Question

For a reaction $\mathrm{A} \rightarrow \text{product}$, rate constant is $2 \times 10^{-2}\ \mathrm{s}^{-1}$. The initial concentration of A is $1.0\ \mathrm{mol\ dm}^{-3}$. What is the value of $\log_{10}\!\left(\frac{1}{[\mathrm{A}]_t}\right)$ after 100 seconds?

• A. $0.430\ \mathrm{mol\ dm}^{-3}$
• B. $0.135\ \mathrm{mol\ dm}^{-3}$
• C. $0.270\ \mathrm{mol\ dm}^{-3}$
• D. $0.868\ \mathrm{mol\ dm}^{-3}$

Hint:

Whenever the initial concentration is 1, $\log\textit{{10}\frac{[\mathrm{A}]_0}{[\mathrm{A}]}t}$ is identical to $\log*{10}\frac{1}{[\mathrm{A}]_t}$. Simply multiply the rate constant $k$ by time $t$ and divide by $2.303$ to arrive at the answer instantly: $\frac{2}{2.303} \approx 0.868$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a chemical reaction $\mathrm{A} \rightarrow \text{product}$ with a rate constant $k = 2 \times 10^{-2}\ \mathrm{s}^{-1}$. The unit of the rate constant ($\mathrm{s}^{-1}$) implies that the reaction follows first-order kinetics. We need to find the value of the term $\log_{10} \frac{1}{[\mathrm{A}]_t}$ at time $t = 100\text{ seconds}$ given that the initial concentration $[\mathrm{A}]_0 = 1.0\ \mathrm{mol\ dm}^{-3}$.

Step 2: Key Formula or Approach:
The integrated rate equation for a first-order chemical reaction using base-10 logarithm is written as:
$$k = \frac{2.303}{t} \log_{10} \frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}$$ Using logarithmic properties, we can break this down:
$$\log_{10} \frac{[\mathrm{A}]_0}{[\mathrm{A}]_t} = \log_{10} [\mathrm{A}]_0 + \log_{10} \frac{1}{[\mathrm{A}]_t}$$ Since the initial concentration $[\mathrm{A}]_0 = 1.0\ \mathrm{mol\ dm}^{-3}$, we have $\log_{10}(1) = 0$. Therefore, the equation simplifies directly to:
$$k = \frac{2.303}{t} \log_{10} \frac{1}{[\mathrm{A}]_t}$$

Step 3: Detailed Explanation:
Rearrange the simplified formula to solve for the target term $\log_{10} \frac{1}{[\mathrm{A}]_t}$:
$$\log_{10} \frac{1}{[\mathrm{A}]_t} = \frac{k \cdot t}{2.303}$$ Substitute the given values into the equation:
$$k = 2 \times 10^{-2}\ \mathrm{s}^{-1}$$ $$t = 100\ \mathrm{s}$$ $$\log_{10} \frac{1}{[\mathrm{A}]_t} = \frac{(2 \times 10^{-2}) \times 100}{2.303}$$ $$\log_{10} \frac{1}{[\mathrm{A}]_t} = \frac{2}{2.303}$$ $$ \log_{10} \frac{1}{[\mathrm{A}]*t} \approx 0.8684$$

Step 4: Final Answer:
The computed value of $\log\textit{{10} \frac{1}{[\mathrm{A}]}t}$ is $0.868$, which perfectly corresponds with option (D).