Chemical Kinetics Question 26
Question: A first order reaction was started with a decimolar solution of the reactant, 8 minutes and 20 seconds later its concentration was found to be $ M/100 $ . So the rate of the reaction is [Kerala PMT 2004]
Options:
A) $ 2.303\times {10^{-5}}\ {{\sec }^{-1}} $
B) $ 2.303\times {10^{-4}}\ {{\sec }^{-1}} $
C) $ 4.606\times {10^{-3}}\ {{\sec }^{-1}} $
D) $ 2.606\times {10^{-5}}\ {{\sec }^{-1}} $
E) $ 2.603\times {10^{-4}}\ {{\sec }^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
For first order reaction $ K=\frac{2.303}{t}\log \frac{a}{a-x} $ Given: $ a=\frac{1}{10}=.1m $ ; $ a-x=\frac{1}{100}=.01m $ ; t = 500 sec
$ \therefore \ \ K=\frac{2.303}{500}\log \frac{.10}{.01}=\frac{2.303}{500}\log 10 $
$ =\frac{2.303}{500}=0.004606=4.6\times {10^{-3}}{{\sec }^{-1}} $ .
BETA
