Chemical Equilibrium — Page 4 | JEE Chemistry

Q31

A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :

A 3 atm

B 0.3 atm

C 0.18 atm

D 1.8 atm

Correct Answer

Option D

Solution

To determine the equilibrium constant $K$ , consider the equilibrium reaction: $\text{CO}_2 + \text{C (graphite)} \rightleftharpoons 2\text{CO}$ Initially, the pressure of $\text{CO}_2$ is 0.5 atm, and there is no $\text{CO}$ : Initial Pressure: $\text{CO}_2 = 0.5 \text{ atm}, \quad \text{CO} = 0$ At equilibrium, assume that $x$ atm of $\text{CO}_2$ is converted into $\text{CO}$ : Final Pressure: $\text{CO}_2 = 0.5 - x \text{ atm}, \quad \text{CO} = 2x \text{ atm}$ The total pressure at equilibrium is given as 0.8 atm: $0.5 - x + 2x = 0.5 + x = 0.8$ Solving for $x$ : $x = 0.8 - 0.5 = 0.3 \text{ atm}$ The equilibrium constant $K_p$ is calculated as: $K_p = \dfrac{(P_{\text{CO}})^2}{P_{\text{CO}_2}}$ Substitute the equilibrium pressures: $K_p = \dfrac{(2x)^2}{0.5 - x} = \dfrac{(2 \times 0.3)^2}{0.5 - 0.3} = \dfrac{0.6^2}{0.2}$ $K_p = \dfrac{0.36}{0.2} = 1.8 \text{ atm}$

Q32

For the reaction Fe2N(s) +

{3 \over 2}

H2(g) ⇌ 2Fe(s) + NH3(g)

A KC = Kp(RT)1/2

B KC = Kp(RT)-1/2

C KC = Kp(RT)

D KC = Kp(RT)3/2

Correct Answer

Option A

Solution

To determine the correct relationship between the equilibrium constants $K_C$ and $K_p$ for the reaction:

\text{Fe}_2\text{N(s)} + \frac{3}{2}\text{H}_2\text{(g)} \rightleftharpoons 2\text{Fe(s)} + \text{NH}_3\text{(g)}

First, consider the general relationship between $K_C$ and $K_p$ :

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

where $\Delta n$ is the change in the number of moles of gas between the products and the reactants, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin.

For the given reaction, calculate $\Delta n$ : Moles of gas in products: NH3(g) = 1 mole Moles of gas in reactants:

{3 \over 2}

H2(g) = 1.5 moles Therefore:

\Delta n = 1 - 1.5 = -0.5

Substitute this value into the equation relating $K_C$ and $K_p$ :

K_p = K_C(RT)^{-0.5}

Rearrange to express $K_C$ :

K_C = K_p(RT)^{0.5}

Thus, the correct option is: Option A:

K_C = K_p(RT)^{1/2}

Q33

Consider the following reaction: N2O4(g) ⇌ 2NO2(g);

\Delta

Ho = +58 kJ For each of the following cases (a, b), the direction in which the equilibrium shifts is : (a) Temperature is decreased. (b) Pressure is increased by adding N2 at constant T.

A (a) towards reactant, (b) towards product

B (a) towards reactant, (b) no change

C (a) towards product, (b) towards reactant

D (a) towards product, (b) no change

Correct Answer

Option B

Solution

$\because$ Given reaction is endothermic.

$\therefore$ On decreasing temperature backward reaction will be favoured.

On adding N2, pressure is increased at constant T, and volume would also be constant so no change is observed.

Q34

In aqueous solution the ionization constants for carbonic acid are K1 = 4.2 x 10–7 and K2 = 4.8 x 10–11 Select the correct statement for a saturated 0.034 M solution of the carbonic acid.

A The concentration of

CO_3^{2−}

is 0.034 M.

B The concentration of

CO_3^{2−}

is greater than that of

HCO_3^{−}

C The concentration of H+ and

HCO_3^−

are approximately equal.

D The concentration of H+ is double that of

CO_3^−

.

