Question

hey, please help me to solve this problem. (no spamming please)​

Answer 1

$\sf \bold {Given\::}$

$\bullet$ rate constant (k) = 60 /s

$\bullet$ amount of reactant left = (1/10) of initial amount

$\sf \bold {To\: find\::}$ = time when (1/10) of initial amount of reactant is left = t (let)

$\sf \bold {Let\::}$

$\bullet$ initial amount of reactant = a

$\bullet$ final amount of reactant = x

$\sf \bold {Formula\:used\::}$

$\bullet$ $\sf k = \dfrac{2.303}{t} \times ( log \dfrac{a}{x} )$

$\bullet$ log 10 = 1

$\sf \bold {Solution\::}$

Final amount (x) = (1/10) of initial amount

$\implies$ x = (1/10)× a

Putting respective values of k , a & x in

$\sf k = \dfrac{2.303}{t} \times ( log \dfrac{a}{x} )$

We get;

$\sf 60 = \dfrac{2.303}{t} \times ( log \dfrac{a}{\dfrac{a}{10} } ) \\\\\\\sf t = \dfrac{2.303}{60} \times ( \sf log \dfrac{a\times10}{a} )\\\\\\\sf t = \dfrac{2.303}{60} \times ( \sf log 10)\\\\\\\sf t = \dfrac{2.303}{60} \times 1$

t = 0.038s

$\sf \bold {ANSWER\::}$ Time taken = $\sf \bold {0.038s\::}$