Order of Reaction from Half-Life
Deduce the reaction order by observing how half-life responds to changes in initial concentration (pressure).
Question:
The half life for the decomposition of gaseous compound $A$ is $240 \text{ s}$ when the gaseous pressure was $500 \text{ Torr}$ initially.
When the pressure was $250 \text{ Torr}$, the half life was found to be $4.0 \text{ min}$.
The order of the reaction is ________.
(Nearest integer)
Detailed Solution
Step 1: Organize the data and ensure consistent units
Let's list the given conditions. We have two cases with different initial pressures ($P_0$) and corresponding half-lives ($t_{1/2}$). We must ensure time is in the same unit.
- Case 1: Initial Pressure $P_1 = 500 \text{ Torr}$ $\rightarrow$ $t_{1/2(1)} = 240 \text{ s}$
- Case 2: Initial Pressure $P_2 = 250 \text{ Torr}$ $\rightarrow$ $t_{1/2(2)} = 4.0 \text{ min} = 4.0 \times 60 \text{ s} = \mathbf{240 \text{ s}}$
The "Aha!" Moment
Look closely at the data. When the initial pressure was halved (from 500 Torr to 250 Torr), the half-life remained exactly the same ($240 \text{ s}$).
For which order of reaction is the half-life completely independent of the initial concentration (or pressure)? A first-order reaction!
Step 2: Mathematical Proof (General Formula)
If you want to prove it mathematically, recall the general relationship between half-life ($t_{1/2}$), initial concentration/pressure ($a$), and reaction order ($n$):
We can set up a ratio for the two cases:
$$ \frac{240}{240} = \left( \frac{250}{500} \right)^{n-1} $$
$$ 1 = \left( \frac{1}{2} \right)^{n-1} $$
Any non-zero number raised to the power of 0 equals 1. Therefore:
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