H\(_3^+\) Deuteration Rate Derivation

This page derives the integrated rate equations for the deuteration of H\(_3^+\) as studied in Hugo et al. (2009). These equations are used in the Week 5 notebook.

Overview

In Hugo et al. 2009, the deuteration of H\(_3^+\) was investigated in an ion trap at 13.5 K. The key reactions and their rates are:

Reaction	Rate (cm\(^{-3}\) s\(^{-1}\))
H\(_3^+\) + HD \(\to\) H\(_2\)D\(^+\) + H\(_2\)	\(k_1^{(2)}[\text{HD}][\text{H}_3^+]\)
H\(_2\)D\(^+\) + HD \(\to\) D\(_2\)H\(^+\) + H\(_2\)	\(k_2^{(2)}[\text{HD}][\text{H}_2\text{D}^+]\)
D\(_2\)H\(^+\) + HD \(\to\) D\(_3^+\) + H\(_2\)	\(k_3^{(2)}[\text{HD}][\text{D}_2\text{H}^+]\)

In these equations, brackets indicate the number density (cm\(^{-3}\)), and the \(k_n^{(2)}\) refer to second-order rate coefficients in units of cm\(^3\) s\(^{-1}\) so that the rate has units of cm\(^{-3}\) s\(^{-1}\). At the low temperature, the reverse reactions are negligible. Furthermore, under the experimental conditions, HD is present in excess, and it is reasonable to treat [HD] as constant. Under these pseudo-first-order conditions, we can redefine the rate coefficients:

\[ k_n \equiv k_n^{(2)}[\text{HD}] \]

Using the rates above, we obtain a set of coupled differential equations describing the time evolution of the number densities:

\[ \frac{\text{d}[\text{H}_3^+]}{\text{d}t} = -k_1[\text{H}_3^+] \]

\[ \frac{\text{d}[\text{H}_2\text{D}^+]}{\text{d}t} = k_1[\text{H}_3^+] - k_2[\text{H}_2\text{D}^+] \]

\[ \frac{\text{d}[\text{D}_2\text{H}^+]}{\text{d}t} = k_2[\text{H}_2\text{D}^+] - k_3[\text{D}_2\text{H}^+] \]

\[ \frac{\text{d}[\text{D}_3^+]}{\text{d}t} = k_3[\text{D}_2\text{H}^+] \]

Solving for [H\(_3^+\)]

Solving for \([\text{H}_3^+](t)\) involves a standard first-order integrated rate equation. Rearranging the first equation above:

\[ \frac{\text{d}[\text{H}_3^+]}{[\text{H}_3^+]} = -k_1\,\text{d}t \]

Integrating both sides:

\[ \ln [\text{H}_3^+](t) = -k_1 t + C \implies [\text{H}_3^+](t) = Ae^{-k_1 t} \]

At \(t=0\), \([\text{H}_3^+] = [\text{H}_3^+]_0\), so \(A = [\text{H}_3^+]_0\) and:

\[ \boxed{[\text{H}_3^+](t) = [\text{H}_3^+]_0\, e^{-k_1 t}} \]

Solving for [H\(_2\)D\(^+\)]

To solve for the time evolution of \([\text{H}_2\text{D}^+]\), we substitute the result above into the second rate equation and rearrange:

\[ \frac{\text{d}[\text{H}_2\text{D}^+]}{\text{d}t} + k_2[\text{H}_2\text{D}^+] = k_1[\text{H}_3^+]_0\, e^{-k_1 t} \]

To make progress, we introduce an integrating factor \(\mu \equiv e^{k_2 t}\), so \(\text{d}\mu/\text{d}t = k_2 e^{k_2 t}\). Multiplying both sides by \(\mu\):

\[ \mu\frac{\text{d}[\text{H}_2\text{D}^+]}{\text{d}t} + [\text{H}_2\text{D}^+]\frac{\text{d}\mu}{\text{d}t} = k_1[\text{H}_3^+]_0\, e^{-(k_1-k_2)t} \]

The left-hand side is recognized as a product rule derivative, so we can write and integrate:

\[ \frac{\text{d}}{\text{d}t}\left(\mu[\text{H}_2\text{D}^+]\right) = k_1[\text{H}_3^+]_0\, e^{-(k_1-k_2)t} \]

\[ \mu[\text{H}_2\text{D}^+](t) = \frac{k_1[\text{H}_3^+]_0}{k_2 - k_1}\,e^{-(k_1-k_2)t} + C \]

