Chemical Equilibrium MCQ Quiz - Objective Question with Answer for Chemical Equilibrium - Download Free PDF

Concept:

Henderson-Hasselbalch Equation for Buffer Solutions

• A buffer solution contains a weak acid and its conjugate base or a weak base and its conjugate acid.
• The pH of a buffer solution is given by the Henderson-Hasselbalch equation:

  \(pH = pKa + log(\frac {[A-]}{[HA]})\)

• Where:
  • \(pKa = -log(Ka)\) of the weak acid
  • [A^-] = concentration of the conjugate base
  • [HA] = concentration of the weak acid

Explanation:

Given data:

• Mass of acetic acid (CH₃COOH) = 3 g
• Mass of sodium acetate (CH₃COONa) = 5 g
• Volume of solution = 1 L
• Molar mass of acetic acid = 60 g/mol
• Molar mass of sodium acetate = 82 g/mol
• \(Ka = 1.8 × 10^{-5}\)

Calculate the moles and concentrations of CH₃COOH and CH₃COONa:

• Moles of CH₃COOH = Mass / Molar mass = 3 g / 60 g/mol = 0.05 mol
• Moles of CH₃COONa = Mass / Molar mass = 5 g / 82 g/mol = 0.061 mol

Since the solution is 1 L, the concentrations are:

• [HA] = [CH₃COOH] = 0.05/1 = 0.05 M
• [A^-] = [CH₃COONa] = 0.061/1 = 0.061 M

Calculate the pKa:

\(pKa = -log(Ka)\)

\(pKa = -log(1.8 × 10^{-5})\)

\(pKa = 4.75\)

Calculate the pH using the Henderson-Hasselbalch equation:

\(pH = pKa + log(\frac {[A-]}{[HA]})\)

\(pH = 4.75 + log(\frac {0.061}{0.05})\)

\(pH = 4.75 + log(1.22)\)

\(pH = 4.75 + 0.086\)

\(pH = 4.8\)

Therefore, the pH of the solution is 4.8.

CONCEPT:

Determination of the Standard Equilibrium Constant Using Electrode Potentials

• The standard equilibrium constant, ( \(K^\circ \)), for a redox reaction can be calculated using the Nernst equation in the form: \( \Delta G^\circ = -RT \ln K^\circ \).
• We also know that \( \Delta G^\circ = -nFE^\circ_{\text{cell}} \), where ( n ) is the number of moles of electrons exchanged, ( F ) is Faraday’s constant (96485 C/mol), and \( E^\circ_{\text{cell}} \) is the standard cell potential.
• Thus, \( \ln K^\circ = \frac{nFE^\circ_{\text{cell}}}{RT} \), where R = 8.314 J/mol K and T = 298 K.

CALCULATION:

• Identify the half-reactions and their standard electrode potentials:
  • Reduction half-reaction for Fe³⁺: \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \, (E^\circ = 0.77 \, \text{V}) \)
  • Oxidation half-reaction for Sn²⁺: \( \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \, (E^\circ = -0.15 \, \text{V}) \)
• Calculate the cell potential, \( E^\circ_{\text{cell}} \):
  • \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \)
  • \( E^\circ_{\text{cell}} = 0.77 - (-0.15) = 0.92 \, \text{V} \)
• Calculate the equilibrium constant using \( K^\circ = e^{\frac{nFE^\circ_{\text{cell}}}{RT}} \):
  • \( n = 2 \) (since 2 electrons are transferred per iron ion in the balanced reaction)
  • \( K^\circ = e^{\frac{(2)(96485)(0.92)}{(8.314)(298)}} \)
  • \( K^\circ = e^{71.62} \approx 10^{21} \)

CONCLUSION:

The correct answer is 10²¹.

