Question

Rate of first order reaction is 6.93×10^-3 calculate the half life period​

Answer 1

Given

Rate of a first order reaction = $6.93\times\:10^{-3}$

To Find

Half life of the reaction

Concepts

Half life period is that time interval in which half of the reactants is consumed.It is related to rate constant k as

$\boxed{\green{k\:=\:\dfrac{ln2}{t_{\frac{1}{2}}}}}$

• ln2 = 0.693

Calculations

We know,

$k\:=\:\dfrac{ln2}{t_{\frac{1}{2}}}\\ \\ =>\: t_{\frac{1}{2}}\:=\:\dfrac{ln2}{k} \\ \\ =>\: t_{\frac{1}{2}}\:=\:\dfrac{0.693}{6.93\:\times\:10^{-3}}.......(\because\:ln2\:=\:0.693)\\ \\ => \: t_{\frac{1}{2}}\:=\: \frac{\cancel{6.93}\times\:10^{-1}}{\cancel{6.93}\:\times\:10^{-3}}\\ \\=>\: t_{\frac{1}{2}}\:=\:{10}^2\:seconds \\ \\ \boxed{\boxed{\green{Hence,t_{\frac{1}{2}}\:of\:the\:reaction\:is\:{10}^2\:seconds}}}$