First Order Gas Phase Kinetics
Relate total system pressure to the concentration of reactants over time.
Question:
$A(g) \rightarrow 2B(g) + C(g)$ is a first order reaction.
The initial pressure of the system was found to be $800 \text{ mm Hg}$ which increased to $1600 \text{ mm Hg}$ after $10 \text{ min}$.
The total pressure of the system after $30 \text{ min}$ will be ________ $\text{mm Hg}$.
(Nearest integer)
Detailed Solution
Step 1: Set up the reaction equation (ICE Table)
Let initial pressure of $A$ be $P_0$. At time $t$, let $x$ amount of $A$ react.
| Time | $A(g)$ | $\rightarrow$ | $2B(g)$ | $+$ | $C(g)$ | Total Pressure ($P_t$) |
|---|---|---|---|---|---|---|
| Initial ($t=0$) | $P_0 = 800$ | $0$ | $0$ | $\mathbf{800}$ | ||
| At time $t$ | $800 - x$ | $2x$ | $x$ | $(800 - x) + 2x + x$ = $\mathbf{800 + 2x}$ |
Step 2: Analyze data at $t = 10 \text{ min}$
We are given that at $t = 10 \text{ min}$, the total pressure $P_t = 1600 \text{ mm Hg}$.
Using our total pressure expression:
Now, calculate the partial pressure of reactant $A$ left at $t = 10 \text{ min}$:
The Half-Life Shortcut!
Notice what just happened! The pressure of $A$ went from $800$ to $400$ in exactly $10 \text{ minutes}$. Because it dropped to exactly half its initial value, we immediately know that $t_{1/2} = 10 \text{ minutes}$.
You do not need to use the complex formula $k = \frac{2.303}{t} \log\frac{P_0}{P_t}$. Use half-lives to jump straight to the answer!
Step 3: Calculate pressure at $t = 30 \text{ min}$
We need to find the state of the system at $t = 30 \text{ min}$.
- Number of half-lives ($n$) = $\frac{\text{Total Time}}{t_{1/2}} = \frac{30}{10} = \mathbf{3} \text{ half-lives}$.
- Amount of $A$ remaining after 3 half-lives: $$ P_A = \frac{P_0}{2^n} = \frac{800}{2^3} = \frac{800}{8} = \mathbf{100 \text{ mm Hg}} $$
We know that $P_A = 800 - x_{new}$. Let's find the new $x$:
Finally, calculate the total pressure using our expression ($P_t = 800 + 2x$):
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