Advanced Chemistry - Mono Mole

November 8, 2019August 3, 2025

1st order consecutive reaction

A 1st order consecutive reaction of the type A → B → C is composed of the reactions:

$A\rightarrow B\; \; \; \; \; \; \; v_1=k_1[A]$

$B\rightarrow C\; \; \; \; \; \; \; v_2=k_2[B]$

The rate laws are:

$\frac{d[A]}{dt}=-k_1[A]\; \; \; \; \; \; \; \; 20$

$\frac{d[B]}{dt}=k_1[A]-k_2[B]$

$\frac{d[C]}{dt}=k_2[B]$

To understand how a 1^st order consecutive reaction proceeds over time, we need to develop equations for $[A]$ , $[B]$ and $[C]$ . The expression for $[A]$ is the solution for eq20, i.e. $[A]=[A_0]e^{-k_1t}$ . Substituting this in the 2nd rate law above and rearranging gives:

$\frac{d[B]}{dt}+k_2[B]=k_1[A_0]e^{-k_1t}\; \; \; \; \; \; \; \; 21$

Eq21 is a linear first order differential equation of the form y’ + P(t)y = f(t). Multiplying eq21 with the integrating factor $e^{k_2t}$ , we have

$e^{k_2t}\frac{d[B]}{dt}+k_2e^{k_2t}[B]=k_1e^{k_2t}[A_0]e^{-k_1t}\; \; \; \; \; \; \; \; 22$

The LHS of eq22 is the derivative of the product of $e^{k_2t}$ and [B], i.e. $\frac{d\left ( e^{k_2t}[B] \right )}{dt}$ . So,

$\frac{d\left ( e^{k_2t}[B] \right )}{dt}=k_1e^{k_2t}[A_0]e^{-k_1t}$

Integrating both sides with respect to time, noting that [B] = 0 at t = 0, and rearranging, we have

$[B]=\frac{k_1}{k_2-k_1}\left ( e^{-k_1t}-e^{-k_2t} \right )[A_0]\; \; \; \; \; \; \; \; eq23$

As t → ∞, [B] = 0.

At all times, [A] + [B] + [C] = [A₀], so from eq23,

$[A_0]-[A]-[C]=\frac{k_1}{k_2-k_1}\left ( e^{-k_1t}-e^{-k_2t} \right )[A_0]$

Substituting $[A]=[A_0]e^{-k_1t}$ in the above equation and rearranging yields:

$[C]=\left( 1+\frac{k_1e^{-k_2t}-k_2e^{-k_1t}}{k_2-k_1} \right )[A_0]$

As t → ∞, [C] = [A₀].

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November 8, 2019May 9, 2025

1st order reversible reaction

A 1st order reversible reaction of the type $A\rightleftharpoons B$ is composed of the reactions:

$A\rightarrow B\; \; \; \; \; \; \; v_1=k_1[A]$

$B\rightarrow A\; \; \; \; \; \; \; v_2=k_2[B]$

The rate law is:

$\frac{d[A]}{dt}=-k_1[A]+k_2[B]$

Substituting [B] = [A₀] – [A], where [A₀] is the initial concentration of A, in the above equation and rearranging yields:

$\frac{d[A]}{dt}=-(k_1+k_2)[A]+k_2[A_0]\; \; \; \; \; \; \; \; 16$

Let

$x=(k_1+k_2)[A]-k_2[A_0]\; \; \; \; \; \; \; \; 17$

Eq16 becomes $\frac{d[A]}{dt}=-x$ . Differentiating eq17 with respect to [A], we have $d[A]=\frac{dx}{k_1+k_2}$ , and differentiating this expression with respect to time, we have $\frac{d[A]}{dt}=\frac{1}{k_1+k_2}\frac{dx}{dt}$ , which is equivalent to

$-x=\frac{1}{k_1+k_2}\frac{dx}{dt}\; \; \Rightarrow \; \; -(k_1+k_2)dt=\frac{dx}{x}$

Let x = x₀when t =0 and integrate the above expression. We have

$-(k_1+k_2)t=lnx-lnx_0\; \; \; \; \; \; \; \; 18$

Substituting eq17 in eq18, noting that the 2nd term on RHS of eq18 refers to concentrations at t = 0, where [A] = [A₀], gives