CONCEPT:

Equilibrium and Degree of Dissociation

• The decomposition reaction is: 2P2Q ⇌ 2P2 + Q2
• Let 4 moles of P2Q be taken initially. Degree of dissociation is α.
• At equilibrium:
  • P2Q = 4(1 – α)
  • P2 = 4α (since 2 × α for each 2 moles)
  • Q2 = 2α
  • Total moles = 4(1 – α) + 4α + 2α = 4 + 2α
• Given that Kp = P (numerically equal to total pressure at equilibrium)
• This leads to a cubic equation in α: 2α³ = (1 – α)²(4 + 2α)
• Solving it gives α = 2/3

EXPLANATION:

\(2P_2Q \rightleftharpoons 2P_2 + Q_2\)

\(t = 0 \quad 4 \quad 0 \quad 0\)

\(t = eq \quad 4(1-\alpha) \quad 4\alpha \quad 2\alpha\)

\(K_p = \dfrac{\left(\dfrac{4\alpha}{4+2\alpha} \times P\right)^2 \left(\dfrac{2\alpha}{4 + 2\alpha}\times P \right)}{\left(\dfrac{4(1-\alpha)}{4+2\alpha} \times P\right)^2} = \dfrac{2\alpha^3}{(1-\alpha)^2 (4+2\alpha)} \times P\)

But \(K_p = P\) (given) \(\therefore 2\alpha^3 = (1-\alpha)^2 (4+2\alpha)\)

\(\therefore - 6\alpha + 4 = 0 \quad \therefore \alpha= \dfrac{2}{3}\)

Total moles at eq \(= 4 + 2\alpha = \dfrac{16}{3}\)

\(n_{P_2 Q} = n_{Q_2} = \dfrac{4}{3}, n_{P_2} = \dfrac{8}{3}\)

∴" id="MathJax-Element-6949-Frame" role="presentation" style="position: relative;" tabindex="0">∴∴" id="MathJax-Element-193-Frame" role="presentation" style="position: relative;" tabindex="0">∴∴ ∴ $\therefore$ Total number of moles of products \(= \dfrac{8}{3} + \dfrac{4}{3} = \dfrac{12}{3} = 4 \)

• Statement 1:  Incorrect.
  • At equilibrium, moles of P₂Q = 4(1 – α) = 4(1 – 2/3) = 4/3 ≠ initial moles (4)
• Statement 2:  Correct.
  • P₂ moles = 4α = 4 × 2/3 = 8/3
• Statement 3: Correct.
  • α = 2/3 as derived from solving the equation
• Statement 4: Correct.
  • Products = P₂ (8/3) + Q₂ (4/3) = 12/3 = 4 moles

Therefore, the correct statements are: Options 2, 3, and 4.

Concept:

Le Chatelier's Principle and Colour Change in Equilibrium Reactions

• Le Chatelier's Principle states that when a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will adjust to counteract the effect of the disturbance.
• In the given equilibrium:

  2NO₂(g) ⇌ N₂O₄(g)

  • \(\Delta n = -1\)

• NO₂ is a brown gas, and N₂O₄ is colourless. The reaction is dynamic, and its position depends on temperature and pressure.
• At high temperatures, the reaction shifts towards the products (N₂O₄), and the equilibrium mixture becomes colourless.
• At low pressures, the reaction favours the formation of more NO₂, leading to a darker colour due to the brownish NO₂ gas.
• The reaction is exothermic, meaning it releases heat when it forms N₂O₄.

Explanation:

• Statement 1: This is incorrect because the reaction is exothermic.
• Statement 2: This is incorrect because the reaction is exothermic.
• Statement 3: The reaction is exothermic, and N₂O₄ is darker in colour than NO₂. This is incorrect because the reaction goes backward.
• Statement 4: The reaction is exothermic, and NO₂ is darker in colour than N₂O₄. This is the correct statement because the reaction goes backward so NO₂ will more produce,

Final Answer: The correct statement is: The reaction is exothermic, and NO₂ is darker in colour than N₂O₄.

