ExamBro
ExamBro
JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

In the reaction of formation of sulphur trioxide by contact process \(2S{O_2} + {O_2} \rightleftharpoons 2S{O_3}\) the rate of reaction was measured as \(\frac{{d\left[ {{O_2}} \right]}}{{dt}} =  - 2.5 \times {10^{ - 4}}\,mol\,{L^{ - 1}}\,{s^{ - 1}}\) . The rate of reaction is terms of \([SO_2]\) in \(mol\,L^{- 1}\, s^{-1}\) will be:

  1. A \(- 1.25\times10^{-4}\)
  2. B \(- 2.50\times10^{-4}\)
  3. C \(- 3.75\times10^{-4}\)
  4. D \(-5.00\times10^{-4}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-5.00\times10^{-4}\)

Step-by-step Solution

Detailed explanation

From rate law \( - \frac{1}{2}\frac{{dS{O_2}}}{{dt}}\, - \,\frac{{d{O_2}}}{{dt}} - \frac{1}{2}\frac{{dS{O_3}}}{{dt}}\) \(\therefore \,\,\frac{{dS{O_2}}}{{dt}}\, = -\,2 \times \frac{{d{O_2}}}{{dt}}\) \(=\,-\,2\times 2.5\times 10^{-4}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app