Chemical Equilibrium

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ; K_C

2NH₃(g) ⇌ N₂(g) + 3H₂(g) ; ${1 \over {{K_C}}}$

Multiplying by ${1 \over 2}$, reaction becomes

NH₃(g) ⇌ ${1 \over 2}$N₂(g) + ${3 \over 2}$H₂(g) ;

$ \therefore $ New K_C = ${\left( {{1 \over {{K_C}}}} \right)^{{1 \over 2}}}$ = ${\left( {{1 \over {64}}} \right)^{{1 \over 2}}}$ = ${1 \over 8}$

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: NH_3(g) = 1 mole

Moles of gas in reactants: ${3 \over 2}$ H_2(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: $K_C = K_p(RT)^{1/2}$

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 $G^o = –RT ln K

T₁ = 25^oC K₁ = 10

$\Delta $G^o at T₁ = –8.314 × 298 × 2.303 × log 10

= –5.71 kJ/mol

$\Delta $G^o at T₂ = –8.314 × 373 × 2.303 × log(100)

= –14.29 kJ/mol

$ \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.

We have two reactions:

$ A \;\rightleftharpoons\; B + C \quad\text{with } K_{eq}^{(1)}$

$ B + C \;\rightleftharpoons\; P \quad\text{with } K_{eq}^{(2)}$

We want the equilibrium constant for the overall reaction

$ A \;\rightleftharpoons\; P. $

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How to combine the equilibrium constants

When two reactions are added to yield a net (overall) reaction, the equilibrium constant of the net reaction is the product of the equilibrium constants of the individual reactions. Formally:

$ \text{(Reaction 1)} + \text{(Reaction 2)} \;\Rightarrow\; \text{(Net Reaction)}, $

and

$ K_{\text{net}} \;=\; K_{\text{(Reaction 1)}} \times K_{\text{(Reaction 2)}}. $

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Applying it here

Reaction 1: $A \to B + C$ has $K_{eq}^{(1)}.$

Reaction 2: $B + C \to P$ has $K_{eq}^{(2)}.$

When we add them,

$ \underbrace{A}_{\text{Reactant 1}} \;\longrightarrow\; \underbrace{B + C}_{\text{Products of Reaction 1}=\text{Reactants of Reaction 2}} \;\longrightarrow\; \underbrace{P}_{\text{Product of Reaction 2}}. $

So the net reaction is $A \to P$, and its equilibrium constant is

$ K_{eq}^{\text{(net)}} \;=\; K_{eq}^{(1)} \times K_{eq}^{(2)}. $

Thus, the correct answer is:

$ \boxed{K_{eq}^{(1)} \, K_{eq}^{(2)}}. $

Hence, the answer (Option D) is correct.

At equilibrium,

rate of forward reaction = Rate of backward reaction

With addition of solute in solvent, surface area for vapourisation decreases causes lowering in vapour pressure.

In the figure, Reactant A is represented by square.
And there are 6 squares. So at equilibrium
[A]_eq = 6

In the figure, Product B is represented by circle.
And there are 11 circles. So at equilibrium
[B]_eq = 11

For reaction, A ⇌ B

K_eq = ${{\left[ B \right]} \over {\left[ A \right]}}$ = ${{11} \over 6}$ $ \approx $ 2

We know, $\Delta {G^o} = - RT\ln K$
Case 1 :
If $\Delta $G^o < 0
$ \Rightarrow $ $- RT\ln K$ < 0
$ \Rightarrow $ $\ln K$ > 0
$ \Rightarrow $ K > 1

Case 2 :
If $\Delta $G^o > 0
$ \Rightarrow $ $- RT\ln K$ > 0
$ \Rightarrow $ $\ln K$ < 0
$ \Rightarrow $ K < 1

Case 2 :
If $\Delta $G^o = 0
$ \Rightarrow $ $- RT\ln K$ = 0
$ \Rightarrow $ $\ln K$ = 0
$ \Rightarrow $ K = 1

We know,
K_p = K_C(RT)^{$\Delta $n_g}

K_p = K_C when $\Delta $n_g = 0
K_p $ \ne $ K_C when $\Delta $n_g $ \ne $ 0 and T $ \ne $ 12 K

In this reaction, 2C(s) + O₂(g) ⇋ 2CO(g)
$\Delta $n_g = 1, so K_p $ \ne $ K_C

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.

S(s) + O₂(g) ⇋ SO₂(g); K₁ = 10⁵²

By reversing the equation, we get

SO₂(g); ⇋ S(s) + O₂(g) ; ${1 \over {{K_1}}} = {1 \over {{{10}^{52}}}}$ ..............(1)

2S(s) + 3O₂(g) ⇋ 2SO₃(g); K₂ = 10¹²⁹ ....................(2)

Performing (2) - 2 $ \times $ (1), we get

2SO₂(g) + O₂(g) ⇋ 2SO₃(g)

$ \therefore $ Equilibrium constant of this reaction is

= ${{{K_2}} \over {K_1^2}}$ = ${{{{10}^{129}}} \over {{{\left( {{{10}^{52}}} \right)}^2}}}$ = 10²⁵

K_p1 = P₁(P₁ + P₂)

K_p2 = P₂(P₁ + P₂)

