Question

A particular reaction has a rate constant $1.15\times 10^{-3}{s}^{-1}$. How long does it take for $6g$ of the reactant of reduce to $3g$ ? $(\log 2=0.301)$

• A. 301 s
• B. 603 s
• C. 840 s
• D. 15 s

Answer 1

Answer: (B)

Given, rate constant $(k)=1.15\times 10^{-3}{s}^{-1}$

Reaction time = ?

Initial amount of reactant $\left(A_o\right)=6g$

Final amount of reactant $\left(A_f\right)=3g$

From rate constant equation,

$k=\frac{2.303}{t}\log \frac{A_o}{A_f}$

$1.15\times 10^{-3}=\frac{2.303}{t}\log \frac{6}{3}$

$t=\frac{2.303}{1.15\times 10^{-3}}\log \frac{6}{3}=\frac{2.303}{1.15\times 10^{-3}}\times .3010$ $t=.6027\times 10^{+3}s$

or $t=603s$