Rate of reaction at \(10\, \mathrm{s}=-0.04 \,\mathrm{mol} \,\mathrm{L}^{-1} \,\mathrm{s}^{-1}\)
Rate of reaction at \(20\, \mathrm{s}=\mathrm{s}=0.03\, \mathrm{mol} \,\mathrm{L}^{-1} \mathrm{s}^{-1}\)
\(\therefore\) Half - life period \(\left(\mathrm{t}_{1 / 2}\right)=?\)
We have the equation for rate \(-\) constant \(k\) in first order reaction
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{A}_{\mathrm{t}}}{\mathrm{A}_{0}}\)
\(=\frac{2.303}{105} \log \frac{0.04}{0.03}\)
\(=\frac{2.303}{105} \times 0.124\)
\(\mathrm{k}=0.028\, \mathrm{s}^{-1}\)
We know that, \(t_{1 / 2}=\frac{0.693}{k}\)
\(=\frac{0.693}{0.028773391}=24.14\, \mathrm{s}=24.1\, \mathrm{s}\)
${O_3} \rightleftharpoons {O_2} + \left[ O \right]$
${O_3} + \left[ O \right] \to 2{O_2}$ (slow)
તો $2{O_3} \to 3{O_2}$ પ્રક્રિયાનો કમ જણાવો.
[અહી આપેલ $\left.\log _{10} 2=0.3010\right]$