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: State the defining feature of zero order.
In a zero-order reaction the rate stays constant no matter how concentration changes, so $-\frac{d[\text{A}]}{dt} = k$.
Step 2: Separate the variables.
Rearranging gives $d[\text{A}] = -k\,dt$.
Step 3: Integrate over the run.
Integrating the left side from $[\text{A}]_0$ to $[\text{A}]_t$ and the right from $0$ to $t$ gives $[\text{A}]_t - [\text{A}]_0 = -kt$.
Step 4: Solve for the rate constant.
Flipping the signs, $kt = [\text{A}]_0 - [\text{A}]_t$, so $k = \frac{[\text{A}]_0 - [\text{A}]_t}{t}$.
Step 5: Note the linear signature.
This is a straight line of $[\text{A}]_t$ against $t$ with slope $-k$, the hallmark of zero order.
Step 6: Reject the log forms.
The options containing $\log_{10}\frac{[\text{A}]_0}{[\text{A}]_t}$ belong to first-order kinetics, so they do not fit here.
Step 7: Choose the answer.
The correct expression is option (3).
\[ \boxed{k = \dfrac{[\text{A}]_0 - [\text{A}]_t}{t}} \]