$-(k_1+k_2)t=ln\frac{(k_1+k_2)[A]-k_2[A_0]}{(k_1+k_2)[A_0]-k_2[A_0]}$

which rearranges to

$[A]=\frac{k_2+k_1e^{-(k_1+k_2)t}}{k_1+k_2}[A_0]\; \; \; \; \; \; \; \; 19$

When t = 0, eq19 becomes [A] = [A₀]. As t → ∞,

$[A_\infty ]=[A_{eqm}]=\frac{k_2}{k_1+k_2}[A_0]$

Since $[B_{eqm}]=[A_0]-[A_\infty ]$ ,

$[B_{eqm}]=[A_0]-\frac{k_2}{k_1+k_2}[A_0]=[A_0]\frac{k_1}{k_1+k_2}$

Therefore, the equilibrium constant for the reaction is:

$K=\frac{[B_{eqm}]}{[A_{eqm}]}=\frac{k_1}{k_2}$

This is the link between chemical kinetics and thermodynamics.

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November 8, 2019October 18, 2024

3rd order reaction of the type (A + B + C → P)

The rate law for the 3rd order reaction of the type A + B + C → P is:

$\frac{d[A]}{dt}=-[A][B][C]$

Using the same logic described in a previous article, we can rewrite the rate law as:

$\frac{dx}{dt}=k(a-x)(b-x)(c-x)$

where a = [A₀], b = [B₀] and c = [C₀].

Integrating the above equation throughout gives

$\int_{0}^{x}\frac{dx}{(a-x)(b-x)(c-x)}=k\int_{0}^{t}dt$

Substituting the partial fraction expression $\small \frac{1}{(a-x)(b-x)(c-x)}=\frac{1}{(a-x)(b-a)(c-a)}+\frac{1}{(b-x)(a-b)(c-b)}+\frac{1}{(c-x)(a-c)(b-c)}$ in the above integral and working out some algebra yields

$kt=\frac{1}{(b-a)(c-a)}ln\frac{a}{a-x}+\frac{1}{(a-b)(c-b)}ln\frac{b}{b-x}+\frac{1}{(a-c)(b-c)}ln\frac{c}{c-x}$

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November 8, 2019October 18, 2024

3rd order reaction of the type (A + 2B → P)

The rate law for the 3rd order reaction of the type A + 2B → P is:

$\frac{d[A]}{dt}=-k[A][B]^2$

Using the same logic described in a previous article, we can rewrite the rate law as:

$\frac{dx}{dt}=k(a-x)(b-2x)^2$

where a = [A₀] and b = [B₀].

Integrating the above equation throughout yields

$\int_{0}^{x}\frac{dx}{(a-x)(b-2x)^2}=k\int_{o}^{t}dt\; \; \; \; \; \; \; \; 15$

Substituting the partial fraction expression $\frac{1}{(a-x)(b-2x)^2}=\frac{2}{(2a-b)(b-2x)^2}+\frac{1}{(2a-b)^2(a-x)}-\frac{2}{(2a-b)^2(b-2x)}$ in eq15 and working out some algebra gives

$kt=\frac{2x}{b(2a-b)(b-2x)}+\frac{1}{(2a-b)^2}ln\frac{a(b-2x)}{b(a-x)}$

Question

How to compute $\int_{0}^{x}\frac{2}{(2a-b)(b-2x)^2}dx$ ?

Answer

$\int_{0}^{x}\frac{2}{(2a-b)(b-2x)^2}dx=\int_{0}^{x}\frac{2b}{b(2a-b)(b-2x)^2}dx$

$=\int_{0}^{x}\frac{2[(b-2x)+2x]}{b(2a-b)(b-2x)^2}dx=\int_{0}^{x}\frac{2[(2a-b)(b-2x)+2x(2a-b)]}{b(2a-b)^2(b-2x)^2}dx$

$=\int_{0}^{x}\frac{2b(2a-b)(b-2x)-4bx(b-2a)}{b^2(2a-b)^2(b-2x)^2}dx=\frac{2x}{b(2a-b)(b-2x)}$

Note that the integrand in the 5th equality can be obtained by differentiating the last term using the quotient rule.