Dividing by \(\mu = e^{k_2 t}\):

\[ [\text{H}_2\text{D}^+](t) = \frac{k_1[\text{H}_3^+]_0}{k_2 - k_1}\,e^{-k_1 t} + Ce^{-k_2 t} \]

Applying the boundary condition \([\text{H}_2\text{D}^+] = 0\) at \(t=0\) gives \(C = -k_1[\text{H}_3^+]_0/(k_2 - k_1)\). Substituting back:

\[ \boxed{[\text{H}_2\text{D}^+](t) = \frac{k_1[\text{H}_3^+]_0}{k_2 - k_1}\left(e^{-k_1 t} - e^{-k_2 t}\right)} \]

Note that if \(k_1 = k_2 \equiv k\), the denominator goes to zero and this form is undefined. In that case, the right-hand side of the equation to integrate simplifies to the constant \(k[\text{H}_3^+]_0\):

\[ \mu[\text{H}_2\text{D}^+](t) = k[\text{H}_3^+]_0\, t + C \implies [\text{H}_2\text{D}^+](t) = [\text{H}_3^+]_0\, kte^{-k t} + Ce^{-kt} \]

Applying \([\text{H}_2\text{D}^+] = 0\) at \(t=0\) gives \(C=0\), so:

\[ \boxed{[\text{H}_2\text{D}^+](t) = [\text{H}_3^+]_0\, kt\,e^{-k t}, \qquad (k = k_1 = k_2)} \]

Solving for [D\(_2\)H\(^+\)]

The procedure is the same as for \([\text{H}_2\text{D}^+]\). Substituting the result for \([\text{H}_2\text{D}^+](t)\) (assuming \(k_1 \neq k_2\)) into the rate equation for \([\text{D}_2\text{H}^+]\) and rearranging:

\[ \frac{\text{d}[\text{D}_2\text{H}^+]}{\text{d}t} + k_3[\text{D}_2\text{H}^+] = \frac{k_1 k_2[\text{H}_3^+]_0}{k_2 - k_1}\left(e^{-k_1 t} - e^{-k_2 t}\right) \]

Introducing the integrating factor \(\mu \equiv e^{k_3 t}\) and applying the same product-rule and integration steps:

\[ \mu[\text{D}_2\text{H}^+](t) = \frac{k_1 k_2[\text{H}_3^+]_0}{k_2 - k_1}\left(\frac{e^{-(k_1-k_3)t}}{k_3 - k_1} - \frac{e^{-(k_2-k_3)t}}{k_3 - k_2}\right) + C \]

Dividing by \(\mu = e^{k_3 t}\):

\[ [\text{D}_2\text{H}^+](t) = \frac{k_1 k_2[\text{H}_3^+]_0}{k_2 - k_1}\left(\frac{e^{-k_1 t}}{k_3 - k_1} - \frac{e^{-k_2 t}}{k_3 - k_2}\right) + Ce^{-k_3 t} \]

Applying the boundary condition \([\text{D}_2\text{H}^+] = 0\) at \(t=0\):

\[ C = -\frac{k_1 k_2[\text{H}_3^+]_0}{k_2 - k_1}\left(\frac{1}{k_3 - k_1} - \frac{1}{k_3 - k_2}\right) \]

Substituting back and collecting terms:

\[ \boxed{[\text{D}_2\text{H}^+](t) = \frac{k_1 k_2[\text{H}_3^+]_0}{k_2 - k_1}\left(\frac{e^{-k_1 t} - e^{-k_3 t}}{k_3 - k_1} - \frac{e^{-k_2 t} - e^{-k_3 t}}{k_3 - k_2}\right)} \]

Note that if any two of \(k_1\), \(k_2\), \(k_3\) are equal, an alternate form must be derived analogously to the \(k_1 = k_2\) case above.

Solving for [D\(_3^+\)]

This result follows directly from conservation of mass. The total number of ions in the trap is constant:

\[ [\text{H}_3^+](t) + [\text{H}_2\text{D}^+](t) + [\text{D}_2\text{H}^+](t) + [\text{D}_3^+](t) = [\text{H}_3^+]_0 \]

Therefore:

\[ \boxed{[\text{D}_3^+](t) = [\text{H}_3^+]_0 - [\text{H}_3^+](t) - [\text{H}_2\text{D}^+](t) - [\text{D}_2\text{H}^+](t)} \]

where the expressions for the other three species are given by the boxed equations above.