Approximate calculations of the product distribution under kinetic control

Question

What is the product distribution when the energy difference between the transition states (∆TS = TS1 – TS2), is at 2 kcal/mol and when it is at 5 kcal/mol (assuming room temperature)?

I know that 1.4 kcal/mol approximately corresponds to factor 10 and that for a difference of, for example, 3 kcal/mol the product distribution would approximately be 100:1 P2:P1.

But I find it hard to know the distribution at a difference of 1, 2, 4, and 5 kcal/mol.

Could anyone please hook me up with some approximate calculations or values and explain to me how to think in order to understand this without having to go through all sorts of equations like Eyring and Ahrrenius?

Answer 1

I am afraid you have to use either the Arrhenius or the Eyring equation to get a rough estimate of the product distribution.

If you assume that the pre-exponential terms in either of these equations are the same or very similar, you will find that

$$ \mathrm{\frac{k_2}{k_1} = \exp \left( \frac{\Delta TS}{RT}\right)} $$

At 298 K, $\mathrm{RT = 0.593} \, \mathrm{kcal.mol^{-1}}$ so

$$ \mathrm{\frac{k_2}{k_1} = \exp \left( \frac{\Delta TS}{0.593}\right)} $$

and substituting in this equation we will obtain the values we are looking for.

Answer 2

Quick Answer

The relative product concentrations are dependent on $\Delta E_a$, $T$, orders, and reversibility of the reactions. It would be inaccurate to specify relative product concentrations simply based on $\Delta E_a$. For competitive reactions of the same order, the relative product concentrations are a simple ratio of the rate constants. Moreover, reversibility is crucial for thermodynamic and kinetic control.

The following values tabulated from Equation (5) should help you estimate the relative product concentrations $\dfrac{\ce{[P_2]}}{\ce{[P_1]}}$ for competitive same-order non-reversible reactions:

$\Delta E_a \pu{kcal mol^{-1}}$	$\pu{100 K}$	$\pu{200 K}$	$\pu{300 K}$	$\pu{400 K}$	$\pu{500 K}$
1	1.53E2	1.24E1	5.35	3.52	2.74
2	2.35E4	1.53E2	2.86E1	1.24E1	7.49
3	3.61E6	1.89E3	1.53E2	4.36E1	2.05E1
4	5.53E8	3.25E4	8.21E2	1.53E2	5.60E1
5	8.48E10	2.91E5	4.39E3	5.40E2	1.53E2

Sigh! I tried, but there is no generalized simple formula to do this for such a wide range of $\Delta E_a$, even approximately so at a fixed temperature. You will have to go through The Fun Part.

The Fun Part

The information provided just isn't enough to claim any results. The following derivation makes some assumptions, that the two:

1. reactions have the same order,
2. products have zero initial concentration,
3. rate constants follow the Arrhenius equation, and
4. pre-exponential factors in the Arrhenius equation are equal.

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Consider two reactions:

$$ \ce{A ->[$k_1$][$n_1$] P1} \tag{1} $$

$$ \ce{A ->[$k_2$][$n_2$] P2} \tag{2} $$

where $k_1$ and $k_2$ and $n_1$ and $n_2$ are the rate constants and orders, respectively, of Reactions (1) and (2). Note that I have assumed that the reactions are not reversible:

$$ \ce{P1 ->[$k\,=\,0$] A}\\ \ce{P2 ->[$k\,=\,0$] A} $$

Same Order Competitive Reactions

Assuming $n_1=n_2=n$, the rates $R_1$ and $R_2$ for reactions (1) and (2), respectively, are given by:

$$ R_1 = \dfrac{\mathrm{d}\ce{P1}}{\mathrm{d}t} = k_1\ce{[A]^n} \tag{3}\\ $$

$$ R_2 = \dfrac{\mathrm{d}\ce{P2}}{\mathrm{d}t} = k_2\ce{[A]^n} \tag{4}\\ $$

Dividing Equation (4) with Equation (3):

$$ \dfrac{R_2}{R_1} = \dfrac{\dfrac{\mathrm{d}\ce{[P_2]}}{\mathrm{d}t}}{\dfrac{\mathrm{d}\ce{[P_1]}}{\mathrm{d}t}} = \dfrac{\mathrm{d}\ce{[P_2]}}{\mathrm{d}\ce{[P_1]}} = \dfrac{k_2}{k_1} $$