Concept:

Henderson-Hasselbalch Equation

• The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:

  \(pH = pK_a + log ([A^-]/[HA])\)

  where pH is the pH of the solution, pKa is the acid dissociation constant of the weak acid (CH3COOH in this case), [A^-] is the concentration of the conjugate base (CH3COO^-), and [HA] is the concentration of the weak acid (CH3COOH).
• The ratio of the concentrations of the base (CH3COO^-) to the acid (CH3COOH) can be determined by rearranging the Henderson-Hasselbalch equation:

  \(log ([A^-]/[HA]) = pH - pK_a\)

  [\([A^-]/[HA] = 10^{(pH - pK_a)}\)

Explanation:

Given:

• pH = 4.80
• pKa of CH₃COOH = 4.75

From the Henderson-Hasselbalch equation:

\(log ([CH_3COO^-]/[CH_3COOH]) = 4.80 - 4.75 = 0.05\)

\([CH_3COO^-]/[CH_3COOH] = 10^{0.05}\)

\([CH_3COO^-]/[CH_3COOH] = 1.12\)

Now-

\(x = {[CH_3COO^-] - [CH_3COOH]\over [CH_3COOH]}\)

\(x =\frac {[CH_3COO^-]}{[CH_3COOH]}-1\)

\(x =1.12-1\)

\(x =0.12\)

Therefore, the quantity of x is 0.12.

Concept:

Law of Mass action and Equilibrium constant:

• At a constant temperature, the rate of a chemical reaction is directly proportional to the product of molar concentrations of reactants present at any given time. This is the law of mass action.
• The equilibrium constant for the reversible reaction of type \(aA + bB \rightleftharpoons cC + dD\) is represented as

\({K_{\rm{c}}}{\rm{ = }}\frac{{{{{\rm{[C]}}}^{\rm{c}}}{{{\rm{[D]}}}^{\rm{d}}}}}{{{{{\rm{[A]}}}^{\rm{a}}}{{{\rm{[B]}}}^{\rm{b}}}}}\)

Characteristics of an equilibrium constant:

• It is a definite value for all chemical reactions.
• ​The constants Kp and Kc are both equilibrium constants.
• Kp is used when the concentration terms are given in partial pressures i.e, in gaseous reactions.
• Kc is used when the reaction terms are expressed in molarities.
• The relation between Kp and Kc is given by:

\({K_p} = {K_c} \times {\left( {RT} \right)^{\Delta n}}\) where R = Universal gas constant, T = Temperature, and \(\triangle n\) = change in moles of gases in the reaction.

Calculation:

Given:

The reaction is:  CO (g) + Cl2 (g) ⇌ COCl2

The partial pressure of CO = .7bar

The partial pressure of CO₂ = 1 bar

The partial pressure of COCl2 (g) at equilibrium = 0.15 bar

• For the given reaction,

• The partial pressure of CO at equilibrium = .7 - .15 = .55
• The partial pressure of CO2 at equilobrium= 1 - .15 = .85
• The partial pressure of COCl2 (g) at equilibrium = 0.15 bar
• The Kp for the reaction is:

\({K_{\rm{p}}}{\rm{ = }}{ p_{COCl_2}\over p_{CO}\times p_{Cl_2}}\)

\(k_p= .{15\over .55 \times .85} = .32\)

Hence, the equilibrium constant for this reaction at 500 °C is .32.

Concept:

Gibb's Free Energy:

• The decrease in free energy is actually the amount of maximum work done by the system excluding the expansion work done when temp and pressure are kept constant.
• Free energy of a system is the difference in energy at the initial state and the equilibrium state energy.
• This free energy can be used to do external work.
• The non-available energy is the equilibrium state energy and is given by T × S, where T = temperature and S = Entropy.

expressions for Gibb's Free energy are:

ΔG = ΔH - TΔS

dG = dq -dq rev

ΔGo = -nFE0​

ΔG = ΔGo + RTlnKc

• Gibbs free energy is denoted by ΔG and has the units J/mol.

Gibbs Free Energy and EMF of a cell:

• In a voltaic cell, the heat energy liberated during the chemical change is converted into electrical energy.
• The electrical energy produced for one equivalent of product is FE, where F = faradays constant 96500
• The electrical energy produced for 'n' equivalents is nFE.
• According to Helmholtz, the decrease in free energy of a reaction is equal to the electrical energy obtained from a galvanic cell. 

