Chemical Equilibrium — Page 3 | JEE Chemistry

Q21

The equilibrium constant at 298 K for a reaction A + B

\leftrightharpoons

C + D is 100. If the initial concentration of all the four species were 1M each, then equilibrium concentration of D (in mol L–1) will be:

A 0.818

B 1.818

C 1.182

D 0.182

Correct Answer

Option B

Solution

Given, $\therefore$

\,\,\,{K_c} = {\left( {{{1 + a} \over {1 - a}}} \right)^2} = 100

$\therefore$

\,\,\,{{1 + a} \over {1 - a}} = 10

On solving

a=0.81

{\left[ D \right]_{At\,eq}} = 1 + a = 1 + 0.81 = 1.81

Q22

For the reaction 2NO2 (g)

\leftrightharpoons

2NO (g) + O2 (g), (Kc = 1.8

\times

10-6 at 184oC) (R = 0.0831 kJ/(mol. K)) When Kp and Kc are compared at 184oC , it is found that :

A Kp is greater than Kc

B Kp is less than Kc

C Kp = Kc

D Whether Kp is greater than, less than or equal to Kc depends upon the total gas pressure

Correct Answer

Option A

Solution

For the reaction : -

2N{O_2}\left( g \right)\rightleftharpoons2NO\left( g \right) + {O_2}\left( g \right)

Given

{K_c} = 1.8 \times {10^{ - 6}}\,\,

at

\,\,{184^ \circ }C

R=0.0831

\,\,kJ/mol.k

{K_p} = 1.8 \times {10^{ - 6}} \times 0.0831 \times 457

= 6.836 \times {10^{ - 6}}

\left[ {} \right.

as

\,\,\,\,\,{184^ \circ }C = \left( {273 + 184} \right) = 457\,k,\,\,

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. {\Delta n = \left( {2 + 1, - 1} \right) = 1} \right]

Hence it is clear that

{K_p} > {K_c}

Q23

A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T. The equilibrium pressure is 10 bar in this vessel. Kp for this reaction is :

A 5

B 10

C 25

D 100

Correct Answer

Option C

Solution

To determine the equilibrium constant, $K_p$ , for the decomposition reaction of the solid compound XY into its gaseous components X and Y, we need to consider the following reaction:

\text{XY (s)} \rightleftharpoons \text{X (g)} + \text{Y (g)}

For a reaction where a solid decomposes into gases, the equilibrium constant $K_p$ is defined in terms of the partial pressures of the gaseous products.

Since XY is a solid, its activity is considered to be 1 and does not appear in the equilibrium expression.

Hence the expression for $K_p$ is:

K_p = P_X \cdot P_Y

Where $P_X$ and $P_Y$ are the partial pressures of the gases X and Y, respectively.

Given that the total pressure at equilibrium is 10 bar, and assuming that X and Y are present in equal amounts due to the stoichiometry of the decomposition reaction (i.e., 1:1), we can write:

P_X = P_Y = \frac{10}{2} = 5 \, \text{bar}

Substituting these partial pressures into the expression for $K_p$ gives:

K_p = P_X \cdot P_Y = 5 \, \text{bar} \cdot 5 \, \text{bar} = 25 \, \text{bar}^2

Therefore, the equilibrium constant $K_p$ for this reaction is 25. Thus, the correct answer is: Option C - 25

Q24

\mathrm{A}_{(\mathrm{g})} \rightleftharpoons \mathrm{B}_{(\mathrm{g})}+\dfrac{\mathrm{C}}{2}(\mathrm{g})

The correct relationship between

\mathrm{K}_{\mathrm{P}}, \alpha

and equilibrium pressure

\mathrm{P}

is

A

K_P=\dfrac{\alpha^{1 / 2} P^{3 / 2}}{(2+\alpha)^{3 / 2}}

B

K_P=\dfrac{\alpha^{3 / 2} P^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}

