If compound A reacts with B following first-order kinetics with rate constant 2.011 10^-3 s^-1, the time taken by A (in

Question

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 $7 \, \text{g}$ to $2 \, \text{g}$ will be _____. (Nearest Integer)

Given: $\log 5 = 0.698, \quad \log 7 = 0.845, \quad \log 2 = 0.301.$

Accepted Answer

Given: The reaction follows first-order kinetics with rate constant $k = 2.011 \times 10^{-3} \, s^{-1}$ . Initial amount of A is $[A]_0 = 7 \, \text{g}$ and final amount is $[A]_t = 2 \, \text{g}$ .

Find: The time required for A to decrease from $7 \, \text{g}$ to $2 \, \text{g}$ .

For a first-order reaction,

t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}

Substituting the given values,

t = \frac{2.303}{2.011 \times 10^{-3}} \log \frac{7}{2}

Now,

\log \frac{7}{2} = \log 7 - \log 2 = 0.845 - 0.301 = 0.544

Therefore,

t = \frac{2.303}{2.011 \times 10^{-3}} \times 0.544

t = \frac{2.303 \times 0.544}{2.011 \times 10^{-3}} = \frac{1.252832}{2.011 \times 10^{-3}}

t = 622.989 \, \text{seconds} \approx 623 \, \text{seconds}

Therefore, the required time is $623$ seconds.

Answer

Step-by-step solution

Common mistakes

Practice more Integrated Rate Laws questions

Related questions