ΔGo = -nFE0​........................1

At Equilibrium, 

Eo =\({RT\over nF} lnK_c\).....................2, where Kc = Equilibrium constant Keq.

Equating Relation 1 and 2, we get:

\(ΔG^o = -nF​​​​{RT\over nF}lnK_{eq}\)

Or, ΔG° = - 2.303 RT Iog Keq

Hence, the relationship between the equilibrium constant K and Δ G is:​ 

ΔG° = - 2.303 RT Iog Keq

Calculation:

Given:

• ΔG° for the reaction 3 NO (g) ⇌ N2O (g) + NO2 (g) = −104.18 kJ = -104180Joules
• Temperature of the reaction = 25C = 298K
• Value of R = 8.314 J mol−1 K−1
• We know that: 

ΔG° = - 2.303 RT Iog Keq

• So substituting the given values in the above equation, we get:

-104180Joules​ = -2.303 × 8.314 J mol−1 K−1× 298 K ×  Iog Keq

or, Iog Keq = 18.255

or, K_eq = 1.8 × 10¹⁸.

Hence, the equilibrium constant for the reaction is Keq = 1.8 × 1018.

Concept:

For the reaction A(g) ⇌ B(g) at a particular temperature with an equilibrium constant greater than one, it indicates that the formation of product B is favored over reactant A at equilibrium. The relationship between the Gibbs free energy change (ΔG) and the equilibrium constant (K) is given by:

\(\Delta G = -RT \ln K \)

• Equilibrium Constant (K): Since ( K > 1 ), it implies that ΔG is negative, suggesting that the reaction is spontaneous in the forward direction.

• Energy Levels: The energy level diagrams for molecules A and B represent their respective stability. A lower energy level indicates a more stable state, which correlates with a favorable reaction forming B.

Explanation: 

• In the energy level diagram, molecule A is at a higher energy state compared to molecule B. This indicates that the conversion from A to B is exothermic, leading to a decrease in energy when moving to the product state.

• The transition from A to B represents a favorable energy change, which aligns with the observed equilibrium constant (K > 1). Thus, the schematic energy levels suggest that products are more stable than the reactants at equilibrium.

• As indicated in the diagram, the energy levels reflect the tendency of the system to favor products B over A, which supports the conclusion that the energy of A is higher than that of B.

Conclusion:

The correct energy level configuration, where B is more stable than A (reflecting K_eq > 1 ), is shown in Option 1.

Concept:

Harmonic Oscillator:

• The harmonic oscillator model is the basis for the vibrational motion in diatomic molecules.
• The energy of a harmonic oscillator is given by,

\({E_n} = (n + {1 \over 2})h\nu \)  for n= 0, 1, 2... 

Explanation:

• The equilibrium dissociation energy of a diatomic molecule is 4.75 eV
• The stretching frequency of the diatomic molecule is 0.5 eV. Now, the stretching frequency is the energy difference between the two successive energy levels is \(h\nu \).
• Now, the energy of zero-point energy is

​ \({E_0} = {1 \over 2}h\nu \)

• So, the value of zero-point energy is

=\({1 \over 2}h\nu \) 

= \({{0.5} \over 2}\)

= \(0.25\).

• Now, for a harmonic oscillator, 

Equilibrium Bond dissociation energy = (Zero point energy + Minimum bond dissociation energy).

• Therefore, Minimum bond dissociation energy = Equilibrium dissociation energy - Zero point energy.
• So, The minimum energy required to dissociate the molecule

= \(\left( {4.75 - 0.25} \right)\) eV

=  \(4.50\) eV. 