Assuming initial concentrations of products is zero, upon integration we obtain:

$$ \dfrac{\ce{[P2]}}{\ce{[P1]}} = \dfrac{k_2}{k_1}\\ $$

Arrhenius Approximation

Assume the rate constants follow the Arrhenius equation:

$$ k_1 = A_1\mathrm{e}^{-\dfrac{(E_a)_1}{RT}}\\ k_2 = A_2\mathrm{e}^{-\dfrac{(E_a)_2}{RT}}\\ $$

Assuming $A_1 = A_2$ and with Equation (3):

$$ \dfrac{\ce{[P2]}}{\ce{[P1]}} = \dfrac{k_2}{k_1} = \mathrm{e}^{-\dfrac{\Delta E_a}{RT}} \tag{5} $$

where $\Delta E_a =(E_a)_2-(E_a)_1$. Notice that we still have a temperature dependence. Plotting $\dfrac{\ce{[P2]}}{\ce{[P1]}}$ with varying $T$ and $\Delta E_a$, we obtain, with $R = \pu{1.987\times 10^{-3} kcal K^{-1} mol^{-1}}$:

Legend

1. DEa: $\Delta E_a/\pu{kcal mol^{-1}}$
2. T: $T/\pu{K}$
3. [P2]/[P1]: $\dfrac{\ce{[P2]}}{\ce{[P1]}}$

Discussion

Higher $\dfrac{\ce{[P2]}}{\ce{[P1]}}$ is obtained at lower $T$ and higher $\Delta E_a$.

I know that 1.4 kcal/mol approximately corresponds to factor $\left( \dfrac{\ce{[P2]}}{\ce{[P1]}} \right)$ 10, 3 kcal/mol the product distribution would approximately be $100:1::\ce{[P_2]}:\ce{[P_1]}$.

From Equation (5), the relative product concentrations are dependent on both $\Delta E_a$ and $T$. It would be inaccurate to specify relative product concentrations simply based on $\Delta E_a$. You can immediately see that your rough estimates are extremely affected by even small changes in $\Delta E_a$ and $T$.

Note: Since reactions are not reversible, there is no thermodynamic or kinetic control here. Relative product concentrations are simply dependent on $\Delta E_a$ and $T$. In other word, reversibility is crucial in establishing thermodynamic and kinetic control.

Further Considerations

Varying the equation for rate constant, orders of the reactions, and reversibility may provide interesting insights, but will require extensive numerical analysis.

Answer 3

I am posting an alternative answer because none provide a general expression for the product distribution. It is impossible to know it without:

• The rate laws that follow both reactions. If you do not know the rate law, to answer the question is impossible.
• Kinetic data regarding the Arrhenius equation for both reactions. If you only know the difference of activation energies, to answer the question is impossible. You also need the ratio of the frequency factors, i.e., $A_2/A_1$.
• The answer depends on the concentration of the reactant $R$. If the final concentration is unknown, to answer the question is impossible.

1. Reactions The chemical reactions are \begin{align} \ce{R -> P\text{1}} \quad r_1 &= k_1 C_\mathrm{R}^\alpha \tag{1} \\ \ce{R -> P\text{2}} \quad r_2 &= k_2 C_\mathrm{R}^\beta \tag{2} \end{align} where we assume power laws for both of them.

2. Fractional Yield In multiple reactions, it is useful to define the fractional yield $\varphi_{X,Y}$. This is defined as the rate of generation/consumption of species $X$ with respect to the rate of generation/consumption of species $Y$. This allows for rapid calculations as we show below.