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November 8, 2019October 16, 2024

2nd order autocatalytic reaction

An autocatalytic reaction is one in which a product catalyses the reaction. An example is the Mn²⁺-catalysed oxidation of oxalic acid by potassium manganate (VII):

$2MnO_4^{\; -}+16H^++5C_2O_4^{\; 2-}\rightarrow 2Mn^{2+}+10CO_2+8H_2O$

The equation can be reduced to A + P → 2P, or simply, A → P, with the rate law:

$\frac{d[A]}{dt}=-k[A][P]\; \; \; \; \; \; \; \; 13a$

Using the same logic described in a previous article, we can rewrite the rate law as:

$\frac{dx}{dt}=k(a-x)(p+x)$

where a = [A₀] and p = [P₀].

Integrating throughout, we have

$\int_{0}^{x}\frac{dx}{(a-x)(p+x)}=k\int_{0}^{t}dt\; \; \; \; \; \; \; \; 14$

Substituting the partial fraction expression $\frac{1}{(a-x)(p+x)}=\frac{1}{a+p}\left ( \frac{1}{a-x}+\frac{1}{p+x} \right )$ in eq14 and working out the algebra yields

$kt=\frac{1}{a+p}ln\frac{a(p+x)}{p(a-x)}$

which is equivalent to

$kt=\frac{1}{[A_0]+[P_0]}ln\frac{[A_0][P]}{[P_0][A]}$

Question

The mechanism of the above reaction is found to include the following steps:

$A+P\rightleftharpoons AP$

$AP\rightarrow P+P\; \; \; \; \; \; (rds)$

With this in mind, derive the rate law and show that it is consistent with eq13a.

Answer

We can write the rate law as:

$\frac{d[P]}{dt}=2k_{rds}[AP]=2k_{rds}K[A][P]=k[A][P]$

where K is the equilibrium constant for the 1st step and k = 2k_rdsK. Furthermore, the overall reaction is A → P, which means that

$\frac{d[P]}{dt}=-\frac{d[A]}{dt}=k[A][P]$

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November 8, 2019September 26, 2024

2nd order reaction of the type (A + 2B → P)

The rate law for the 2nd order reaction of the type A + 2B → P is:

$\frac{d[A]}{dt}=-k[A][B]$

Using the same logic described in the previous article, we can rewrite the rate law as:

$\frac{dx}{dt}=k(a-x)(b-2x)$

where a = [A₀] and b = [B₀].

Rearranging the above equation and integrating throughout, we have

$\int_{0}^{x}\frac{dx}{(a-x)(\frac{b}{2}-x)}=2k\int_{0}^{t}dt\; \; \; \; \; \; \; \; 13$

Substituting the partial fraction expression $\frac{1}{(a-x)(\frac{b}{2}-x)}=\frac{1}{a-\frac{b}{2}}\left ( \frac{1}{\frac{b}{2}-x}-\frac{1}{a-x} \right )$ in eq13 and after some algebra, we have,

$kt=\frac{1}{b-2a}ln\frac{a(b-2x)}{b(a-x)}$

which is equivalent to

$kt=\frac{1}{[B_0]-2[A_0]}ln\frac{[A_o]([B])}{[B_0]([A])}$

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November 7, 2019September 20, 2025

Transition state theory

The transition state theory provides a theoretical way to calculate the pre-exponential factor A and the activation energy E_a, and therefore the rate constant k of an elementary chemical reaction. This is in contrast with the Arrhenius equation, where A and E_a are obtained empirically. The theory is based on statistical thermodynamics and can be illustrated for a gas-phase bimolecular reaction using the following assumptions:

1) The reacting system, expressed as $A+B\rightleftharpoons C^{\ddagger}\rightarrow P$ , proceeds along an energy path that includes a point of highest potential energy called the saddle point, where a distinct species known as the activated complex $C^{\ddagger}$ is formed. The conversion of the activated complex to the product P can be seen as an asymmetric vibrational stretch:

2) A rapid pre-equilibrium is established between the activated complex and the reactants: $A+B\rightleftharpoons C^{\ddagger}$ .

3) The rate of the reaction is attributed to the rate-determining step, which is the rate of conversion of the activated complex to the product: $rate=k^{\ddagger}[C^{\ddagger}]$ , where $k^{\ddagger}$ is proportional to the vibrational frequency $v$ of the activated complex.