$ \therefore $ K_p1 + K_p2 = (P₁ + P₂)²

$ \Rightarrow $ x + y = (P₁ + P₂)²

$ \Rightarrow $ (P₁ + P₂) = $\sqrt {x + y} $ .....(1)

Now total pressure

P_T = P_C + P_B + P_E

= (P₁ + P₂) + P₁ + P₂

= 2(P₁ + P₂)

= $2\sqrt {x + y} $     [From equation (1)]

At equilibrium [A] = [B]

$ \Rightarrow $ a - x = 1.5a - 2x

$ \Rightarrow $ x = 0.5a

K_c = ${{{{\left[ C \right]}^2}\left[ D \right]} \over {\left[ A \right]{{\left[ B \right]}^2}}}$

= ${{{{\left( a \right)}^2}\left( {0.5a} \right)} \over {\left( {0.5a} \right){{\left( {0.5a} \right)}^2}}}$ = 4

N₂(g) + 3H₂(g) $\rightleftharpoons$ 2NH₃(g) ; K_eq = K_p

Write this equation reverse way,

2NH₃(g)  $\rightleftharpoons$ N₂(g) + 3H₂(g) ; K_eq = ${1 \over {{K_p}}}$

	2NH3(g)	⇌	N2(g)	+	3H2(g)
At t = 0	Po		0		0
At t = teq	PNH3		p		3p

At equillibrium

P_Total = P_NH3 + P_N2 + P_H2

= P_NH3 + p + 3p

(As P_NH3 << P_total so we can ignore P_NH3)

$ \therefore $ P_Total = 4p

$ \Rightarrow $ p = ${{{{P_{total}}} \over 4}}$

Formula of
K_eq = ${{{p_{{N_2}}} \times {{\left( {{p_{{H_2}}}} \right)}^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$ = ${1 \over {{K_p}}}$

$ \Rightarrow $ ${1 \over {{K_p}}}$ = ${{p \times 27{p^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$

$ \Rightarrow $ ${{{\left( {{p_{N{H_3}}}} \right)}^2}}$ = K_p $ \times $ 27 $ \times $ ${{{\left( {{{{P_{total}}} \over 4}} \right)}^4}}$

$ \Rightarrow $ P_NH3 = $\sqrt {{K_p}} \times {\left( {27} \right)^{{1 \over 2}}} \times {\left( {{{{P_{Total}}} \over 4}} \right)^{{4 \over 2}}}$

= ${{{3^{{3 \over 2}}}K_p^{{1 \over 2}}P_{Total}^2} \over {16}}$

   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

P_total $ \times $ 3 lit = (.03 + .03) $ \times $ .082 $ \times $ 600

P_total = .984 atm

At equilibrium

<sup>P_NH3 = P_H2S = ${{{P_{total}}} \over 2}$ = .492

So k_p = P_NH3 . P_H2S = (.492) (.492)

k_p = .242 atm2</sup>

2NO₂ (g) $\buildrel \, \over \longrightarrow $ N₂O₄ (g); $\Delta $H = $-$ ve

At equilibrium, N₂O₄ $\rightleftharpoons$ 2NO₂ ; $\Delta $H = + ve

On adding the inert gas at constant pressure, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases. Therefore, in above case, reaction will move towards forward direction which will lead to the decomposition of dinitrogen tetraoxide (N₂O₄).

	CO	+	Cl2	$\rightleftharpoons$	COCl2
Initially number of moles	2		3		0
At equilibrium number of moles	1		2		1

The equilibrium constant,

K_c = ${{\left[ {COC{l_2}} \right]} \over {\left[ {CO} \right]\left[ {C{l_2}} \right]}}$

=  ${{\left( {{1 \over 5}} \right)} \over {\left( {{1 \over 5}} \right) \times \left( {{2 \over 5}} \right)}}$

=   ${5 \over 2}$

=  2.5

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

Addition of a solid component to a system at constant pressure has no effect on the equilibrium. Therefore, addition of Fe₂O₃ will not disturb the equilibrium.

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

Given,



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

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

On solving

$a=0.81$

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

$\Delta {G^ \circ } = 2494.2J$

$2A\,\rightleftharpoons\,B + C$

$R = 8.314\,J/K/mol.$

$e = 2.718$

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

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

$ = {{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}.$

$S{O_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right)\,\rightleftharpoons\,S{O_3}\left( g \right)$

${K_p} = {K_C}{\left( {RT} \right)^x}$

where $x = \Delta {n_g} = $ number of gaseous moles in product

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$-$ number of gaseous moles in reactant

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

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

For the reaction

${N_2} + {O_2} \to 2NO$

$\,K = 4 \times {10^{ - 4}}$

Hence for the reaction

$NO \to {1 \over 2}{N_2} + {1 \over 2}{O_2}$

$K' = {1 \over {\sqrt K }}$

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

$ = 50$

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 = \frac{(P_{\text{CO}})^2}{P_{\text{CO}_2}} $

Substitute the equilibrium pressures:

$ K_p = \frac{(2x)^2}{0.5 - x} = \frac{(2 \times 0.3)^2}{0.5 - 0.3} = \frac{0.6^2}{0.2} $

$ K_p = \frac{0.36}{0.2} = 1.8 \text{ atm} $

$\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

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

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