For the first order reaction A $$\rightarrow$$ 2B, 1 mole of reactant A gives 0.2 moles of $$B$$ after 100 minutes. The half life of the reaction is _________ min. (Round off to nearest integer).
[$$Use: ln 2 = 0.69, ln 10 = 2.3$$
Properties of logarithms: $$\ln x^y = y \ln x$$
$$\ln\left(\frac{x}{y}\right)=\ln x-\ln y$$]
(Round off to the nearest integer)

We have a first-order decomposition $$\text A \;\longrightarrow\; 2\text B.$$

Initially 1 mol of A is present, i.e. $$n_0(\text A)=1\text{ mol}.$$ After 100 min, 0.2 mol of B has appeared. Because the stoichiometry is $$1\text{ A}\;\to\;2\text{ B},$$ every mole of A that disappears produces twice that amount of B. Hence the moles of A actually consumed are

$$n_{\text{consumed}}(\text A)=\frac{0.2\text{ mol B}}{2}=0.1\text{ mol A}.$$

So the moles of A still left after 100 min are

$$n_t(\text A)=n_0(\text A)-n_{\text{consumed}}(\text A)=1-0.1=0.9\text{ mol}.$$

For a first-order reaction the rate constant is related to the concentrations by the equation

$$k=\frac{1}{t}\ln\!\left(\frac{[\text A]_0}{[\text A]_t}\right).$$

Concentration is directly proportional to the number of moles when the volume is constant, so we may write

$$k=\frac{1}{100\;\text{min}}\ln\!\left(\frac{1}{0.9}\right).$$

We first evaluate the natural logarithm. Using $$\ln(0.9)=-0.105$$ (from a calculator or a logarithm table) we get

$$\ln\!\left(\frac{1}{0.9}\right)=\ln(1.111\dots)=0.105.$$

Substituting this value,

$$k=\frac{0.105}{100}=1.053\times10^{-3}\;\text{min}^{-1}.$$

The half-life for a first-order reaction is given by the formula

$$t_{1/2}=\frac{\ln 2}{k}.$$

We use the value given in the question, $$\ln 2 = 0.69,$$ so

$$t_{1/2}=\frac{0.69}{1.053\times10^{-3}}\;\text{min}.$$

Carrying out the division,

$$t_{1/2}=6.548\times10^{2}\;\text{min}\;\approx\;655\;\text{min}.$$

So, the answer is $$655\;\text{min}.$$