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Ground state (quantum mechanics)

The ground state of a system is the lowest energy state of the system.

The state of a system is dependent on the way electrons are distributed in a chemical species. A distribution that results in the system having the lowest energy is the ground state of the system. Every other distribution configuration  is associated with a higher energy state known as an excited state. For example, the ground state of carbon is given by the term symbol 3P, while the term symbols 1D and 1S denote excited states of carbon.

 

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

\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]

Substitute [A]=[A_0]e^{-k_1t} in the above equation and rearranging,

[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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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 = x0 when 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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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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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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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 = 2krdsK. 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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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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Transition state theory

The transition state theory provides a theoretical way to calculate the pre-exponential factor A and the activation energy Ea, and therefore the rate constant k of an elementary chemical reaction. This is in contrast with the Arrhenius equation, where A and Ea 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

Substitute eq1 in the rate equation , rate=k^{\ddagger}[C^{\ddagger}],

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}} where v_j is the respective stoichiometric coefficient of the 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

Substitute eq3 in eq2

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}}

Substitute eq4 in the rate constant equation k=\frac{RT}{p^o}k^{\ddagger}K^{\ddagger}

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.

Substitute eq6 in eq5,

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 in eq7,

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. Substitute eq8 in the definition of activation energy E_a=RT^2\frac{\partial lnk}{\partial T} and differentiate, noting that \Delta^{\ddagger}S^o and \Delta^{\ddagger}H^o are standard states, which are temperature-independent, we have

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

Substitute eq9 in eq8

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,

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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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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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 lnxa +lnxb +… = ln(xaxb …), 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.

 

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