For the reaction, 3A + 2B → C + D The differential rate law can be written as :

Chemistry · JEE Main · NTA Exams — Chemical Kinetics

For the reaction,
3A + 2B → C + D
The differential rate law can be written as :

\(+\frac{1}{3} \frac{\mathrm{~d}[\mathrm{~A}]}{\mathrm{dt}}=-\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}\)
\(-\frac{1}{3} \frac{\mathrm{~d}[\mathrm{~A}]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}\)
\(-\frac{\mathrm{d}[\mathrm{~A}]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{~A}]^{\mathrm{m}}[\mathrm{~B}]^{\mathrm{m}}\)
\(\frac{1}{3} \frac{\mathrm{~d}[\mathrm{~A}]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}\)

Answer

(B) - 1 3 ~d [ ~A ] dt = d [ C ] dt = k [ ~A ]^ n [ ~B ]^ m

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