The expression for the gas-phase rapid equilibrium from assumption 2 is:

$K^{\ddagger}=\frac{\frac{p_{c^{\ddagger}}}{p^o}}{\frac{p_A}{p^o}\frac{p_B}{p^o}}=\frac{p^op_{C^{\ddagger}}}{p_Ap_B}$

which in terms of molar concentration is:

$K^{\ddagger}=\frac{p^o}{RT}\frac{[C^{\ddagger}]}{[A][B]}\; \; \; \; \; \; \; \; 1$

Substituting eq1 into the rate equation , $rate=k^{\ddagger}[C^{\ddagger}]$ , yields:

$rate=k[A][B]\; \; \; \; \; \; \; where\; k=\frac{RT}{p^o}k^{\ddagger}K^{\ddagger}$

In general, the equilibrium constant, when written in terms of standard molar partition functions, is $K=\left [ \prod _j\left ( \frac{q^o_{j,m}}{N_A} \right )^{v_j} \right ]e^{-\frac{\Delta_rE_0}{RT}}$ (see this article for derivation), where $v_j$ is the respective stoichiometric coefficient of the J-th species. For our bimolecular reaction,

$K^{\ddagger}=\frac{N_A\, q^o_{C^{\ddagger},m}}{q^o_{A,m}q^o_{B,m}}e^{-\frac{\Delta_rE_0}{RT}}\; \; \; \; \; \; \; \; 2$

where $\Delta_rE_0=E_0(C^{\ddagger})-E_0(A)-E_0(B)$ .

The activated complex in our example is a linear molecule with N = 3 atoms and therefore has 3N – 5 = 4 modes of vibration (2 bending, 1 symmetric stretch and 1 asymmetric stretch). The standard molar partition function for the activated complex is the product of the partition functions of different modes of motion: $q^o_{C^{\ddagger},m}=q^{o\; \; \; \; \; \; T}_{C^{\ddagger},m}q^{o\; \; \; \; \; \; V}_{C^{\ddagger},m}q^{o\; \; \; \; \; \; R}_{C^{\ddagger},m}$ , which we can rewrite in the form: $q^o_{C^{\ddagger},m}=q^{o\; \; \; \; \; V_{asym}}_{C^{\ddagger},m}\, \bar{q}^{\, o}_{C^{\ddagger},m}$ , where $\bar{q}^{\, o}_{C^{\ddagger},m} =q^{o\; \; \; \; \; T}_{C^{\ddagger},m}\, q^{o\; \; \; \; \; V_{others}}_{C^{\ddagger},m}\, q^{o\; \; \; \; \; R}_{C^{\ddagger},m}$ .

The standard molar partition function for the asymmetric vibration is $q^{o\; \; \; \; \; \; V_{asym}}_{C^{\ddagger},m}=\frac{1}{1-e^{-\frac{hv}{kT}}}$ , where $v$ is the vibrational frequency that leads to the conversion of the activated complex to the product. Assuming $\frac{hv}{kT}\ll 1$ , $q^{o\; \; \; \; \; \; V_{asym}}_{C^{\ddagger},m}=\frac{1}{1-e^{-\frac{hv}{kT}}}\approx\frac{kT}{hv}$ . Therefore,

$q^o_{C^{\ddagger},m}=q^{o\; \; \; \; \; V_{asym}}_{C^{\ddagger},m}\, \bar{q}^{\, o}_{C^{\ddagger},m}\approx\frac{kT}{hv}\bar{q}^{\, o}_{C^{\ddagger},m}\; \; \; \; \; \; \; \; 3$

Substituting eq3 into eq2 gives:

$K^{\ddagger}=\frac{kT}{hv}\bar{K}^{\ddagger}\; \; \; \; \; \; \; \; 4$

where $\bar{K}^{\ddagger}=\frac{N_A\, \bar{q}^{\, o}_{C^{\ddagger},m}}{q^o_{A,m}q^o_{B,m}}e^{-\frac{\Delta_rE_0}{RT}}$

Substituting eq4 into the rate constant equation $k=\frac{RT}{p^o}k^{\ddagger}K^{\ddagger}$ yields:

$k=\frac{RT}{p^o}k^{\ddagger}\frac{kT}{hv}\bar{K}^{\ddagger}\; \; \; \; \; \; \; \; 5$

From assumption 3, $k^{\ddagger}$ is proportional to the vibrational frequency $v$ of the activated complex,

$k^{\ddagger}=\kappa v\; \; \; \; \; \; \; \; 6$

where $\kappa$ is the proportionality constant called the transmission coefficient, which accounts for the notion that not every oscillation of the activated complex leads to its conversion to the product.

