Question

For a reaction \(\text{A}⟶\) product, \(k = 2\times 10^{-2} \text{s}^{-1}\). If the initial concentration of A is \(1.0 \text{mol dm}^{-3}\) find the value of \(log\frac{1}{[A]_t}\) after 100 second ?

• A. \(0.135\)
• B. \(0.270\)
• C. \(0.430\)
• D. \(0.868\)

Hint:

A reaction with unit of k in s^-1 is first order. Use log[A]0/[A] = kt/2.303.

Answer 1

The Correct Option is D

Step 1: Integrate the rate law:
For first order, $\ln[A]_t = \ln[A]_0 - kt$.
With $[A]_0 = 1$, $\ln[A]_t = -kt = -(2\times10^{-2})(100) = -2$.

Step 2: Switch to base-10 logs:
$\log[A]_t = -2/2.303 = -0.868$.
Therefore $\log(1/[A]_t) = -\log[A]_t = 0.868$.

Step 3: Avoid the trap:
The concentration itself is $e^{-2} = 0.135$ mol dm$^{-3}$. That is option (A), but the question asks for the logarithm of its reciprocal.

Final Answer:
The value is $0.868$, option (D). \[ \boxed{0.868 \text{ (D)}} \]