Q.Show that, time required for 99.9% completion of a first order reaction is three times the time required for 90% completion. Give electronic configuration of Gd (Z=64). Write the name of nano structured material used in car tyres to increase the life of tyres.

Plugging x=99.9 and x=90 into the first-order time formula $t_x=(2.303/k)\log[100/(100-x)]$ gives a ratio of exactly 3.

Proof that $t_{99.9\%} = 3\times t_{90\%}$: for a first-order reaction, the time for $x\%$ completion is $t_x = \dfrac{2.303}{k}\log_{10}\dfrac{100}{100-x}$.

For 99.9% completion: $t_{99.9} = \dfrac{2.303}{k}\log_{10}\dfrac{100}{0.1} = \dfrac{2.303}{k}\log_{10}(1000) = \dfrac{2.303}{k}\times3 = \dfrac{6.909}{k}$

For 90% completion: $t_{90} = \dfrac{2.303}{k}\log_{10}\dfrac{100}{10} = \dfrac{2.303}{k}\log_{10}(10) = \dfrac{2.303}{k}\times1 = \dfrac{2.303}{k}$

$\dfrac{t_{99.9}}{t_{90}} = \dfrac{6.909/k}{2.303/k} = \dfrac{6.909}{2.303} = 3$

Hence $t_{99.9\%} = 3\times t_{90\%}$, proved.

…