Substituting eq6 into eq5 results in:

$k=\kappa\frac{RT}{p^o}\frac{kT}{h}\bar{K}^{\ddagger}\; \; \; \; \; \; \; \; 7$

The standard Gibbs energy is $\Delta G^o=-RTlnK$ . By analogy, we define the standard Gibbs activation energy as $\Delta^{\ddagger}G^o=-RTln\bar{K}^{\ddagger}$ , which we shall substitute into eq7 to give:

$k=\kappa\frac{RT}{p^o}\frac{kT}{h}e^{-\frac{\Delta^{\ddagger}G^o}{RT}}=\kappa\frac{RT}{p^o}\frac{kT}{h}e^{\frac{\Delta^{\ddagger}S^o}{R}}e^{-\frac{\Delta^{\ddagger}H^o}{RT}}\; \; \; \; \; \; \; \; 8$

Question

Is the expression for standard Gibbs activation energy still valid when the equilibrium constant now excludes the factor $q_{C^{\ddag},m}^{o}\, ^{V_{asym}}$ ?

Answer

It is an approximation that we make, which delivers an expression for the rate constant that is in good agreement with experimental values.

Eq7 and eq8 are different forms of the Eyring equation. Substituting eq8 in the definition of activation energy $E_a=RT^2\frac{\partial lnk}{\partial T}$ and differentiating, noting that $\Delta^{\ddagger}S^o$ and $\Delta^{\ddagger}H^o$ are standard states, which are temperature-independent, gives:

$E_a=\Delta^{\ddagger}H^o+2RT\; \; \; \; \; \; \; \; \; 9$

Substituting eq9 into eq8 yields:

$k=\kappa e^2\frac{RT}{p^o}\frac{kT}{h}e^\frac{\Delta^{\ddagger}S^o}{R}e^{-\frac{E_a}{RT}}\; \; \; \; \; \; \; \; 10$

Comparing eq10 and the Arrhenius equation, we have:

$A=\kappa e^2\frac{RT}{p^o}\frac{kT}{h}e^\frac{\Delta^{\ddagger}S^o}{R}$

Repeating all the above steps for a gas-phase unimolecular reaction, we arrive at an activation energy of $E_a=\Delta^{\ddagger}H^o+RT$ instead of eq9. Therefore, the transition state theory predicts that the activation energy for a gas-phase elementary reaction is

$E_a=\Delta^{\ddagger}H^o+xRT$

where $x=1$ for a unimolecular reaction and $x=2$ for a bimolecular reaction.

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November 7, 2019October 16, 2024

2nd order reaction of the type (A + B → P)

The rate law for the reaction A + B → P is:

$\frac{d[A]}{dt}=-k[A][B]\; \; \; \; \; \; \; \; 11$

Unlike the 2nd order reaction of the type A → P with a rate law of $\frac{d[A]}{dt}=-k[A]^2$ , eq11 cannot be integrated unless we express [B] in terms of [A], or [A] and [B] in terms of a third variable x. Since A and B react in a ratio of 1:1, the remaining concentrations of A and B at any time of the reaction are:

$[A]=[A_0]-x\; \; \; and\; \; \; [B]=[B_0]-x$

where [A₀] and [B₀] are the initial concentrations of A and B respectively.

Differentiating both sides of [A] = [A₀] – x with respect to time yields

$\frac{d[A]}{dt}=-\frac{dx}{dt}$

Therefore, we can rewrite eq11 as

$\frac{dx}{dt}=k\left ( a-x \right )\left ( b-x \right )\; \; \; \; \; \; \; \; 12$

where a = [A₀] and b = [B₀]

Integrating eq12 using the partial fraction expression of $\frac{1}{(a-x)(b-x)}=\frac{1}{b-a}\left ( \frac{1}{a-x}-\frac{1}{b-x} \right )$ gives

$\int_{0}^{x}\frac{1}{b-a}\left ( \frac{1}{a-x}-\frac{1}{b-x} \right )dx=k\int_{0}^{t}dt$

After some algebra, we have,

$kt=\frac{1}{b-a}ln\frac{a(b-x)}{b(a-x)}$

which is equivalent to

$kt=\frac{1}{[B_0]-[A_0]}ln\frac{[A_0][B]}{[B_0][A]}$

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October 27, 2019September 12, 2025

Equilibrium constant (derivation)

The equilibrium constant of a chemical reaction is the reaction quotient of the reaction at dynamic equilibrium.

The derivation of the equilibrium constant involves the following steps:

1. Derive the expression for the Gibbs energy of a multi-component reaction.
2. Derive the chemical potential of a multi-component system.
3. Combine the two expressions.

