Question bank/JEE Main/Chemistry

JEE MainChemistryToughChemical Kinetics

The substance undergoes first order decomposition. The
decomposition follows two parallel first order reactions as:
[IMAGE] K1 = 1.26 × 10-4 sec-1 and K2 = 3.8 × 10-5 sec-1
The percentage distribution of B and C

Show answer and solution
% of B=k1×100k1+k2=1.26×10−4×10012.6×10−5+3.8×10−5=76.83%\%\ of\ B = \frac{k_{1} \times 100}{k_{1} + k_{2}} = \frac{1.26 \times 10^{- 4} \times 100}{12.6 \times 10^{- 5} + 3.8 \times 10^{- 5}} = 76.83\%
% of C=k2×100k1+k2=3.8×10−5×10012.6×10−5+3.8×10−5=23.17%\%\ of\ C = \frac{k_{2} \times 100}{k_{1} + k_{2}} = \frac{3.8 \times 10^{- 5} \times 100}{12.6 \times 10^{- 5} + 3.8 \times 10^{- 5}} = 23.17\%

Source: Grass-G / CH-15 Chemistry Paper 11 Nov..docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans