Chemical Equilibrium — Page 4 | NEET Chemistry

Q31

For the reaction : CH4 (g) + 2O 2(g)

\rightleftharpoons

CO 2(g) + 2H 2 O (l) ,

\Delta

H r =

-

170.8 kJ mol

-

1 . Which of the following statements is not true?

A The reaction is exothermic.

B At equilibrium, the concentrations of CO 2(g) and H 2 O (l) are not equal.

C The equilibrium constant for the reaction is given by K p =

{{\left[ {C{O_2}} \right]} \over {\left[ {C{H_4}} \right]\left[ {{O_2}} \right]}}

D Addition of CH 4(g) or O 2(g) at equilibrium will cause a shift to the right.

Correct Answer

Option C

Solution

Option C is incorrect as the value of K P given is wrong. It should have been K P =

{{{P_{C{O_2}}}} \over {{P_{C{H_4}}} \times {{\left[ {{P_{{O_2}}}} \right]}^2}}}

Q32

Equilibrium constants K 1 and K 2 for the following equilibriam: are related as

A K 2 = 1/K 1 2

B K 2 = K 1 2

C K 2 = 1/K 1

D K 2 = K 1 /2

Correct Answer

Option A

Solution

NO (g) + O 2(g) ⇌ NO 2(g) , K 1 Reverse the above equation NO 2(g) ⇌ NO (g) + O 2(g) ,

{1 \over {{K_1}}}

Multiply the above equation by 2, we get NO 2(g) ⇌ NO (g) + O 2(g),

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

Q33

The reaction quotient (Q) for the reaction N 2(g) + 3H 2(g)

\rightleftharpoons

2NH 3(g) is given by

Q = {{{{\left[ {N{H_3}} \right]}^2}} \over {\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}

. The reaction will proceed from right to left if

A Q = K c

B Q < K c

C Q > K c

D Q = 0

Correct Answer

Option C

Solution

For any reaction to process in forward direction the reaction quotient (Q) must be less than equilibrium constant K C .

Q < K c

Q34

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}}}

Q35

Reaction BaO 2(g)

\rightleftharpoons

BaO (s) + O 2(g) ;

\Delta

H = +ve. In equilibrium condition, pressure of O 2 depends on

A increase mass of BaO 2

B increase mass of BaO

C increase temperature on equilibrium

D increase mass of BaO 2 and BaO both.

Correct Answer

Option C

Solution

For the reaction BaO 2(g) $\rightleftharpoons$ BaO (s) + O 2(g) ; H = +ve.

At equilibrium K p = P O 2 [For solid and liquids concentration term is taken as unity] Hence, the value of equilibrium constant depends only upon partial pressure of O 2 .

Further on increasing temperature formation of O 2 increases as this is an endothermic reaction.

Q36

For any reversible reaction, if we increase concentration of the reactants, then effect on equilibrium constant

A depends on amount of concentration

B unchange

C decrease

D increase

Correct Answer

Option B

Solution

Equilibrium constant of a reaction is independent of the concentration of species involved in the reaction but dependent only on the temperature.

Q37

Equilibrium constant K p for following reaction MgCO 3(s)

\rightleftharpoons

MgO (s) + CO 2(g)

A K p = P CO 2

B K p = P CO 2

\times

{{{P_{C{O_2}}} \times {P_{MgO}}} \over {{P_{MgC{O_3}}}}}

C K p =

{{{P_{C{O_2}}} + {P_{MgO}}} \over {{P_{MgC{O_3}}}}}

D K p =

{{{P_{MgC{O_3}}}} \over {{P_{C{O_2}}} + {P_{MgO}}}}

Correct Answer

Option A

Solution

K p = P CO 2 As solids do not exert pressure, so their partial pressure is taken as unity.

Q38

At temperature T, compound

AB_{2(g)}

dissociates as

AB_{2(g)} \rightleftharpoons AB_{(g)} + \dfrac{1}{2} B_{2(g)}

having degree of dissociation

x

(small compared to unity). The correct expression for

x

in terms of

K_p

and

p

is:

A

\sqrt{K_p}

B

\sqrt[3]{\dfrac{2 K_{\mathrm{p}}^2}{\mathrm{p}}}

C

\sqrt[3]{\dfrac{2 K_p}{p}}

D

\sqrt[4]{\dfrac{2 K_p}{p}}

Correct Answer

Option B

Solution

$A B_2(g) \rightleftharpoons A B_{(g)}+\dfrac{1}{2} B_2({g})$ Degree of dissociation $\rightarrow x$ (small compared to unity) Equilibrium constant interms of pressure $\rightarrow k_p$ Partial pressure of each gas $\rightarrow P_{A B_2}, P_{A B}$ and $P_{B_2}$ Total pressure $\rightarrow P$ ICE Table: $\begin{aligned} & \text{ partial pressure } \\ & \text{ at equilibrium }\end{aligned}=\dfrac{\text{ moles at equilibrium }}{\begin{array}{l}\text{ total moles at } \\ \text{ equilibrium }\end{array}} \times$ total pressure Partial pressure of $A B_2(g)$