We will consider the case for $P1$. In here, $R$ can be consumed by two reactions, so its rate has two terms. Using Eqs. (1-2) yields \begin{align} \varphi_\mathrm{P1/R} &= \frac{\mathrm{d}C_\mathrm{P1}}{-dC_\mathrm{R}} \tag{3} \\ \varphi_\mathrm{P1/R} &= \frac{r_\mathrm{1,P}}{-(r_\mathrm{1,R} + r_\mathrm{2,R})} \\ \varphi_\mathrm{P1/R} &= \frac{k_1 C_\mathrm{R}^\alpha}{k_1 C_\mathrm{R}^\alpha + k_2 C_\mathrm{R}^\beta} \\ \varphi_\mathrm{P1/R} &= \frac{1}{1 + (k_2/k_1) C_\mathrm{R}^{\beta - \alpha}} \tag{4} \\ \end{align} Two comments regarding Eq. (4):

• Only the ratio of the frequency factors $A_2/A_1$ and the difference between the activation energies $E_\mathrm{a2} - E_\mathrm{a1}$ is needed. Let me call $ k := k_2/k_1$.
• Only the difference of partial reaction orders $\alpha - \beta$ is needed. Let me call $\nu := \beta - \alpha $. As an example, I will consider $\nu = 1$.

Now we integrate Eq. (4) by noting the equality with Eq. (3), under the assumption that initially $P1$ was not present \begin{align} \frac{\mathrm{d}C_{P1}}{-\mathrm{d}C_R} &= \frac{1}{1 + k C_\mathrm{R}^\nu} \\ \int_{0}^{C_\mathrm{P1}} \mathrm{d}C_\mathrm{P1} &= - \int_{C_\mathrm{R0}}^{C_\mathrm{R}} \frac{\mathrm{d}C_\mathrm{R}}{1 + k C_\mathrm{R}} \rightarrow \boxed{C_\mathrm{P1}(C_\mathrm{R}) = -\frac{1}{k} \ln\left(\frac{1 + kC_\mathrm{R}}{1 + kC_\mathrm{R0}}\right)} \tag{5} \end{align}

Now, we could do the same for the fractional yield of $\varphi_\mathrm{P2/R}$. However, we have an easier path looking at Eqs. (1-2). The stoichiometry and the extents of reactions $X_1$ and $X_2$ (in terms of molarity) tells us \begin{align} C_\mathrm{R} &= C_\mathrm{R0} - X_1 - X_2 \\ C_\mathrm{P1} &= X_1 \\ C_\mathrm{P2} &= X_2 \\ \tag{6,7,8} \end{align} but \begin{align} \require{cancel} &C_\mathrm{R} + C_\mathrm{P1} + C_\mathrm{P2} = C_\mathrm{R0} \cancel{- X_1 - X_2} + \cancel{X_1 + X_2} \\ &C_\mathrm{R} + C_\mathrm{P1} + C_\mathrm{P2} = C_\mathrm{R0} \\ &C_\mathrm{P2} = C_\mathrm{R0} - C_\mathrm{R} - C_\mathrm{P1} \rightarrow \boxed{C_\mathrm{P2}(C_\mathrm{R}) = C_\mathrm{R0} - C_\mathrm{R} + \frac{1}{k} \ln\left(\frac{1 + kC_\mathrm{R}}{1 + kC_\mathrm{R0}}\right)} \tag{9} \end{align}

Eqs. (5) and (9) are the final expressions that give the distribution of the products. If other order of reactions take place, the process is the same, but the primitives are different. An analytical expression always exists provided that $\nu$ is an integer.

Thus, to know the distribution we need:

• The initial and final concentration of the reactant $R$.
• The value of $k$, that at temperature $T$ is $$ k = \frac{k_2}{k_1} = \frac{A_2}{A_1} \exp\left(-\frac{E_{a2} - E_{a1}}{RT}\right) \tag{10} $$

3. Example Eqs (5) and (9) may be plotted as a function of $C_\mathrm{R}$ at a fixed temperature that fixes $k$. I will do an example with an initial concentration of $C_\mathrm{R0} = \pu{1 mol/L}$ with two different values of $k$. This situation is funny, because you should read the graph from right to left because $C_\mathrm{R}$ decreases in that way:

Note how the product distribution changes with $k$:

1. If $k<1$ (circles) $\text{P1}$ dominates.
2. If $k>1$ (squares) $\text{P2}$ dominates.

If you have some way of controlling the temperature, and enlarging the differences between the distribution of products, you can play with the equations in order to obtain the value of $k$. This means, ending the reaction with essentially $\text{P1}$ or $\text{P2}$. This is only true if you do not activate other reactions when you change temperature, which unfortunately, happens a lot.