Question

The rate constant for a first order reaction is \(2.2\times 10^{-5}\) s\({}^{-1}\) at \(320^{\circ }\text{C}\). Calculate the time required for \(15\%\) completion?

• A. \(639\) S
• B. \(7391\) S
• C. \(3971\) S
• D. \(9371\) S

Hint:

Use the first order integrated rate law with the remaining fraction 0.85.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a first order reaction the integrated rate law is
\(t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}\).

Step 2: Data
\(k = 2.2\times 10^{-5}\text{ s}^{-1}\)
For \(15\%\) completion, if \([A]_0 = 100\) then \([A] = 85\).

Step 3: Calculation
\(t = \dfrac{2.303}{2.2\times 10^{-5}}\log\dfrac{100}{85}\)
\(\log 1.1765 = 0.0706\)
\(t = \dfrac{2.303\times 0.0706}{2.2\times 10^{-5}}\approx 7391\text{ s}\)

Step 4: Compare the options
The time is about \(7391\) s, the second option.

Final Answer:
The time required is about 7391 s. \[ \boxed{\text{(B) }7391\text{ s}} \]