After a certain pesticide compound is applied to crops, its decomposition is a first-order reaction — Kinetics Chemistry Question

STEPS:

1. Identify the reaction order from the problem:
The prompt explicitly states that the decomposition of the pesticide compound follows first-order kinetics.

2. Retrieve the standard kinetic equation for a first-order reaction:
From the AP Chemistry Equations and Constants sheet, the relationship between the half-life ($t_{1/2}$) and the rate constant ($k$) of a first-order process is represented by the formula:
$$t_{1/2} = \frac{0.693}{k}$$

3. Rearrange the equation to isolate the rate constant ($k$):
Algebraically solve for $k$ by multiplying both sides by $k$ and dividing by $t_{1/2}$:
$$k = \frac{0.693}{t_{1/2}}$$

4. Substitute the given half-life into the equation:
The problem specifies that the pesticide has a half-life of $56\text{ days}$. Substituting this value yields:
$$k = \frac{0.693}{56\text{ days}}$$

5. Perform decimal division (without a calculator):
Since calculators are not allowed on Section I of the exam, use mental estimation or long division to find the value:
* Estimation method: Treat $0.693$ as approximately $0.70$.
$$\frac{0.70}{56} = \frac{70 \times 10^{-2}}{56}$$
Recognize that both $70$ and $56$ are divisible by $7$:
$$\frac{70}{56} = \frac{10}{8} = 1.25$$
Now re-apply the power of ten:
$$1.25 \times 10^{-2} = \mathbf{0.0125}$$
* Exact division: $0.693 \div 56 = 0.012375\text{ day}^{-1}$.
* Rounded to two significant figures, this equals $0.012\text{ day}^{-1}$.

6. Identify the matching option:
The calculated rate constant of $0.012\text{ day}^{-1}$ aligns perfectly with Option A.

*

WHY_OTHERS_WRONG:

• Option B is incorrect ($0.018\text{ day}^{-1}$): This value is a mathematical error that arises if a student sets up the ratio incorrectly or misremembers the constant in the numerator as $1.0$ instead of $0.693$ (as $\frac{1.0}{56} \approx 0.018$).
• Option C is incorrect ($56\text{ day}^{-1}$): This option simply duplicates the numerical value of the half-life. It fails to recognize that the rate constant and half-life are inversely proportional, meaning they cannot share the same numerical value under these conditions.
• Option D is incorrect ($81\text{ day}^{-1}$): This value represents a division error where a student mistakenly multiplies the half-life by the reciprocal constant (calculating $56 \times 1.44 \approx 81$) instead of dividing $0.693$ by the half-life.