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: Use the integrated rate law for zero order reaction.
For a zero order reaction,
\[ [A]_t=[A]_0-kt \]
where
\[ [A]_0=\text{initial concentration} \] \[ [A]_t=\text{concentration after time }t \] \[ k=\text{rate constant} \]

Step 2: Substitute the given values.
Given,
\[ [A]_0=0.10\ \mathrm{M} \] \[ [A]_t=0.075\ \mathrm{M} \] \[ k=0.0030\ \mathrm{mol\ L^{-1}\ s^{-1}} \]
Substituting into the equation,
\[ 0.075=0.10-0.0030\,t \]

Step 3: Solve for time.
\[ 0.0030\,t=0.10-0.075 \]
\[ 0.0030\,t=0.025 \]
\[ t=\frac{0.025}{0.0030} \]
\[ t=8.33\ \mathrm{s} \]

Step 4: Final conclusion.
Hence, the required time is
\[ \boxed{8.33\ \mathrm{s}} \]