Step 1

From eq14 of the article on the Nernst equation, the reaction Gibbs energy for a reaction is

$\Delta_rG=\sum_j\mu_jv_j\; \; \; \; \; \; \; \; 1$

and at standard state

$\Delta_rG^{\: o}=\sum_j\mu_j^{\; \; o}v_j\; \; \; \; \; \; \; \; 2$

Step 2

From eq24 of the article on the Nernst equation, the chemical potential of a multi-component system is

$\mu_j=\mu_j^{\; \, o}+RTlna_j\; \; \; \; \; \; \; \; 3$

Step 3

Combining eq1 and eq3

$\Delta_rG=\sum_j\left ( \mu_j^{\; o}+RTlna_j \right )v_j=\sum_j\mu_j^{\; o}v_j+RT\sum_jv_jlna_j\; \; \; \; \; \; \; \; 4$

Substitute eq2 in eq4

$\Delta_rG=\Delta_rG^{\; o}+RT\sum_jlna_j^{\: v_j}\; \; \; \; \; \; \; \; 5$

Since lnx^a+lnx^b +… = ln(x^ax^b …), eq5 becomes

$\Delta_rG=\Delta_rG^{\; o}+RTln\prod _ja_j^{\: v_j}\; \; \; \; \; \; \; \; 6$

Let $Q=\prod_ja_j^{\; v_j}$ , where Q is the reaction quotient.

$\Delta_rG=\Delta_rG^{\; o}+RTlnQ\; \; \; \; \; \; \; \; 7$

At equilibrium, a reversible reaction is spontaneous in neither direction. Hence the change in Gibbs energy with respect to the change in the extent of the reaction, i.e. the reaction Gibbs energy $\Delta_rG=\left ( \frac{\partial G}{\partial\xi} \right )_{T,p}$ , is zero. Eq7 becomes

$\Delta_rG^{\circ}=-RTlnK\; \; \; \; \; \; \; \; 8$

where K is the reaction quotient at equilibrium, i.e.

$K=\left ( \prod _ja_j^{\; v_j} \right )_{eqm}\; \; \; \; \; \; \; \; 9$

Since the value K of does not change for a particular reaction at constant temperature, we call it the equilibrium constant.

But how does $K$ , which is dimensionless, relate to $K_p$ and $K_c$ ? $K_p$ and $K_c$ are approximations of the thermodynamic equilibrium constant $K$ . For an ideal gas,

$K=\prod_j\biggr$\frac{p_j}{p^o}\biggr$^{v_j}=\frac{\prod_j(p_j)^{v_j}}{(p^o)^{\Delta v}}=\frac{K_p}{(p^o)^{\Delta v}}\; \; \; \; \; \; \; \; 10$

where $\Delta v=v_{\alpha}+v_{\beta}+\cdots+v_a+v_b+\cdots$ .

Similarly, for solutes in dilute solutions,

$K=\prod_j\biggr$\frac{c_j}{c^o}\biggr$^{v_j}=\frac{\prod_j(c_j)^{v_j}}{(c^o)^{\Delta v}}=\frac{K_c}{(c^o)^{\Delta v}}\; \; \; \; \; \; \; \; 11$

Eq21a and eq21b show that $K_p$ and $K_c$ are only unitless when $\Delta v=0$ . Equating the two expressions gives:

$K_p=K_c(RT^o)^{\Delta v}\; \; \; \; \; \; \; \; 12$

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October 22, 2019September 26, 2024

Nernst equation (derivation)

In 1887, Walther Nernst, a German chemist, developed the Nernst equation, which describes the relationship between an electrochemical reaction’s potential and its standard electrode potential under standard and non-standard conditions.

The derivation of the Nernst equation involves the following steps:

1. Derive the formula for the reversible open circuit potential , E^o, of an electrochemical cell at constant temperature and pressure under standard conditions.
2. Derive the expression for E^o in terms of chemical potentials of the reaction species at constant temperature and pressure.
3. Derive a ‘correction factor’ for E^oin step2. This final expression, the Nernst equation, describes how the resultant E^ovaries with activities of chemical species.

