Question

A first order reaction has the rate constant, $k =46 \times 10^{-3} s ^{-1}$ The number of correct statement/s from the following is/are _______ 
Given: $\log 3=0.48$
A. Reaction completes in $1000 \,s$
B. The reaction has a half-life of $500 \, s$ 
C. The time required for $10 \%$ completion is 25 times the time required for $90 \%$ completion 
D. The degree of dissociation is equal to $\left(1-e^{-k t}\right)$ 
E. The rate and the rate constant have the same unit

Answer 1

Correct Answer: 2

The correct answer is 2.
$t_{10\%}=\frac{1}{K}\ln \left(\frac{a}{a-x}\right)=\frac{1}{K}\ln \left(\frac{100}{90}\right)$
$t_{10\%}=\frac{2.303}{K}(\log 10-\log 9)$
$t_{10\%}=\frac{2.093}{K}\times (0.04)$
Similarly
$t_{90\%}=\frac{1}{K}\ln \left(\frac{100}{10}\right)$
$t_{90\%}=\frac{2.303}{K}$
$\frac{t_{90\%}}{t_{10\%}}=\frac{1}{0.04}=25$
$e^{kt}=\frac{a}{a-x}$
$\frac{a-x}{a}=e^{-kt}$
$1-\frac{x}{a}=e^{-kt}$
$x=a\left(1-e^{-kt}\right)$
$\alpha =\frac{x}{a}=\left(1-e^{-kt}\right)$

Answer 2

\[t_{10\%} = \frac{1}{K} \ln \left( \frac{a}{a-x} \right) = \frac{1}{K} \ln \left( \frac{100}{90} \right)\] \[t_{10\%} = \frac{2.303}{K} (\log 10 - \log 9)\] \[t_{10\%} = \frac{2.093}{K} \times (0.04)\] Similarly \[t_{90\%} = \frac{1}{K} \ln \left( \frac{100}{10} \right)\] \[t_{90\%} = \frac{2.303}{K}\] \[\frac{t_{90\%}}{t_{10\%}} = \frac{1}{0.04} = 25\] \[e^{kt} = \frac{a}{a-x}\] \[\frac{a-x}{a} = e^{-kt}\] \[1 - \frac{x}{a} = e^{-kt}\] \[x = a(1 - e^{-kt})\] \[ \alpha = \frac{x}{a} = 1 - e^{-kt} \] where \(\alpha\) is the degree of dissociation, \(x\) is the amount dissociated, and \(a\) is the initial amount.