Question

Which of the following equations represents integrated rate law for zero order reaction?

• A. $k = \frac{[\text{A}]_t - [\text{A}]_0}{t}$
• B. $k = \frac{1}{t} \log_{10} \frac{[\text{A}]_0}{[\text{A}]_t}$
• C. $k = \frac{[\text{A}]_0 - [\text{A}]_t}{t}$
• D. $k = \frac{t}{2.303} \times \log_{10} \frac{[\text{A}]_0}{[\text{A}]_t}$

Hint:

To avoid mixing up the concentrations in the numerator, remember that as a reaction runs, concentration drops over time. Therefore, the initial value $[\text{A}]_0$ is always larger than the later value $[\text{A}]_t$. To keep $k$ positive, it must be $(\text{Larger} - \text{Smaller})$, or $[\text{A}]_0 - [\text{A}]_t$!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to identify the mathematically correct integrated rate equation representing a zero-order chemical kinetics reaction.

Step 2: Key Formula or Approach:
For a zero-order reaction ($\text{A} \rightarrow \text{Products}$), the rate of the reaction is completely independent of the reactant concentration:
$$-\frac{\text{d}[\text{A}]}{\text{d}t} = k[\text{A}]^0 = k$$ Separating variables and integrating both sides yields the linear integrated expression.

Step 3: Detailed Explanation:
Let's perform the straightforward calculus derivation step-by-step:
$$\text{d}[\text{A}] = -k\,\text{d}t$$ Integrating both sides from time $0$ (initial concentration $[\text{A}]_0$) to time $t$ (remaining concentration $[\text{A}]_t$):
$$\int_{[\text{A}]_0}^{[\text{A}]_t} \text{d}[\text{A}] = -k \int_{0}^{t} \text{d}t$$ $$[\text{A}]_t - [\text{A}]_0 = -kt$$ Rearranging this equation to solve explicitly for the rate constant $k$:
$$kt = [\text{A}]_0 - [\text{A}]_t$$ $$k = \frac{[\text{A}]_0 - [\text{A}]_t}{t}$$ This matches option (C). Note that options (B) and (D) are modifications of the first-order integrated rate law expression, not zero-order.

Step 4: Final Answer:
The correct integrated rate law for a zero-order reaction is $k = \frac{[\text{A}]_0 - [\text{A}]_t}{t}$, which corresponds to option (C).