Step 1

From the definition of enthalpy, H = U + pV, we have

$dH=dU+d(pV)\; \; \; \; \; \; \; \; 1$

Substitute the definition of the internal energy of a system undergoing reversible change, dU = dq_rev+ dw_rev in eq1

$dH=dq_{rev}+dw_{rev}+d(pV)\; \; \; \; \; \; \; \;2$

Substitute the definition of the change in Gibbs energy of a system, dG = dH – d(TS) in eq2

$dG+TdS+SdT=dq_{rev}+dw_{rev}+pdV+Vdp$

At constant temperature and pressure, dT = dp = 0

$dG=dq_{rev}+dw_{rev}+pdV-TdS\; \; \; \; \; \; \; \; 3$

Substitute the definition of the change in entropy of a system, dS = dq_rev/T in eq3

$dG=dw_{rev}+pdV\; \; \; \; \; \; \; \; 4$

Substitute the definition of total reversible work, w_rev = w_ex+ w_add, where w_ex is reversible expansion work and w_add is reversible additional work other than reversible expansion work, in eq4

$dG=dw_{ex}+dw_{add}+pdV\; \; \; \; \; \; \; \; 5$

Substitute the definition of the change in reversible expansion work, dw_ex= –pdV in eq5

$dG=dw_{add}\; \; \; \; \; \; \; \; 6$

Substitute the definition of the reversible electrical work between two points in an electric circuit, dw_add= –nFEdξ in eq6

$\Delta _rG=-nFE\; \; \; \; \; \; \; \; 7$

where n is the number of moles of electrons, F is the Faraday constant, E is the open circuit potential of the electrochemical cell and Δ_rG = dG/dξ is the Gibbs energy of the electrochemical cell reaction.

When the open circuit potential of an electrochemical cell is measured at standard conditions (1 bar, 298K, 1M), we write eq7 as:

$\Delta _rG^{\, o}=-nFE^{\, o}\; \; \; \; \; \; \; \; 8$

Step 2

The chemical potential of the j-th species, in a multi-component system is

$\mu_j=\left (\frac{\partial G}{\partial n_j} \right )_{T,p,n'}\; \; \; \; \; \; \; 9$

where n‘ is all n other than nj.

The total differential of the Gibbs energy of the system at constant temperature and pressure is

$dG=\left (\frac{\partial G}{\partial n_a} \right )_{T,p,n'}dn_a+\left (\frac{\partial G}{\partial n_b} \right )_{T,p,n'}dn_b+...\; \; \; \; \; \; \; 10$

Substituting eq9 in eq10

$dG=\mu_adn_a+\mu_bdn_b+...=\sum_{j}\mu_jdn_j\; \; \; \; \; \; \; \; 11$

Eq11 can be expressed in another way by considering the reversible reaction

$v_aa+v_bb+...\rightleftharpoons v_{\alpha}\alpha+v_{\beta}\beta+...$

with the change in Gibbs energy at constant temperature and pressure of the system given by eq11. Next, let’s define

$dn_j=v_jd\xi\; \; \; \; \; \; \; \; 12$

where vj is the stoichiometric number of the j-th species in the reversible reaction and dξ is the amount of substance that is being changed in the reaction.

ξ is called the extent of a reaction and has units of amount in moles. Note that the stoichiometric numbers for reactants are negative by convention and those for products are positive. Substituting eq12 in eq11

$\left ( \frac{\partial G}{\partial \xi} \right )_{T,p}= \sum_{j}\mu_jv_j\; \; \; \; \; \; \; \; 13$

Substitute the definition of the Gibbs energy of an electrochemical cell reaction, $\Delta _rG=\left ( \frac{\partial G}{\partial \xi} \right )_{T,p}$ , into eq13

$\Delta _rG=\sum_{j}\mu_jv_j\; \; \; \; \; \; \; \; 14$

Substitute eq7 from Step1 in eq14

$E=-\frac{1}{nF}\sum_{j}\mu_jv_j\; \; \; \; \; \; \; \; 15$

When the open circuit potential of an electrochemical cell is measured at standard conditions (1 bar, 298K, 1M), we write eq15 as:

$E^{\, o}=-\frac{1}{nF}\sum_{j}\mu^o_{\, j}v_j\; \; \; \; \; \; \; \; 16$

Step 3

Firstly, let’s consider a system with a single pure substance. Substitute the definition of the change in Gibbs energy of a system, dG = dH – d(TS), in the change of enthalpy of the system dH = dU + d(pV)

$dG+SdT+TdS=dU+pdV+Vdp\; \; \; \; \; \; \; \; 17$

The reversible change in internal energy of a pure substance (a system with a single pure substance only does expansion work) can be expressed as

