –10 CN $-$ + CH 3 COOH $\rightleftharpoons$ HCN + CH 3 COO $-$ K =

{{\left[ {HCN} \right]\left[ {C{H_3}CO{O^ - }} \right]} \over {\left[ {C{N^ - }} \right]\left[ {C{H_3}COOH} \right]}}

$\Rightarrow$ K =

{{{K_1}} \over {{K_2}}}

=

{{1.5 \times {{10}^{ - 5}}} \over {4.5 \times {{10}^{ - 10}}}}

= 3.33 × 10 4

\simeq

3.0 × 10 4

Q26

The values of for the reactions, X

\rightleftharpoons

Y + Z . . . .(i) A

\rightleftharpoons

2B . . . .(ii) are in the ratio 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (i) and (ii) are in the ratio

A 36 : 1

B 1 : 1

C 3 : 1

D 1 : 9

Correct Answer

Option A

Solution

Given <br><br>X $\rightleftharpoons$ Y + Z . . . .(i) <br><br>A $\rightleftharpoons$ 2B . . . .(ii) <br><br>Let the total pressure for reaction (i) and (ii) be P<sub>1</sub> and P<sub>2</sub> respectively, then <br><br>

{{{K_{{P_1}}}} \over {{K_{{P_2}}}}} = {9 \over 1}

<br><br> <table class=tg> <tbody><tr> <th class=tg-c3ow></th> <th class=tg-9wq8>X</th> <th class=tg-c3ow>⇌ </th> <th class=tg-c3ow>Y</th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>Z</th> </tr> <tr> <td class=tg-c3ow>Initial mole</td> <td class=tg-9wq8>1</td> <td class=tg-c3ow></td> <td class=tg-c3ow>0</td> <td class=tg-c3ow></td> <td class=tg-c3ow>0</td> </tr> <tr> <td class=tg-c3ow>At equilibrium</td> <td class=tg-c3ow>1 - $\alpha$ </td> <td class=tg-c3ow></td> <td class=tg-c3ow> $\alpha$ </td> <td class=tg-c3ow></td> <td class=tg-c3ow> $\alpha$ </td> </tr> </tbody></table> <br><br>Total number of moles at equilibrium <br><br>= 1 - $\alpha$ + $\alpha$ + $\alpha$ = 1 + $\alpha$ <br><br> $\therefore$ K<sub>P<sub>1</sub></sub> =

{{{P_Y} \times {P_Z}} \over {{P_X}}}

=

{{{\alpha \over {1 + \alpha }} \times {P_1} \times {\alpha \over {1 + \alpha }} \times {P_1}} \over {{{1 - \alpha } \over {1 + \alpha }} \times {P_1}}}

<br><br> <table class=tg> <tbody><tr> <th class=tg-c3ow></th> <th class=tg-9wq8>A</th> <th class=tg-c3ow>⇌</th> <th class=tg-c3ow>2B</th> </tr> <tr> <td class=tg-c3ow>Initial mole</td> <td class=tg-9wq8>1</td> <td class=tg-c3ow></td> <td class=tg-c3ow>0</td> </tr> <tr> <td class=tg-c3ow>At equilibrium</td> <td class=tg-c3ow>1 - $\alpha$ </td> <td class=tg-c3ow></td> <td class=tg-c3ow>2 $\alpha$ </td> </tr> </tbody></table> <br><br>Total number of moles at equilibrium <br><br>= 1 - $\alpha$ + 2 $\alpha$ = 1 + $\alpha$ <br><br> $\therefore$ K<sub>P<sub>2</sub></sub> =

{{{{\left( {{P_B}} \right)}^2}} \over {{P_A}}}

=

{{{{\left( {{{2\alpha } \over {1 + \alpha }} \times {P_2}} \right)}^2}} \over {{{1 - \alpha } \over {1 + \alpha }} \times {P_2}}}

<br><br> $\therefore$

{{{K_{{P_1}}}} \over {{K_{{P_2}}}}} = {{{P_1}} \over {4{P_2}}}

<br><br> $\Rightarrow$

{{{P_1}} \over {4{P_2}}} = {9 \over 1}

<br><br> $\Rightarrow$

{{{P_1}} \over {{P_2}}} = {{36} \over 1}

Q27

The value of equilibrium constant of the reaction HI (g)

\rightleftharpoons

{1 \over 2}

H 2(g) +

{1 \over 2}

I 2(g) is 8.0. The The equilibrium constant of the reaction H 2(g) + I 2(g)

\rightleftharpoons

2HI (g) will be

A 16

B 1/8

C 1/16

D 1/64

Correct Answer

Option D

Solution

HI (g) $\rightleftharpoons$

{1 \over 2}

H 2(g) +

{1 \over 2}

I 2(g) , For this reaction, K = 8.0 Reversing the equation, $\therefore$

{1 \over 2}

H 2(g) +

{1 \over 2}

I 2(g) $\rightleftharpoons$ HI (g) ......(1) For this reaction, K 1 =

{1 \over 8}

H 2(g) + I 2(g) $\rightleftharpoons$ 2HI (g) We get this equation by multiplying equation (1) by 2, $\therefore$ For this reaction, K 2 =