C

K_P=\dfrac{\alpha^{1 / 2} P^{1 / 2}}{(2+\alpha)^{3 / 2}}

D

K_P=\dfrac{\alpha^{1 / 2} P^{1 / 2}}{(2+\alpha)^{1 / 2}}

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{A}_{(\mathrm{g})} \rightleftharpoons \mathrm{B}_{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{~g}) \\ & \mathrm{t}=\mathrm{t}_{\mathrm{eq}} \quad(1-\alpha) \quad \alpha \quad \frac{\alpha}{2} \end{aligned}

\mathrm{P}_{\mathrm{B}}=\frac{\alpha}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}_{\mathrm{A}}=\frac{(1-\alpha)}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}_{\mathrm{C}}=\frac{\frac{\alpha}{2}}{\left(1+\frac{\alpha}{2}\right)} . \mathrm{P}

\mathrm{K_P=\frac{P_B \cdot P_C^{\frac{1}{2}}}{P_A}}

\mathrm{=\frac{(\alpha)^{\frac{3}{2}}(P)^{\frac{1}{2}}}{(1-\alpha)(2+\alpha)^{\frac{1}{2}}}}

Q25

Consider the reaction equilibrium 2 SO2 (g) + O2 (g)

\leftrightharpoons

2 SO3 (g);

\Delta H^o

= -198 kJ One the basis of Le Chatelier's principle, the condition favourable for the forward reaction is :

A increasing temperature as well as pressure

B lowering the temperature and increasing the pressure

C any value of temperature and pressure

D lowering temperature as well as pressure

Correct Answer

Option B

Solution

Due to exothermicity of reaction low or optimum temperature will be required. Since

3

moles are changing to

2

moles. $\therefore$ High pressure will be required.

Q26

The standard Gibbs energy change at 300 K for the reaction 2A

\leftrightharpoons

B + C is 2494.2 J. At a given time, the composition of the reaction mixture is [A] = 1/2, [B] = 2 and [C] = 1/2. The reaction proceeds in the: [R = 8.314 J/K/mol, e = 2.718]

A reverse direction because Q > Kc

B forward direction because Q < Kc

C reverse direction because Q < Kc

D forward direction because Q > Kc

Correct Answer

Option A

Solution

\Delta {G^ \circ } = 2494.2J

2A\,\rightleftharpoons\,B + C

R = 8.314\,J/K/mol.

e = 2.718

\left[ A \right] = {1 \over 2},\left[ B \right] = 2,

\left[ C \right] = {1 \over 2};Q = {{\left[ B \right]\left[ C \right]} \over {{{\left[ A \right]}^2}}}

= {{2 \times 1/2} \over {{{\left( {{1 \over 2}} \right)}^2}}} = 4

\Delta {G^ \circ } = - 2.303\,\,RT\,\log \,{K_c}.

2494.2J = - 2.303 \times \left( {8.314J/K/mol} \right)

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left( {300K} \right)\log {K_c}

\Rightarrow \log \,{K_c}

= - {{2494.2\,J} \over {2.303 \times 8.314\,J/K/mol \times 300\,K}}

\Rightarrow \log \,{K_c} = - 0.4341;\,\,{K_c} = 0.37;\,\,Q > {K_c}.

Q27

The variation of equilibrium constant with temperature is given below : .tg .tg Temperature Equilibrium Constant T1 = 25oC K1 = 10 T2 = 100oC K2 = 100 The values of

\Delta

Ho,

\Delta

Go at T1 and

\Delta

Go at T2 (in kJ mol–1) respectively, are close to : [Use R = 8.314 J K–1 mol–1]

A 28.4, –5.71 and –14.29

B 0.64, –7.14 and –5.71

C 28.4, –7.14 and –5.71

D 0.64, –5.71 and –14.29

Correct Answer

Option A

Solution

ln

\left[ {{{{k_2}} \over {{k_1}}}} \right]

=

{{\Delta H^\circ } \over R}\left\{ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right\}