$dU=dq_{rev}+dw_{ex}\; \; \; \; \; \; \; \; 18$

Substitute the definitions of entropy, dS = dq_rev/T, and expansion work, –pdV, in eq18

$dU=TdS-pdV\; \; \; \; \; \; \; \; 19$

Substitute eq19 in eq17

$dG=Vdp-SdT\; \; \; \; \; \; \; \; 20$

At constant temperature, dT = 0 and eq20 becomes dG = Vdp. Integrating this new expression on both sides from $p_i$ to $p_f$ ,

$G\left ( p_f \right )=G\left ( p_i \right )+nRTln\frac{p_f}{p_i}$

Let p_i = 1 bar = p^o(i.e. standard conditions) and p_f = p

$G=G^{\, o}+nRTln\frac{p}{p^{\, o}}$

Secondly, we extend the above working to a multi-component system of ideal gases. Since G is an extensive property, we add the Gibbs energies of all components to give the total Gibbs energy of the system:

$G_{total}=\left ( G_a^{\: o}+n_aRTln\frac{p_a}{p_a^{\, o}} \right )+\left ( G_b^{\: o}+n_bRTln\frac{p_b}{p_b^{\, o}} \right )+...$

Let’s assume the reference pressures of the component gases are the same since they are all ideal gases, i.e. p_a^o= p_b^o= p^o

$G_{total}=\left ( G_a^{\, o}+G_b^{\, o}+...\right )+\left ( n_aRTln\frac{p_a}{p^o}+n_bRTln\frac{p_b}{p^o}+...\right )$

$G_{total}=G_{total}^{\, o}+\left ( n_aRTln\frac{p_a}{p^o}+n_bRTln\frac{p_b}{p^o}+...\right )$

For simplicity, let G_Totaland G^o_Total be G and G^orespectively.

$G=G^{\, o}+\left ( n_aRTln\frac{p_a}{p^o}+n_bRTln\frac{p_b}{p^o}+...\right )$

Taking the partial derivative of the above with respect to n_j, at constant temperature, pressure and the amount of the other components, n‘, we have

$\left ( \frac{\partial G}{\partial n_j} \right )_{T,p,n'}=\left ( \frac{\partial G^{\, o}}{\partial n_j} \right )_{T,p,n'}+\left [ \frac{\partial \left ( n_aRTln\frac{p_a}{p^{\, o}}+ n_bRTln\frac{p_b}{p^{\, o}}+...\right )}{\partial n_j} \right ]_{T,p,n'}$

For the last partial derivative of the above equation, only the j-th term survives.

$\left ( \frac{\partial G}{\partial n_j} \right )_{T,p,n'}=\left ( \frac{\partial G^{\, o}}{\partial n_j} \right )_{T,p,n'}+RTln\frac{p_j}{p^{\, o}}\; \; \; \; \; \; \; \; 21$

Substitute eq9 from Step 2 in eq21

$\mu_j=\mu_j^{\, o}+RTln\frac{p_j}{p^{\, o}}\; \; \; \; \; \; \; \; 22$

Similarly, for a solute in a dilute solution satisfying Henry’s law, we can write

$\mu_j=\mu_j^{\, o}+RTln\frac{c_j}{c^{\, o}}\; \; \; \; \; \; \; \; 23$

where c_jand c^oare the j-th solute’s concentration and molar concentration respectively.

We can fine-tune eq22 and eq23 by replacing p_j/p^oand c_j/c^owith γ_j(p_j/p^o) and γ_j(c_j/c^o) where γ_j is a factor called the activity coefficient that can account for both ideal and non-ideal fluids (γ = 1 and γ < 1 for ideal and non-ideal fluids respectively). Next, we combine the modified equations into a single equation by letting a_j= γ_j(p_j/p^o) and a_j= γ_j(c_j/c^o), where a_j is the activity of species j.

$\mu_j=\mu_j^{\, o}+RTlna_j\; \; \; \; \; \; \; \; 24$

Substituting eq24 in eq15 from Step 2,

$E=-\frac{1}{nF}\sum_{j}\left ( \mu_j^{\, o}+RTlna_j \right )v_j=-\frac{1}{nF}\sum_{j} \mu_j^{\, o}v_j-\frac{RT}{nF}\sum_{j}v_jlna_j \; \; \; \; \; \; \; \; 25$