For a first order reaction A $$\to$$ B, the rate constant, $$k = 5.5 \times 10^{-14}$$ s$$^{-1}$$. The time required for $$67\%$$ completion of reaction is $$x \times 10^{-1}$$ times the half life of reaction. The value of $$x$$ is ______ Nearest integer) (Given : $$\log 3 = 0.4771$$)

We need to find the value of $$x$$ such that the time for 67% completion of a first-order reaction equals $$x \times 10^{-1}$$ times the half-life.

For a first-order reaction, the time for a given fraction of completion is given by $$t = \frac{2.303}{k} \log\left(\frac{1}{1 - \text{fraction}}\right).$$ When 67% of the reaction is complete, the fraction remaining is $$1 - 0.67 = 0.33 = \frac{1}{3},$$ so $$t_{67\%} = \frac{2.303}{k} \log\left(\frac{1}{1/3}\right) = \frac{2.303}{k} \log 3.$$

The half-life for a first-order reaction is $$t_{1/2} = \frac{0.693}{k}.$$ Hence, $$\frac{t_{67\%}}{t_{1/2}} = \frac{2.303 \times \log 3}{0.693}.$$ Substituting $$\log 3 = 0.4771$$ gives $$\frac{t_{67\%}}{t_{1/2}} = \frac{2.303 \times 0.4771}{0.693}.$$

Since $$2.303 \times 0.4771 = 1.0988,$$ we get $$\frac{1.0988}{0.693} = 1.5855 \approx 1.6.$$ Writing $$t_{67\%} = x \times 10^{-1} \times t_{1/2}$$ then implies $$x \times 10^{-1} = 1.6,$$ so $$x = 16.$$

Therefore, the answer is 16.