Question

aA + bB$\longrightarrow$ Product, dx/dt = k [A]a [B]b . If

concentration of A is doubled, rate is four times. If concentration of B is made four times, rate is doubled. What is relation between rate of disappearance of A and that of B ?

• A. $\text{- }\{\text{d }\lbrack A\rbrack\text{ / dt}\}\text{ = - }\{\text{d }\lbrack B\rbrack\text{ / dt}\}$
• B. $\text{- }\{\text{d }\lbrack A\rbrack\text{ / dt}\}\text{ = - }\{\text{4 d }\lbrack B\rbrack\text{ / dt}\}$
• C. $\text{- }\{\text{4 d }\lbrack A\rbrack\text{ / dt}\}\text{ = - }\{\text{d }\lbrack B\rbrack\text{/ dt}\}$
• D. None of these

Answer 1

Answer: (B)

$\text{- }\{\text{d }\lbrack A\rbrack\text{ / dt}\}\text{ = - }\{\text{4 d }\lbrack B\rbrack\text{ / dt}\}$

Answer 2

$\frac{dx}{dt} = k\lbrack A\rbrack^{a}\lbrack B\rbrack^{b}$

(i) As on doubling concentration of A rate become four time so a = 2.

(ii) On four time concentration of B rate become double so b = $\frac{1}{2}$ .

So, Given equation :$2a + \frac{1}{2}b\overset{\quad\quad}{\rightarrow}\Pr 莀$

$- \frac{1}{2}\frac{d\lbrack A\rbrack}{dt} = - 2\frac{d\lbrack B\rbrack}{dt}$

$- \frac{d\lbrack A\rbrack}{dt} = - 4\frac{d\lbrack B\rbrack}{dt}$