MHT CET · Chemistry · Chemical Kinetics
In the reaction \(2 \mathrm{~N}_{2} \mathrm{O}_{5(\mathrm{~g})} \longrightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\) the rate of formation of \(\mathrm{NO}_{2(\mathrm{~g})}\) and \(\mathrm{O}_{2(\mathrm{~g})}\) are in
the ratio of
- A \(1: 4\)
- B \(1: 1\)
- C \(6: 1\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(A) \(1: 4\)
Step-by-step Solution
Detailed explanation
By Differential Rate law;
\(\frac{-1}{2} \frac{d\left(N_{2} O_{5}\right)}{d t}=\frac{1}{4} \frac{d\left(N O_{2}\right)}{d t}=\frac{d\left(O_{2}\right)}{d t}\)
d \(\frac{\text { Rate of formation of } \mathrm{NO}_{2}}{\text { Ratc of formation of } \mathrm{O}_{2}}=\frac{d\left(\mathrm{NO}_{2}\right) / d t}{d\left(\mathrm{O}_{2} / d t\right.}=4: 1\)
\(\frac{-1}{2} \frac{d\left(N_{2} O_{5}\right)}{d t}=\frac{1}{4} \frac{d\left(N O_{2}\right)}{d t}=\frac{d\left(O_{2}\right)}{d t}\)
d \(\frac{\text { Rate of formation of } \mathrm{NO}_{2}}{\text { Ratc of formation of } \mathrm{O}_{2}}=\frac{d\left(\mathrm{NO}_{2}\right) / d t}{d\left(\mathrm{O}_{2} / d t\right.}=4: 1\)
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