P_{A B_2}=\left(\frac{1-x}{1+\frac{x}{2}}\right) P

Partial pressure of $A B(g)$

P_{A B}=\left(\frac{x}{1+\frac{x}{2}}\right) P

Partial pressure of $B_2(g)$

P_{B_2}=\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}\right)^P

$K_p$ for the equation $A_2(g) \rightleftharpoons A B_{(g)}+\dfrac{1}{2} B_2(g)$ can be written as $k_p=\dfrac{P_{A B} P_{B_2}^{1 / 2}}{P_{A B_2}}$ $\begin{aligned} & =\dfrac{\dfrac{x}{1+\dfrac{x}{2}} P 1 \times\left(\dfrac{\dfrac{x}{2}}{1+\dfrac{x}{2}} P\right)^{1 / 2}}{\dfrac{1-x}{1+\dfrac{x}{2}} P} \\ & =\dfrac{x \times\left(\dfrac{x}{1+\dfrac{x}{2}} P\right)^{1 / 2}}{1-x}\end{aligned}$

\begin{aligned} &\text{ given, } x \lll 1\\ &\begin{aligned} & \text{ So, } 1-x \approx 1 \\ & K_p=\frac{x \times\left(\frac{\frac{x}{2}}{1+\frac{x}{2}} p\right)^{1 / 2}}{1} \end{aligned} \end{aligned}

$=x \times \dfrac{\left(\dfrac{x}{2}\right)^{1 / 2}}{\left(\dfrac{2+x}{2}\right)^{1 / 2}} p^{1 / 2}$ $=\dfrac{x \times \dfrac{x^{1 / 2}}{2^{1 / 2}} \times p^{1 / 2}}{\dfrac{(2+x)^{1 / 2}}{2^{1 / 2}}}$ $=\dfrac{x^{3 / 2} \times p^{1 / 2}}{2^{1 / 2}} \times \dfrac{2^{1 / 2}}{(2+x)^{1 / 2}}$

\begin{gathered} x \lll 1 \\ 2+x \approx 2 \end{gathered}

\begin{aligned} & =\frac{x^{3 / 2} \times p^{1 / 2}}{2^{1 / 2}} \\ & =\left(\frac{x^3 p}{2}\right)^{1 / 2} \end{aligned}

\begin{aligned} &k_p=\left(\frac{x^3 p}{2}\right)^{1 / 2}\\ &\text{ Square on both sides, } \end{aligned}

\begin{aligned} k_p^2 & =\frac{x^3 p}{2} \\ x^3 p & =2 k_p^2 \\ x^3 & =\frac{2 k_p^2}{p} \\ x & =\sqrt[3]{\frac{2 k_p^2}{p}} \end{aligned}

Correct answer option (2) $\sqrt[3]{\dfrac{2 K_p^2}{p}}$

Q39

For the following three reactions a, b and c, equilibrium constants are given: a. CO (g) + H2O (g)

\leftrightharpoons

CO2(g) + H2 (g) ; K1 b. CH4 (g) + H2O (g)

\leftrightharpoons

CO(g) + 3H2 (g) ; K2 c. CH4 (g) + 2H2O (g)

\leftrightharpoons

CO2(g) + 4H2 (g) ; K3

A

{K_1}\sqrt {{K_2}} = {K_3}

B K2K3 = K1

C K3 = K1K2

D K3.

K_2^3

=

K_1^2

Correct Answer

Option C

Solution

Reaction

(c)

can be obtained by adding reactions

(a)

and

(b)

therefore

{K_3} = {K_1}.{K_2}

Hence

(c)

is the correct answer.

Q40

For the reaction CO (g) + (1/2) O2 (g)

\leftrightharpoons

CO2 (g), Kp/Kc is :

A RT

B (RT)-1

C (RT)-1/2

D (RT)1/2

Correct Answer

Option C

Solution

{K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};

\Delta n = 1 - \left( {1 + {1 \over 2}} \right)

= 1 - {3 \over 2} = - {1 \over 2}.

$\therefore$

\,\,\,\,{{{K_p}} \over {{K_c}}} = {\left( {RT} \right)^{ - 1/2}}