Question

A substance having a half-life period of 30 minutes decomposes according to the first order rate law. The fraction decomposed, the balance remaining after 1.5 hours and time for $60\%$ decomposition on its doubling the initial concentration will be.

• A. $87.4; 0.126; 39.7$ min
• B. $80.6; 0.135; 40.8$ min
• C. $90.5; 0.144; 2829$ min
• D. $802; 0.135; 26.6$ min

Answer 1

Answer: (A)

We know that $K=\frac{0.693}{t_{1/2}}=\frac{0.693}{30}=0.023/min^{-1}$
Let the initial concentration of the substance $= 100$ and the substance decomposed in 1.5 hours (90 min) = .t then, $a - x= 100$ Substituting these values in
$K=\frac{2.303}{t}log \left(\frac{a}{a-x}\right)$
$\Rightarrow 0.0231=\frac{2.303}{90}log \frac{100}{100-x}$
log$\frac{100}{100-x}=\frac{0.0231\times90}{2.303}$
$\Rightarrow x=87.4$
So the fraction decomposed in 1.5 hours$=\frac{87.4}{100}=0.874$
Fraction remaining behind after 1.5 hours $= 1 - 0.874 = 0.126$ .The time required for $60\%$ of decomposition
$K=\frac{2.303}{t}log \left(\frac{a}{a-x}\right)$
$\Rightarrow 0.023=\frac{2.303}{t}log \left(\frac{100}{40}\right)$
$t=\frac{2.303}{0.0231}log\left(\frac{10}{4}\right)=39.7$ minutes
Since the reaction is of first order the time required to complete specific fraction is independent of initial concentration (or pressure). Hence $60\%$ of the reaction will decompose in 39.7 mins in this case also.