For the given first order reaction A -> B the half life of the reaction is 0.3010 min. The ratio of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to ___________. (Nearest integer)

t_1 / 2=frac0.693 K t_1 / 2 given =0.3010 K=(0.693 / 0.3010) \ K=2.30 K=frac2.303 t log frac( A_0)( A_ t) A_0 -> initial concentration of reactant A_t -> concentration of reactant at time t 2.303=(2.303 / 2) log frac(A_0)(A_t) 2=log frac(A_0)(A_t) 100=fracA_0A_t

For the given first order reaction A arrow B the half l... | null (Chemistry)

\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \quad\quad \mathrm{t}_{1 / 2}

given

=0.3010

\begin{aligned} &K=\frac{0.693}{0.3010} \\\\ &K=2.30 \end{aligned}

\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left(\mathrm{A}_{0}\right)}{\left(\mathrm{A}_{\mathrm{t}}\right)}

A_{0} \rightarrow

initial concentration of reactant

A_{t} \rightarrow

concentration of reactant at time

t

2.303=\frac{2.303}{2} \log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}

2=\log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}

100=\frac{A_{0}}{A_{t}}

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