Conclusion:-

So, The minimum energy required to dissociate the molecule in eV is 4.50 eV

Concept:

• Thermodynamic equilibrium is an axiomatic concept of thermodynamics. It is an internal state of a single thermodynamic system or a relation between several thermodynamic systems.
• Systems in mutual thermodynamic equilibrium are simultaneously in mutual thermal, mechanical, chemical, and radiative equilibrium

K_p is the equilibrium constant it's calculated from the ratio of partial pressure of products to reactants.

Explanation:-

In the reaction, 2A(g) + 3B(g) ⇌ 2C(g)

the equilibrium constant K_p will be,

\(K_p=\frac{P_c^{2}}{P_A^{2}P_{B}^{3}}\)

• Total number of moles

= 1.2+0.8+0.8

=2.8 mol

\(mole fraction=\frac{no. of moles\: of\: that \:compound}{total mole}\)

P_A= X_A.P_T,
P_B=X_B.P_T,
P_C=X_C.PT, 
P_D=X_D.PT

where, P_A, PB, P_C, and P_D are the partial pressure of gas A, B, C, and D respectively.

X_A, XB, XC, and XD are the mole fraction of gas A, B, C, and D respectively.

P_T is the total pressure of the system.

\(K_p=\frac{P_c^{2}}{P_A^{2}P_{B}^{3}}\)

\(K_P= \frac{(\frac{0.8}{2.8}\times2)^{2}}{(\frac{1.2}{2.8}\times2)^{2}\times(\frac{0.8}{2.8}\times2)^{3}}\)

= 2.4

Conclusion:-

The value of the equilibrium constant, KP of this reaction at the given temperature is closest to 2.4. Hence, option 3 is correct.

Concept:

• Thermodynamic equilibrium is an axiomatic concept of thermodynamics. It is an internal state of a single thermodynamic system or a relation between several thermodynamic systems.
• Systems in mutual thermodynamic equilibrium are simultaneously in mutual thermal, mechanical, chemical, and radiative equilibrium.

Explanation:

• The equilibrium constant (K_p) depends on stoichiometry representation.
• The equilibrium constant (Kp) also depends on ΔG° which depends upon the standard state chosen for the reactant and products.
• For the chemical reaction N2(g) + 3H2(g) → 2NH3(g), the value of the ratio of mole fractions at equilibrium is 

\(\frac{{{\rm{x}}_{{\rm{N}}{{\rm{H}}_3}}^{\rm{2}}}}{{{x_{{N_2}}}x_{{H_2}}^3}}\).

• Now, With increases in pressure (P), the equilibrium will shift in the direction, where the number of moles will decrease, to keep the equilibrium constant (Kp) constant.

​Conclusion:

Hence, for the reaction N2(g) + 3H2(g) → 2NH3(g), the thermodynamic quantity which depends upon the pressure at which equilibrium is the ratio of mole fractions,

 \(\frac{{{\rm{x}}_{{\rm{N}}{{\rm{H}}_3}}^{\rm{2}}}}{{{x_{{N_2}}}x_{{H_2}}^3}}\), at equilibrium.

Concept: 

Gibbs Free Energy and Standard State

Standard molar Gibbs free energy of formation  (ΔrG⁰ ) is generally referred to as the free energy change when one mole of compound is made from pure elements in their standard states. 

\( {\vartriangle _r}\mathop G\nolimits^0 = \,\sum \,{\vartriangle _r}\mathop G\nolimits_{{\text{Products}}}^0 - \sum \,{\vartriangle _r}\mathop G\nolimits_{{\text{Reactants}}}^0 \)

If we know the equilibrium constant for a reaction, we can calculate its standard molar free energy change 

ΔrG⁰ = -RT ln K_p

Where, 

• R-gas constant
• T-temperature
• K- equilibrium constant

If we know the standard molar free energy change for a reaction, we can therefore calculate its equilibrium constant 

\( K\, = \,\mathop e\nolimits^{ - \frac{{{\vartriangle _r}{G^0}}}{{RT}}} \)

Explanation:

Given reaction,

SO2(g) + \(\frac{1}{2}\) O2 \(\leftrightharpoons\) SO3(g)

