Easy INTEGER +4 / -1 PYQ · JEE Mains 2021

For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _________ $\times$ 10$-$3 s$-$1. (Round off to the Nearest Integer). [Given : log102 = 0.301, ln10 = 2.303]

Answer (integer) 2

Solution

$k = {1 \over t}\ln \left[ {{a \over {a - x}}} \right]$<br><br>$k = {{2.303} \over {570}}\log \left( {{{100} \over {32}}} \right)$<br><br>$k = {{2.303} \over {570}}\left[ {\log ({{10}^2}) - \log {2^5}} \right]$<br><br>$k = {{2.303} \over {570}} \times 0.5$<br><br>$k = 2 \times {10^{ - 3}}{s^{ - 1}}$

About this question

Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction

This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Chemical Kinetics are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →