### Chemical Kinetics Question 8

#### Question 8 - 30 January - Shift 1

If compound A reacts with B following first order kinetics with rate constant $2.011 \times 10^{-3} s^{-1}$. The time taken by $A$ (in seconds) to reduce from $7 g$ to $2 g$ will be (Nearest Integer)

$[\log 5=0.698, \log 7=0.845, \log 2=0.301]$

## Show Answer

#### Answer: (623)

#### Solution:

$ A+B \to P $

$t=0 \quad 7 g$

$t=t \quad 2 g$

at constant volume

$ \begin{aligned} t & =\frac{2.303}{K} \log \frac{[A]_0}{[A] _t} \\ & =\frac{2 \cdot 303}{2 \cdot 011 \times 10^{-3}} \log \frac{7}{2} \\ & =\frac{2 \cdot 303 \times 0 \cdot 544}{2 \cdot 011 \times 10^{-3}} \\ & =622.989 \\ & \approx 623 \end{aligned} $