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Question: The differential rate law equation for the elementary reaction A + 2B \(\overset{\quad K\quad}{\righ...

The differential rate law equation for the elementary reaction A + 2B K\overset{\quad K\quad}{\rightarrow}3C, is :

A

d[A]dt=d[B]dt=d[C]dt- \frac{d\lbrack A\rbrack}{dt} = - \frac{d\lbrack B\rbrack}{dt} = \frac{d\lbrack C\rbrack}{dt}= k [A] [B]2

B

d[A]dt=12d[B]dt=13d[C]dt- \frac{d\lbrack A\rbrack}{dt} = - \frac{1}{2}\frac{d\lbrack B\rbrack}{dt} = \frac{1}{3}\frac{d\lbrack C\rbrack}{dt}= k [A]2 [B]

C

d[A]dt=12d[B]dt=13d[C]dt- \frac{d\lbrack A\rbrack}{dt} = - \frac{1}{2}\frac{d\lbrack B\rbrack}{dt} = \frac{1}{3}\frac{d\lbrack C\rbrack}{dt} = k [A] [B]2

D

None of these

Answer

d[A]dt=12d[B]dt=13d[C]dt- \frac{d\lbrack A\rbrack}{dt} = - \frac{1}{2}\frac{d\lbrack B\rbrack}{dt} = \frac{1}{3}\frac{d\lbrack C\rbrack}{dt} = k AA BB2

Explanation

Solution

A + 2B K\overset{\quad K\quad}{\rightarrow} 3C [elementary reaction]

Rate = – d[A]dt=12d[B]dt=13d[C]dt\frac{d\lbrack A\rbrack}{dt} = - \frac{1}{2}\frac{d\lbrack B\rbrack}{dt} = \frac{1}{3}\frac{d\lbrack C\rbrack}{dt}= k (1) (2)2