$\Rightarrow$ ln(10) =

{{\Delta H^\circ } \over R}\left\{ {{1 \over {298}} - {1 \over {373}}} \right\}

$\Rightarrow$

{\Delta H^\circ }

= 28.37 kJ/mol

\Delta

Go = –RT ln K T1 = 25oC K1 = 10

\Delta

Go at T1 = –8.314 × 298 × 2.303 × log 10 = –5.71 kJ/mol

\Delta

Go at T2 = –8.314 × 373 × 2.303 × log(100) = –14.29 kJ/mol

Q28

For a reaction at equilibrium A(g)

\rightleftharpoons

B(g) +

{1 \over 2}

C(g) the relation between dissociation constant (K), degree of dissociation (

\alpha

) and equilibrium pressure (p) is given by :

A

K = {{{\alpha ^{{1 \over 2}}}{p^{{3 \over 2}}}} \over {{{\left( {1 + {3 \over 2}\alpha } \right)}^{{1 \over 2}}}(1 - \alpha )}}

B

K = {{{\alpha ^{{3 \over 2}}}{p^{{1 \over 2}}}} \over {{{\left( {2 + \alpha } \right)}^{{1 \over 2}}}(1 - \alpha )}}

C

K = {{{{(\alpha \,p)}^{{3 \over 2}}}} \over {{{\left( {1 + {3 \over 2}\alpha } \right)}^{{1 \over 2}}}(1 - \alpha )}}

D

K = {{{{(\alpha \,p)}^{{3 \over 2}}}} \over {{{\left( {1 + \alpha } \right)}}{{(1 - \alpha )}^{{1 \over 2}}}}}

Correct Answer

Option B

Solution

Now,

{K_P}

or

K = {{{P_B} \times {{\left( {{P_C}} \right)}^{{1 \over 2}}}} \over {{P_A}}}

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

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

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

= {{{\alpha ^{{3 \over 2}}}\,.\,{P^{{1 \over 2}}}} \over {(1 - \alpha ){{(2 + \alpha )}^{{1 \over 2}}}}}

Q29

5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327ºC. 30% of the solid NH4SH decomposed to NH3 and H2S as gases . The Kp of the reaction at 327oC is (R = 0.082 L atm mol–1 K–1, Molar mass of S = 32 g mol–1 molar mass of N = 14 g mol–1)

A 0.242

\times

10

-

4 atm2

B 1

\times

10–4 atm2

C 4.9

\times

10

-

3 atm2

D 0.242 atm2

Correct Answer

Option D

Solution

NH4SH(s) $\rightleftharpoons$ NH3(g) + H2S(g)

n = {{5.1} \over {51}} = .1\,mole\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,

.1\left( { - 1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha

\alpha \,\, = \,\,30\% = .3

so number of moles at equilibrium

\,\,\,\,\,\,\,.1\,(1 - .3)\,\,\,\,\,.1\,\, \times \,\,.3\,\,\,\,\,\,\,\,\,.1\,\, \times \,.3

= \,\,\,\,.07\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = .03\;\,\,\,\,\,\,\,\,\, = .03

Now use PV = nRT at equilibrium Ptotal $\times$ 3 lit = (.03 + .03) $\times$ .082 $\times$ 600 Ptotal = .984 atm At equilibrium PNH3 = PH2S =

{{{P_{total}}} \over 2}

= .492 So kp = PNH3 . PH2S = (.492) (.492) kp = .242 atm2

Q30

In which of the following reactions, an increase in the volume of the container will favour the formation of products?

A 2NO2(g)

\rightleftharpoons

2NO(g) + O2(g)

B H2(g) + I2(g)

\rightleftharpoons

2HI(g)

C 4NH3(g) + 5O2(g)

\rightleftharpoons

4NO(g) + 6H2O(1)

D 3O2 (g)

\rightleftharpoons

2O3(g)

Correct Answer

Option A

Solution

From Boyle's law, we know when volume increase then pressure decreases, and to keep the pressure on original value the reaction should proceeds in the direction where the number of moles of gases increases.

In this reaction, only satisfy this.

2 NO2(g) $\rightleftharpoons$ 2 NO(g) + O2 (g)

\therefore\,\,\,\,

\Delta

ng = (2 + 1) $-$ 2 = 1