• The standard molar Gibbs energies of formation, ΔrG0, of SO2 (g) =  -300  kJ mol-1
• The standard molar Gibbs energies of formation, ΔrG⁰ of SO3 (g) =  -371 kJ mol-1
• ( Oxygen has zero Gibbs free energy of formation )
• R= 8.31 J mol-1 K-1
• T= 298 K

Therefore, ΔrG0 for the given reaction is given by, 

\(\begin{align*} {\vartriangle _r}\mathop G\nolimits^0 &= \,\sum \,{\vartriangle _r}\mathop G\nolimits_{{\text{Products}}}^0 - \sum \,{\vartriangle _r}\mathop G\nolimits_{{\text{Reactants}}}^0 \\ &= \, - 371\,KJmo{l^{ - 1}} - ( - 300KJmo{l^{ - 1}}) \\ &= - 71KJmo{l^{ - 1}} \\ \end{align*} \)

If we know the standard molar free energy change for a reaction, we can therefore calculate its equilibrium constant from the equation 

ΔrG0 = -RT ln Kp

\( \begin{align*} - 71 \times {10^{ - 3}}Jmo{l^{ - 1}}\, &= \,\left( {8.31\,Jmo{l^{ - 1\,}}{K^{ - 1}} \times 298K} \right)\,\ln \,{K_p} \\ {K_p}\, &= \,\mathop e\nolimits^{\frac{{ - \left( {71 \times {{10}^{ - 3}}} \right)}}{{\left( {8.31\,Jmo{l^{ - 1\,}}{K^{ - 1}} \times 298K} \right)}}} \\ &= 283 \times {10^{10}} \\ \end{align*} \)

The value of the equilibrium constant, KP, at this temperature is  283 × 1010. 

Concept:

Law of Mass action and Equilibrium constant:

• At a constant temperature, the rate of a chemical reaction is directly proportional to the product of molar concentrations of reactants present at any given time. This is the law of mass action.
• The equilibrium constant for the reversible reaction of type \(aA + bB \rightleftharpoons cC + dD\) is represented as

\({K_{\rm{c}}}{\rm{ = }}\frac{{{{{\rm{[C]}}}^{\rm{c}}}{{{\rm{[D]}}}^{\rm{d}}}}}{{{{{\rm{[A]}}}^{\rm{a}}}{{{\rm{[B]}}}^{\rm{b}}}}}\)

Characteristics of an equilibrium constant:

• It is a definite value for all chemical reactions.
• ​The constants Kp and Kc are both equilibrium constants.
• Kp is used when the concentration terms are given in partial pressures i.e, in gaseous reactions.
• Kc is used when the reaction terms are expressed in molarities.
• The relation between Kp and Kc is given by:

\({K_p} = {K_c} \times {\left( {RT} \right)^{\Delta n}}\) where R = Universal gas constant, T = Temperature, and \(\triangle n\) = change in moles of gases in the reaction.

Calculation:

Given:

The reaction is:  CO (g) + Cl2 (g) ⇌ COCl2

The partial pressure of CO = .7bar

The partial pressure of CO₂ = 1 bar

The partial pressure of COCl2 (g) at equilibrium = 0.15 bar

• For the given reaction,

• The partial pressure of CO at equilibrium = .7 - .15 = .55
• The partial pressure of CO2 at equilobrium= 1 - .15 = .85
• The partial pressure of COCl2 (g) at equilibrium = 0.15 bar
• The Kp for the reaction is:

\({K_{\rm{p}}}{\rm{ = }}{ p_{COCl_2}\over p_{CO}\times p_{Cl_2}}\)

\(k_p= .{15\over .55 \times .85} = .32\)

Hence, the equilibrium constant for this reaction at 500 °C is .32.

Concept:-

Reversible first-order reaction or opposing reaction:-

• The variation of the concentration with time close to equilibrium, considering the reaction in which A forms B and both forward and backward reactions are first order.

where k₁ and k₂ are the rate constants of forward and backward reactions.