Correct Answer

Option C

Solution

\mathop {{H_2}C{O_3}\left( {aq} \right)}\limits_{0.034 - x} \, + \,{H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {HCO_3^ - \left( {aq} \right)}\limits_x \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_x

{K_1} = {{\left[ {HCO_3^ - } \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {{H_2}C{O_3}} \right]}}

= {{x \times x} \over {0.034 - x}}

\Rightarrow 4.2 \times {10^{ - 7}} = {{{x^2}} \over {0.034}}

\Rightarrow x = 1.195 \times {10^{ - 4}}

As

{H_2}C{O_3}

is a weak acid so the concentration of

{H_2}C{O_3}

will remain

0.034\,\,

as

0.034 > > x.

x = \left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right]

= 1.195 \times {10^{ - 4}}

Now,

\mathop {HCO_3^ - }\limits_{x - y} \left( {aq} \right) + {H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {CO_3^{2 - }\left( {aq} \right)}\limits_y \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_y

As

HCO_3^ - \,

is again a weak acid (weaker than

{H_2}C{O_3})

with

x > > y.

{K_2} = {{\left[ {CO_3^{{2^ - }}} \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {HCO_3^ - } \right]}}

= {{y \times \left( {x + y} \right)} \over {\left( {x - y} \right)}}

Note :

\left[ {{H_3}{O^ + }} \right] = {H^ + }\,\,

from first step

(x)

and from second step

\left( y \right) = \left( {x + y} \right)

[As

\,\,\,x > > y\,\,

so

\,\,\,x + y \simeq x\,\,\,

and

x - y \simeq x

] So,

\,\,\,{K_2} \simeq {{y \times x} \over x} = y

\Rightarrow {K_2} = 4.8 \times {10^{ - 11}}

= y = \left[ {CO_3^{{2^ - }}} \right]

So the concentration of

\,\,\,\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = \,\,\,

concentrations obtained from the first step.

As the dissociation will be very low in second step so there will be no change in these concentrations.

Thus the final concentrations are

\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}}

\& \,\,\,\left[ {CO_3^{2 - }} \right] = 4.8 \times {10^{ - 11}}

Q35

The equilibrium constants KP1 and KP2 for the reactions X

\leftrightharpoons

2Y and Z

\leftrightharpoons

P + Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at these equilibria is :

A 1 : 36

B 1 : 1

C 1 : 3

D 1 : 9

Correct Answer

Option A

Solution

Let the initial moles of

X

be

'a'

and that of

Z

be

'b'

the for the given reactions, we have

X\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,2Y

\begin{aligned}& Initial\,\,\,\,\,\,\,\,\,\,\,\,a\,\,moles\,\,\,\,\,\,\,0 \\ & At\,\,equi.\,\,\,\,\,\,\,a\left( {1 - \alpha } \right)\,\,\,\,\,\,\,2a\alpha \\ & (moles)\end{aligned}

Total no. of moles

= a\left( {1 - \alpha } \right) + 2a\alpha

= a - a\alpha + 2a\alpha

= a\left( {1 + \alpha } \right)

Now,

\,\,\,{K_{{p_1}}} = {{{{\left( {{n_y}} \right)}^2}} \over {{n_x}}} \times {\left( {{{{P_{{T_1}}}} \over {\sum n }}} \right)^{\Delta n}}

or,

\,\,\,{K_{{p_1}}} = {{{{\left( {2a\alpha } \right)}^2}.{P_{{T_1}}}} \over {\left[ {a\left( {1 - \alpha } \right)} \right]\left[ {a\left( {1 + \alpha } \right)} \right]}}

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

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

Z\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,P+Q

\begin{aligned}& Initial\,\,\,\,\,\,\,\,\,\,\,b\,\,moles\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0 \\ & Ar\,\,equi.\,\,\,\,\,\,b\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,b\alpha \,\,\,\,\,\,\,b\alpha \\ & (moles)\end{aligned}

Total no. of moles

= b\left( {1 - \alpha } \right) + b\alpha + b\alpha

= b - b\alpha + b\alpha + b\alpha

= b\left( {1 + \alpha } \right)

Now

\,\,\,{K_{{P_2}}} = {{{n_Q} \times {n_P}} \over {{n_z}}} \times {\left[ {{{{P_{{T_2}}}} \over {\sum\nolimits_n \, }}} \right]^{\Delta n}}

or

\,\,\,{K_{{P_2}}} = {{\left( {b\alpha } \right)\left( {b\alpha } \right).{P_{{T_2}}}} \over {\left[ {b\left( {1 - \alpha } \right)} \right]\left[ {b\left( {1 + \alpha } \right)} \right]}}

or

\,\,\,

\,\,\,{{{K_{{P_1}}}} \over {{K_{P2}}}} = {{4{\alpha ^2}.{P_{{T_1}}}} \over {\left( {1 - {\alpha ^2}} \right)}} \times {{{{\left( {1 - \alpha } \right)}^2}} \over {{P_{{T_2}}}.{\alpha ^2}}}

