Ask Questions, Get Answers

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

What is the half-life of a reaction with $[A]_o = 109\;M$, $[B]_o = 1\;M$, $k' = 45M^{-1}s^{-1}$ ?

$\begin{array}{1 1} 1.41 \times 10^{-4}\;s \\ 4905\;s \\ 0.141 \times 10^{-4}\;s \\ 490.5\;s \end{array}$
Can you answer this question?

1 Answer

0 votes
Answer: $1.41 \times 10^{-4}\;s$
Given: $[A]_o = 109\;M$, $[B]_o = 1\;M$, $k' = 45M^{-1}s^{-1}$
Because $[A]_0 \gt \gt [B]_0$, this is a pseudo first order reaction and we can multiply $k'$ with $[A]_0$ to find $k$.
$\Rightarrow 109\;M \times 45M^{-1}s^{-1} = 4905s^{-1}$
Half life of a pseudo first order reaction is $ t_{1/2} = \large\frac{ln \;0.5}{-k}$
$\Rightarrow t_{1/2} = \large\frac{ln\;0.5}{4905s^{-1}}$$ = 1.41 \times 10^{-4}\;s$
answered Dec 13, 2013 by sreemathi.v
edited 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