{\left( {{1 \over 8}} \right)^2}

=

{1 \over {64}}

Q28

The dissociation equilibrium of a gass AB 2 can be represented as : 2AB 2(g)

\rightleftharpoons

2AB (g) + B 2(g) The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K p and total pressure P is

A (2K p /P) 1/2

B (K p /P)

C (2K p /P)

D (2K p /P) 1/3

Correct Answer

Option D

Solution

<table class=tg> <tbody><tr> <th class=tg-c3ow></th> <th class=tg-9wq8>2AB2(g)</th> <th class=tg-c3ow>⇌</th> <th class=tg-c3ow>2AB(g)</th> <th class=tg-c3ow>+</th> <th class=tg-c3ow>B2(g)</th> </tr> <tr> <td class=tg-c3ow>Initial mole</td> <td class=tg-9wq8>2</td> <td class=tg-c3ow></td> <td class=tg-c3ow>0</td> <td class=tg-c3ow></td> <td class=tg-c3ow>0</td> </tr> <tr> <td class=tg-c3ow>At equilibrium</td> <td class=tg-c3ow>2(1 - x)</td> <td class=tg-c3ow></td> <td class=tg-c3ow>2x</td> <td class=tg-c3ow></td> <td class=tg-c3ow>x</td> </tr> </tbody></table> <br><br>Amount of moles at equilibrium = 2(1 – x) + 2x + x = 2 + x <br><br>

{K_p} = {{{{\left[ {{p_{AB}}} \right]}^2}\left[ {{p_{{B_2}}}} \right]} \over {{{\left[ {{p_{A{B_2}}}} \right]}^2}}}

<br><br>=

{{{{\left[ {{{2x} \over {2 + x}} \times P} \right]}^2}\left[ {{x \over {2 + x}} \times P} \right]} \over {{{\left[ {{{2\left( {1 - x} \right)} \over {2 + x}} \times P} \right]}^2}}}

<br><br>=

{{\left[ {{{4{x^3}} \over {2 + x}} \times P} \right]} \over {4{{\left( {1 - x} \right)}^2}}}

<br><br> $\Rightarrow$ K<sub>p</sub> =

{{{4{x^3} \times P} \over 2} \times {1 \over 4}}

<br><br>( $\because$ 1 – x ≈ 1 and 2 + x ≈ 2) <br><br> $\Rightarrow$ x =

{\left( {{{8{K_p}} \over {4P}}} \right)^{{1 \over 3}}}

<br><br> $\Rightarrow$ x =

{\left( {{{2{K_p}} \over P}} \right)^{{1 \over 3}}}

Q29

If the concentration of OH

-

ions in the reaction Fe(OH) 3(s)

\rightleftharpoons

Fe 3+ (aq) + 3OH

-

(aq) is decreased by 1/4 times, then equilibrium concentration of Fe 3+ will increase by

A 64 times

B 4 times

C 8 times

D 16 times

Correct Answer

Option A

Solution

Fe(OH) 3(s) $\rightleftharpoons$ Fe 3+ (aq) + 3OH $-$ (aq) Equilibrium constant, K c =

{{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}} \over {\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}

We know, equilibrium constant remains same at constant temperature.

Now, let the increase in concentration of Fe 3+ be x times.

$\therefore$ K c =

{{\left[ {x \times F{e^{3 + }}} \right]{{\left[ {{1 \over 4} \times O{H^ - }} \right]}^3}} \over {\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}

=

{x \over {64}}{{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}} \over {\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}

$\Rightarrow$ K c =

{x \over {64}}

K c $\Rightarrow$

{x \over {64}}

= 1 $\Rightarrow$ x = 64

Q30

The equilibrium constants of the following are N 2 + 3H 2

\rightleftharpoons

2NH 3 ; K 1 N 2 + O 2

\rightleftharpoons

2NO ; K 2 H 2 +

{1 \over 2}

O 2

\rightleftharpoons

H 2 O; K 3 The equilibrium constant (K) of the reaction : 2NH 3 +

{5 \over 2}

O 2

\rightleftharpoons

2NO + 3H 2 O will be

A K 2 K 3 3 /K 1

B K 2 K 3 /K 1

C K 2 3 K 3 /K 1

D K 1 K 3 3 /K 2

Correct Answer

Option A

Solution

2NH 3 $\rightleftharpoons$ N 2 + 3H 2 ;

{1 \over {{K_1}}}

N 2 + O 2 $\rightleftharpoons$ 2NO ; K 2 3H 2 +

{3 \over 2}

O 2 $\rightleftharpoons$ 3H 2 O; (K 3 ) 3 By adding all equations, we get 2NH 3 +

{5 \over 2}

O 2 $\rightleftharpoons$ 2NO + 3H 2 O $\therefore$ K =

{{{K_2} \times {{\left( {{K_3}} \right)}^3}} \over {{K_1}}}