Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Calculate half-life for first-order reaction if 68% of a substance is reacted within 66 s.

$\begin{array}{1 1} 39.3\;s \\ 45.3\;s \\ 22.67\;s \\ 17.45\;s \end{array}$

Can you answer this question?

1 Answer

0 votes
Answer: 39.3 s
Before we calculate the half-life, we need to calculate $k$ the rate constant in order to get the half-life.
Given 68% reacted $\rightarrow$ 32% remains:
$ln\;A = -kt + ln\;A_0 \rightarrow ln 0.32 = -k \times 66\;s + ln \; 1 \rightarrow k - 0.01726s^{-1}$
For a first order reaction, $t_{1/2} = \large\frac{ln\;2}{k}$$ =\large\frac{0.693}{0.01762}$$s = 39.3\;s$
answered Jul 25, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App