SATHEE: Chemical Kinetics - Result Question 68

Chemical Kinetics - Result Question 68

####67. At $380^{\circ} C$ , the half-life period for the first order decomposition of $H_{2} O_{2}$ is $360 m i n$ . The energy of activation of the reaction is $200 k J m o l^{- 1}$ . Calculate the time required for $75$ decomposition at $450^{\circ} C$ .

(1995, 4M)

Show Answer

Answer:

Correct Answer: 67. $(1.386 m i n)$

Solution:

For 1 st order reaction :

$\begin{matrix} k \propto \frac{1}{t_{1 / 2}} \Rightarrow \ln \frac{k (450^{\circ} C)}{k (380^{\circ} C)} = \ln \frac{t_{1 / 2} (380^{\circ} C)}{t_{1 / 2} (450^{\circ} C)} = \frac{E_{a}}{R} (\frac{450 - 380}{727 \times 653}) \Rightarrow \ln \frac{360}{t_{1 / 2} (450^{\circ} C)} = \frac{200 \times 10^{3}}{8.314} \times \frac{70}{727 \times 653} = 3.54 \Rightarrow t_{1 / 2} (450^{\circ} C) = 10.37 m i n \Rightarrow Time for 75 = 2 \times t_{1 / 2} = 2 \times 10.37 = 20.74 m i n \end{matrix}$

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Chemical Kinetics - Result Question 68

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu