Chemical Equilibrium — Page 7 | NEET Chemistry

Q61

\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}}}}

Q62

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.

Q63

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}.

Q64

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

Q65

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}}}}}

Q66

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

Q67

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

Q68

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: $K_p = \dfrac{(2x)^2}{0.5 - x} = \dfrac{(2 \times 0.3)^2}{0.5 - 0.3} = \dfrac{0.6^2}{0.2}$ $K_p = \dfrac{0.36}{0.2} = 1.8 \text{ atm}$

Q69

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}

Q70

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.