Explanation:-

• There are two simultaneous reactions:

\(A\overset{K_1}{\rightarrow}B\)

and, rate of forward reaction = K1[A]

\(B\overset{K_2}{\rightarrow}A\)

and, rate of backward reaction = K2[B]

• Thus, the rate of the reaction (\(\frac{dx}{dt}\)),

= (Rate of forward reaction)-(Rate of backward reaction)

= K1[A] - K2[B]...........(i)

• Let, the initial concentration of [A] is a
• Let, after time t, the amount of B formed is x.
• So, after time t, the concentration of A is

[A] = a - x

and, the amount of B formed is

[B] = x

• Substituting the values in equation (i) we get,

\(\frac{dx}{dt}\)=K1 (a - x) - K2x..........(ii)

• Now, at equilibrium \(\frac{dx}{dt}\)=0, and x=x_e
• Thus, from equation (ii) we get,

​K1 (a - xe) = K2xe

\(K_2=\frac{K_1\left ( a-x_{e}] \right )}{x_{e}}\)

• Now, substituting the value of K₂ in equation (ii) we get,

\(\frac{dx}{dt}=K_1\left ( a-x\right )-\left [ \frac{K_1\left ( a-x_{e} \right )}{x_{e}} \right ]x\)

\(\frac{dx}{dt}=K_1 a-K_1x-\left [ \frac{K_1\left ( a-x_{e} \right )}{x_{e}}x \right ]\)

\(\frac{dx}{dt}=\frac{K_1ax_{e}-K_1xx_{e}-K_1[A]_ox+K_1xx_{e}}{x_{e}}\)

\(\frac{dx}{dt}=\frac{K_1ax_{e}-K_1[A]_ox}{x_{e}}\)

\(\frac{dx}{dt}=\rm \frac{K_1a}{x_e}(x_e-x)\)

Conclusion:-

• Hence, option 3 is correct.

Concept:

Gibb's Free Energy:

• The decrease in free energy is actually the amount of maximum work done by the system excluding the expansion work done when temp and pressure are kept constant.
• Free energy of a system is the difference in energy at the initial state and the equilibrium state energy.
• This free energy can be used to do external work.
• The non-available energy is the equilibrium state energy and is given by T × S, where T = temperature and S = Entropy.

expressions for Gibb's Free energy are:

ΔG = ΔH - TΔS

dG = dq -dq rev

ΔGo = -nFE0​

ΔG = ΔGo + RTlnKc

• Gibbs free energy is denoted by ΔG and has the units J/mol.

Gibbs Free Energy and EMF of a cell:

• In a voltaic cell, the heat energy liberated during the chemical change is converted into electrical energy.
• The electrical energy produced for one equivalent of product is FE, where F = faradays constant 96500
• The electrical energy produced for 'n' equivalents is nFE.
• According to Helmholtz, the decrease in free energy of a reaction is equal to the electrical energy obtained from a galvanic cell. 

ΔGo = -nFE0​........................1

At Equilibrium, 

Eo =\({RT\over nF} lnK_c\).....................2, where Kc = Equilibrium constant Keq.

Equating Relation 1 and 2, we get:

\(ΔG^o = -nF​​​​{RT\over nF}lnK_{eq}\)

Or, ΔG° = - 2.303 RT Iog Keq

Hence, the relationship between the equilibrium constant K and Δ G is:​ 

ΔG° = - 2.303 RT Iog Keq

Calculation:

Given:

• ΔG° for the reaction 3 NO (g) ⇌ N2O (g) + NO2 (g) = −104.18 kJ = -104180Joules
• Temperature of the reaction = 25C = 298K
• Value of R = 8.314 J mol−1 K−1
• We know that: 

ΔG° = - 2.303 RT Iog Keq

• So substituting the given values in the above equation, we get:

-104180Joules​ = -2.303 × 8.314 J mol−1 K−1× 298 K ×  Iog Keq

or, Iog Keq = 18.255

or, K_eq = 1.8 × 10¹⁸.

Hence, the equilibrium constant for the reaction is Keq = 1.8 × 1018.