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:
- A \(- 1.25\times10^{-4}\)
- B \(- 2.50\times10^{-4}\)
- C \(- 3.75\times10^{-4}\)
- D \(-5.00\times10^{-4}\)
Answer & Solution
Correct Answer
(D) \(-5.00\times10^{-4}\)
Step-by-step Solution
Detailed explanation
From rate law \( - \frac{1}{2}\frac{{dS{O_2}}}{{dt}}\, - \,\frac{{d{O_2}}}{{dt}} - \frac{1}{2}\frac{{dS{O_3}}}{{dt}}\) \(\therefore \,\,\frac{{dS{O_2}}}{{dt}}\, = -\,2 \times \frac{{d{O_2}}}{{dt}}\) \(=\,-\,2\times 2.5\times 10^{-4}\)…
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