A first-order reaction has a rate constant of $2.303 imes 10^{-3} ext{ s}^{-1}$. The time required for $40

Question

A first-order reaction has a rate constant of $2.303 	imes 10^{-3} 	ext{ s}^{-1}$. The time required for $40 	ext{ g}$ of this reactant to reduce to $10 	ext{ g}$ will be [Given that $\log_{10} 2 = 0.3010$]

Accepted Answer

For a first-order reaction, the integrated rate equation is given by:
$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$
Given:
Rate constant, $k = 2.303 	imes 10^{-3} 	ext{ s}^{-1}$
Initial amount, $[R]_0 = 40 	ext{ g}$
Final amount, $[R] = 10 	ext{ g}$
Substituting the values into the formula:
$t = \frac{2.303}{2.303 	imes 10^{-3}} \log \left(\frac{40}{10}ight)$
$t = 10^3 	imes \log 4$
$t = 1000 	imes 2 \log 2$
Given $\log_{10} 2 = 0.3010$,
$t = 1000 	imes 2 	imes 0.3010 = 1000 	imes 0.6020 = 602 	ext{ s}$.

Step-by-Step Solution

Practice All CHEMISTRY Questions