Question

The rate constant for a zero order reaction \(A \rightarrow\) products is \(0.0030\ \mathrm{mol\ L^{-1}\ s^{-1}}\). How long will it take for the initial concentration of \(A\) to fall from \(0.10\ \mathrm{M}\) to \(0.075\ \mathrm{M}\)?

• A. \(10\ \mathrm{s}\)
• B. \(20\ \mathrm{s}\)
• C. \(8.33\ \mathrm{s}\)
• D. \(1.33\ \mathrm{s}\)

Hint:

For zero order reactions, concentration decreases linearly with time according to \([A]_t=[A]_0-kt\).

Answer 1

The Correct Option is C

Step 1: Recall the rate law for a zero order reaction.
For a zero order reaction the rate is independent of concentration, and the amount reacted grows linearly with time: \[ [A]_0 - [A]_t = k\,t \]
Step 2: List the given quantities.
Here $[A]_0 = 0.10$ M, $[A]_t = 0.075$ M and $k = 0.0030$ mol L$^{-1}$ s$^{-1}$.
Step 3: Find the concentration that has reacted.
The drop in concentration is \[ [A]_0 - [A]_t = 0.10 - 0.075 = 0.025 \text{ M} \]
Step 4: Rearrange the rate law for time.
Solving for $t$, \[ t = \frac{[A]_0 - [A]_t}{k} \]
Step 5: Substitute the numbers.
\[ t = \frac{0.025}{0.0030} \] \[ t = 8.33 \text{ s} \]
Step 6: State the answer.
It takes about $8.33$ seconds for the concentration to fall from $0.10$ M to $0.075$ M, which is option 3.
\[ \boxed{8.33 \text{ s}} \]