Question

44. The rate constant for a first order reaction is 6.909 min-1. Therefore, the time
required in minutes for the participation of 75% of the initial reactant is
(A) log2
(B) log4 (C) log2 (D) log4​

Answer 1

Solution :

⏭ Given:

✏ Rate constant for a first order reaction = 6.909$\sf \: min^{-1}$

⏭ To Find:

✏ Time required in minutes for the participation of 75% of the initial reactant.

⏭ Formula:

✏ Formula of rate constant for first order reaction is given by

$\star \: \boxed{ \sf{ \pink{k = \frac{2.303}{t} log\frac{a_o}{a_t} }}} \: \star$

⏭ Terms indication:

✏ k denotes rate constant

✏ t denotes time

✏ $a_o$ denotes initial amount of reactant

✏ $a_t$ denotes amount of reactant after t time

⏭ Calculation:

$\rightarrow \sf \: a_t = a_o - 75\%a_o \\ \\ \rightarrow \sf \: \blue{a_t = 25\% \: of \: a_o = \frac{1}{4} a_o}$

• Calculation of time

$\rightarrow \sf \: 6.909 = \frac{2.303}{t} log \frac{4 \cancel{a_o}}{ \cancel{a_o}} \\ \\ \rightarrow \sf \: t = \frac{2.303}{6.909} log4 \\ \\ \rightarrow \boxed{ \sf{ \purple{t = \frac{1}{3} log 4 \: min}}}$