\(∵ kt = ln \frac {A_0}{A}\)
\(\frac {ln2}{t_{\frac 12}}\) \(t\)67% \(=ln\frac {A_0}{0.33A_0}\)
\(\frac {log\ 2}{t_{\frac 12}}\)\(t\)67% = \(log \frac {1}{0.33}\)
\(t\)67% \(= 1.566 t_{\frac 12}\)
\(x = 15.66\)
Nearest Integer \(= 16\)
So, the answer is \(16\).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
The cycloalkene (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which C : Br ratio is \(3:1\). The percentage of bromine in the product (Y) is _________ % (Nearest integer).
Given:
\[ \text{H} = 1,\quad \text{C} = 12,\quad \text{O} = 16,\quad \text{Br} = 80 \]

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate equation of the reaction will given in the question.