Question

$t_{1/4}$ can be taken as the time taken for the concentration of a reactant to drop to $\frac{3}{4}$ of its value. If the rate constant for a first order reaction is $K$, the $t_{1/4}$ can be written as:

• A. $\frac{10}{K}$
• B. $\frac{0.29}{K}$
• C. $\frac{0.69}{K}$
• D. $\frac{0.75}{K}$

Answer 1

Answer: (B)

$\frac{0.29}{K}$

Let initial concentration be $a_o$

After concentration decreases by $\frac{1}{4}=a_o(1-\frac{1}{4})=\frac{3a_o}{4}$

Using first order kinetic equation,

$t_{1/4}\times k=ln\left(\frac{a_o}{3a_o/4}\right)=ln\left(\frac{4}{3}\right)$

$t_{1/4}=\frac{0.29}{K}$