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

= {{4{P_{{T_1}}}} \over {{P_{{T_2}}}}}

or

\,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over 9}\,\,\,\,\,

[ as

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

given ] or

\,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over {36}}

or

\,\,\,1:36

Q36

What is the equilibrium expression for the reaction P4 (s) + 5O2

\leftrightharpoons

P4O10 (s)?

A Kc = [P4O10] / 5[P4] [O2]

B Kc = 1/[O2]5

C Kc = [P4O10] / [P4] [O2]5

D Kc = [O2]5

Correct Answer

Option B

Solution

For

{P_4}\left( s \right) + 5{O_2}\left( g \right)\,\rightleftharpoons\,{P_4}{O_{10}}\left( 8 \right)

{K_c} = {1 \over {{{\left( {{O_2}} \right)}^5}}}.

The solids have concentration unity

Q37

Change in volume of the system does not alter which of the following equilibria?

A N2(g) + O2(g)

\leftrightharpoons

2NO (g)

B PCl5(g)

\leftrightharpoons

PCl3 (g) + Cl2 (g)

C N2(g) + 3H2(g)

\leftrightharpoons

2NH3 (g)

D SO2Cl2 (g)

\leftrightharpoons

SO2 (g) + Cl2 (g)

Correct Answer

Option A

Solution

In this reaction the ratio of number of moles of reactants to products is same

i.e.\,\,2:2,

hence change in volume will not alter the number of moles.

Q38

For the reaction, 2SO2(g) + O2(g) = 2SO3(g),

\Delta

H = –57.2 kJ mol–1 and KC = 1.7 × 1016 Which of the following statement is incorrect ?

A The equilibrium will shift in forward direction as the pressure increase.

B The addition of inert gas at constant volume will be not affect the equilibrium constant.

C The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required.

D The equilibrium constant decreases as the temperature increase.

Correct Answer

Option C

Solution

Equilibrium constant has no relation with catalyst.

Catalyst only affects the rate with which a reaction proceeds.

Here we use catalyst V2O5 to speed up the reaction.

Q39

The equilibrium constant for the reaction N2(g) + O2(g)

\leftrightharpoons

2NO(g) at temperature T is 4

\times

10-4. The value of Kc for the reaction NO(g)

\leftrightharpoons

1 \over 2

N2 (g) +

1 \over 2

O2 (g) at the same temperature is :

A 2.5

\times

102

B 4

\times

10-4

C 50

D 0.02

Correct Answer

Option C

Solution

{K_c} = {{{{\left[ {NO} \right]}^2}} \over {\left[ {{N_2}} \right]\left[ {{O_2}} \right]}} = 4 \times {10^{ - 4}}

K{'_c} = {{{{\left[ {{N_2}} \right]}^{1/2}}{{\left[ {{Q_2}} \right]}^{1/2}}} \over {\left[ {NO} \right]}}

= {1 \over {\sqrt {{K_c}} }}

= {1 \over {\sqrt {4 \times {{10}^{ - 4}}} }}

= 50

Q40

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel PCl5 (g)

\leftrightharpoons

PCl3 (g) + Cl2 (g) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be

A

\left( {{x \over {x + 1}}} \right)P

B

\left( {{2x \over {1 - x}}} \right)P

C

\left( {{x \over {x - 1}}} \right)P

D

\left( {{x \over {1 - x}}} \right)P

Correct Answer

Option A

Solution

\mathop {PC{l_5}\left( g \right)}\limits_{1 - x} \mathop {\,\rightleftharpoons\,PC{l_3}\left( g \right)}\limits_x \,\, + \,\,\mathop {C{l_2}\left( g \right)}\limits_x

Total moles after dissociation

1 - x + x + x = 1 + x

{P_{PC{l_3}}} =

mole fraction of

PCl{}_3 \times

Total pressure

= \left( {{x \over {1 + x}}} \right)P