Rate Constant and Rate Law – Lecture 9.4
Rate constant: key concept and purpose
The rate constant, denoted as k, is a proportionality constant for a chemical reaction that remains the same as long as the temperature is constant. It ties together the factors that affect the rate (concentrations, reaction order, etc.) so the rate remains consistent under those conditions. It is analogous in role to the ideal gas constant in linking different parts of a model.
Each rate constant is specific to a particular reaction at a given temperature; changing the temperature generally changes k.
The rate law (rate expression) has the general form for a reaction with reactants A, B, …: \text{rate} = k [A]^x [B]^y \ldots where x, y, etc. are the reaction orders with respect to each reactant.
Temperature dependence and practical use
The rate constant is temperature-dependent. When temperature is fixed, k is constant for that reaction.
You can determine k from experimental data once you know the rate law (i.e., the orders x, y, …).
You can use k to predict rates at other concentrations once the rate law is known.
Units of the rate constant and how they are determined
The units of k depend on the overall order of the reaction (the sum of the exponents in the rate law).
Remember that the rate is expressed in molarity per second (M s^{-1}). In terms of concentration [A] in M:
Zero-order: rate = k ⇒ units of k are M s^{-1} (e.g., mol L^{-1} s^{-1}).
First-order: rate = k[A] ⇒ k has units s^{-1}.
Second-order (overall 2): two common forms:
If rate = k[A]^2, then k has units M^{-1} s^{-1} (equivalently L mol^{-1} s^{-1}).
If rate = k[A][B] (two different reactants), k also has units M^{-1} s^{-1}.
Third-order (overall 3): e.g., rate = k[A]^2[B] or rate = k[A][B][C] ⇒ k has units M^{-2} s^{-1} (equivalently L^2 mol^{-2} s^{-1}).
In practical terms, you can see the pattern by solving for k:k = \frac{\text{rate}}{[A]^x[B]^y\ldots} where the denominator carries the corresponding molarity powers, so k’s units cancel to leave the rate’s units (mol L^{-1} s^{-1}).
Example unit mappings (in terms of M and s):
Zero-order: k in M s^{-1} (mol L^{-1} s^{-1}).
First-order: k in s^{-1}.
Second-order (two-reactant or single 2nd-order): k in M^{-1} s^{-1} (L mol^{-1} s^{-1}).
Third-order: k in M^{-2} s^{-1} (L^2 mol^{-2} s^{-1}).
Practical note: since concentrations are often expressed in mol/L (M), you’ll often see k quoted as L mol^{-1} s^{-1} for second-order, and L^2 mol^{-2} s^{-1} for third-order.
How to determine the rate law and k from experimental data
Step 1: Write the generic rate law for the system with two or more reactants: \text{rate} = k [\text{NO}]^x [\text{O}_2]^y \quad\text{(example with two reactants)}
Step 2: Use data from experiments where you vary one reactant at a time (holding others constant) to determine the orders x, y, …
Compare rates when you double the concentration of one reactant while keeping others constant. The factor by which the rate changes gives you the corresponding order via:
\frac{\text{Rate}2}{\text{Rate}1} = \left(\frac{[\text{A}]2}{[\text{A}]1}\right)^\text{order} ]
Do this systematically for each reactant to find each exponent.
Step 3: Once you have the orders, write the full rate law with the identified exponents.
Step 4: Solve for k by rearranging the rate law: k = \frac{\text{rate}}{[A]^x [B]^y \ldots}
Step 5: Use the calculated k with any set of concentrations to predict the rate; in practice, you can also plot data to determine k from the slope if you linearize appropriately (depends on the overall order).
Step 6: Expect experimental error: different trials may yield slightly different k values, but the rate constant should be the same for the same temperature.
Example walk-through: two reactants (NO and O2)
Target: determine the rate law orders x and y and the rate constant k, then predict a rate for new concentrations.
Generic rate law hypothesized: \text{rate} = k [\text{NO}]^x [\text{O}_2]^y
Determining x (NO order):
Hold [H2] (or the other reactant) constant and vary [NO]. When [NO] doubles from 0.1 to 0.2 M, the rate increases by a factor around 3.75.
Solve: \frac{\text{Rate}2}{\text{Rate}1} = \left(\frac{[\text{NO}]2}{[\text{NO}]1}\right)^x \Rightarrow 3.75 \approx 2^x \Rightarrow x \approx \log_2(3.75) \approx 1.9 \approx 2
Conclusion: NO is approximately second order in NO (x ≈ 2).
