Question

For the reaction : aA¾®bB ;

log$\left\lbrack \frac{- dA}{dt} \right\rbrack$ = log$\left\lbrack \frac{dB}{dt} \right\rbrack$ + 0.6020, then a : b is -

• A. 4
• B. 2
• C. 1.5
• D. 0

Answer 1

Answer: (A)

4

Answer 2

A ¾® bB

$\frac{\left( \frac{- dA}{dt} \right)}{a}$ = $\frac{\left( \frac{dB}{dt} \right)}{b}$ or $\frac{- dA}{dt}$= $\frac{a}{b}$× $\frac{dB}{dt}$

log$\left( \frac{- dA}{dt} \right)$ = log$\left( \frac{a}{b} \right)$ + log$\left( \frac{dB}{dt} \right)$

Given : log $\left( \frac{- dA}{dt} \right)$ = log$\left( \frac{dB}{dt} \right)$ + 0.6020

\ log$\left( \frac{a}{b} \right)$= 0.6020 = 2 × 0.3010

= 2 × log 2 = log 4

\ a : b = 4