11.1 Gas-phase reaction kinetics

Gas-phase reactions center on molecular collisions. Collision theory describes how the energy, orientation, and frequency of those collisions govern whether a reaction actually occurs. Understanding these factors lets you predict and manipulate reaction rates, which is essential for everything from industrial synthesis to atmospheric chemistry.

Temperature and pressure are the two main levers you can pull. Higher temperatures give molecules more kinetic energy and increase collision frequency; higher pressures pack molecules closer together, raising concentrations. For reactions that proceed through multiple elementary steps, the slowest step controls the overall rate.

Collision Theory and Elementary Gas-Phase Reactions

Rate expressions in collision theory

Collision theory says a reaction happens when two molecules collide with (1) enough energy to overcome the activation energy barrier and (2) the right spatial orientation. Not every collision is productive, so both factors matter.

Bimolecular example: $\ce{H2 + I2 -> 2HI}$

Because this is an elementary bimolecular step, the rate law comes directly from the stoichiometry:

$\text{rate} = k[\ce{H2}][\ce{I2}]$

The rate constant $k$ encodes all the collision physics:

$k = P \, Z_{AB} \, e^{-E_a / RT}$

• $Z_{AB}$ is the collision frequency factor, which depends on the collision cross-section $\sigma_{AB}$, temperature $T$, and the reduced mass $\mu$ of the two molecules:

$Z_{AB} = \sigma_{AB}\sqrt{\frac{8\pi k_B T}{\mu}}$

• $P$ is the steric factor, a number between 0 and 1 that accounts for the fraction of collisions with the correct orientation. A small, simple molecule like $\ce{H2}$ has a relatively large $P$; a bulky molecule with a specific reactive site has a much smaller one.
• $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature. The exponential term gives the fraction of collisions that have sufficient energy.

Unimolecular example: $\ce{N2O5 -> 2NO2 + \frac{1}{2}O2}$

For an elementary unimolecular step, the rate depends on only one species:

$\text{rate} = k[\ce{N2O5}]$

The molecule must first acquire enough internal energy (through collisions with other molecules) to reach the transition state before it decomposes.

Factors Affecting Gas-Phase Reaction Rates

Temperature and pressure effects on reactions

Temperature effects

Raising the temperature does two things simultaneously: it increases the average kinetic energy of molecules and it increases the frequency of collisions. The more important effect is the energy increase, because the fraction of molecules exceeding $E_a$ grows exponentially with temperature.

This relationship is captured by the Arrhenius equation:

$k = A \, e^{-E_a / RT}$

• $A$ is the pre-exponential (frequency) factor, related to collision frequency and orientation.
• The exponential term is the fraction of collisions with energy $\geq E_a$.

A useful rule of thumb: for many reactions near room temperature, a 10 K increase roughly doubles the rate constant. The exact factor depends on $E_a$.

Pressure effects

For ideal gases, concentration is proportional to pressure ($[\text{gas}] = P / RT$). Increasing total pressure therefore increases reactant concentrations, which raises the collision rate.

For a rate law $\text{rate} = k[\ce{A}]^a[\ce{B}]^b$, doubling the pressure doubles every concentration, so the rate increases by a factor of $2^{(a+b)}$. For example, if $a + b = 2$, doubling the pressure quadruples the rate.

Multi-Step Gas-Phase Reaction Mechanisms

Rate-determining steps in reaction mechanisms

Most real gas-phase reactions don't happen in a single collision. They proceed through a sequence of elementary steps, and the rate-determining step (RDS) is the slowest one in that sequence. The overall reaction can't go faster than its slowest step.

How to identify the RDS and derive the overall rate law:

1. Write out each elementary step and its individual rate expression.
2. Identify the step with the highest activation energy (or, experimentally, the one whose rate expression matches the observed overall rate law). That step is the RDS.
3. Write the rate law for the RDS. Species that appear only in fast steps before the RDS won't show up directly in the overall rate law.

If the RDS involves a reactive intermediate (a species produced in an earlier step and consumed in the RDS), you need to eliminate that intermediate from the rate law. Two common approaches:

• Pre-equilibrium approximation: If a fast, reversible step precedes the RDS, set the forward and reverse rates of that step equal and solve for the intermediate's concentration.
• Steady-state approximation: Assume the concentration of each intermediate stays roughly constant over time (its rate of formation equals its rate of consumption). This is more general and works even when no step is clearly in equilibrium.

Both methods let you express the overall rate law in terms of measurable reactant concentrations only.

Experimental Determination of Reaction Order

Experimental determination of reaction order

The reaction order with respect to a given reactant is the exponent on its concentration in the experimentally determined rate law. For $\text{rate} = k[\ce{A}]^m[\ce{B}]^n$, $m$ is the order in $\ce{A}$ and $n$ is the order in $\ce{B}$. These exponents must be found experimentally; they don't necessarily match the stoichiometric coefficients (unless the reaction is elementary).

Method 1: Initial rates

Run multiple experiments, changing the concentration of one reactant at a time while holding the others constant. Then compare:

$\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[\ce{A}]_2}{[\ce{A}]_1}\right)^m$

Solve for $m$. For instance, if doubling $[\ce{A}]$ quadruples the rate, then $2^m = 4$, so $m = 2$.

Method 2: Log-log plot

Take the log of the rate law: $\log(\text{rate}) = m\log[\ce{A}] + \text{const}$. Plot $\log(\text{rate})$ vs. $\log[\ce{A}]$; the slope gives $m$ directly.

Method 3: Integrated rate laws

Track concentration vs. time and test which integrated form gives a straight line:

• Zero order: $[\ce{A}]$ vs. $t$ is linear.
• First order: $\ln[\ce{A}]$ vs. $t$ is linear.
• Second order: $1/[\ce{A}]$ vs. $t$ is linear.

The plot that yields the best linear fit tells you the order. This method is especially useful when you have continuous concentration-time data from a single experiment.