Determining y (O2 order):
Vary [O2] while keeping [NO] constant and observe rate changes. If doubling [O2] doubles the rate, then y ≈ 1 (first order in O2).
Resulting rate law for the example: \text{rate} = k [\text{NO}]^2 [\text{O}_2]^1
Solve for k using a data point (example from the transcript):
Use the first line’s values: [NO] = 0.100 M, [O2] = 0.100 M, rate = 1.23 × 10^{-3} M s^{-1}.
Rearranged: k = \frac{\text{rate}}{[\text{NO}]^2 [\text{O}_2]} = \frac{1.23 \times 10^{-3}}{(0.100)^2 (0.100)} = \frac{1.23 \times 10^{-3}}{1.0 \times 10^{-4}} = 12.3 \text{ (units dependent on form)}
Note: the transcript shows a different intermediate result (k ≈ 0.813 L^2 mol^{-2} s^{-1}) due to the way the algebra was presented in that narration. The conventional rearrangement gives k ≈ 1.23 L^2 mol^{-2} s^{-1} for these numbers. The discrepancy highlights the importance of consistent algebra when solving for k.
The transcript’s following calculation (as given) for a new set of concentrations:
With k ≈ 0.813 L^2 mol^{-2} s^{-1}, [NO] = 0.050 M, [O2] = 0.150 M, the rate would be:
\text{rate} = k [\text{NO}]^2 [\text{O}_2] = (0.813) (0.050)^2 (0.150) \ = 0.813 \times 2.5\times 10^{-3} \times 0.150 \ = 3.05\times 10^{-4} \text{ M s}^{-1}
This matches the transcript’s stated rate of 3.05 \times 10^{-4}\ \text{M s}^{-1} for that set of concentrations.
If you perform the calculation with the alternative/k conventional k ≈ 1.23 L^2 mol^{-2} s^{-1}, the predicted rate for the same concentrations would instead be roughly \text{rate} \approx 1.23 \times (0.050)^2 \times (0.150) \approx 4.6\times 10^{-4}\ \text{M s}^{-1}, illustrating how small differences in k propagate to rate predictions.
Practical interpretation:
With two reactants, the rate law can be found by comparing how rates change when you change one reactant at a time while holding others constant.
Once the orders are known, you can compute k from any dataset and then use the rate law to predict rates at other concentrations.
You can also plot data to obtain k from the slope of an appropriate linearized graph, depending on the overall order.
For more than two reactants, the same approach applies: vary one reactant at a time while holding the others constant, which may require multiple runs (potentially 3–4 or more) to isolate all orders.
General takeaways and practical considerations
The rate constant is a temperature-specific proportionality constant that ties together concentration terms to give a rate in mol L^{-1} s^{-1}.
The units of k depend on the overall reaction order; understanding the dimensional analysis helps remember the units for any given order.
To determine k in practice, you must first determine the rate law (the orders) and then solve for k using experimental data.
The rate constant is a fundamental link between microscopic reaction steps and macroscopic observed rates, and it enables predictive modeling across the reaction’s progress when the temperature is fixed.
Quick reference: key formulas to remember
Generic rate law: \text{rate} = k [A]^x [B]^y \ldots
To solve for k: k = \frac{\text{rate}}{[A]^x [B]^y \ldots}
For two reactants where x is the order in A and y in B: determine x, y via experiments, then compute k from any data point using the above formula.
Units summary (in terms of M = mol/L and s = seconds):
0th order: k \sim \text{M s}^{-1}
1st order: k \sim \text{s}^{-1}
2nd order: k \sim \text{M}^{-1} ext{s}^{-1} \quad (\text{or } \text{L mol}^{-1} \text{s}^{-1})
3rd order: k \sim \text{M}^{-2} ext{s}^{-1} \quad (\text{or } \text{L}^2 \text{mol}^{-2} \text{s}^{-1})
Example conclusion from the transcript-oriented walkthrough:
Determined NO is second-order in NO and the other reactant is first-order: \text{rate} = k [\mathrm{NO}]^2 [\mathrm{O}_2]$$
Found k from data, then used the rate law to predict rates at new concentrations, illustrating the practical workflow for kinetic analysis.