\( \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \)
\( \mathrm{~K}=\frac{0.693}{36}=0.01925 \mathrm{hr}^{-1} \)
\(1^{\text {st }}\) order rxn kinetic equation
\( t=\frac{2.303}{K} \log \frac{a}{a-x} \)
\( \log \frac{a}{a-x}=\frac{t \times K}{2.303} \quad(t=1 \text { day }=24 \mathrm{hr}) \)
\( \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\frac{24 \mathrm{hr} \times 0.01925 \mathrm{hr}^{-1}}{2.303} \)
\( \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=0.2006 \)
\( \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\operatorname{anti} \log (0.2006) \)
\( \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=1.587 \)
\( \text { If } \mathrm{a}=1 \)
\( \frac{1}{1-\mathrm{x}}=1.587 \Rightarrow 1-\mathrm{x}=0.6301=\text { Fraction remain } \) after one day
$2N_2O_5 (g) \to 4NO_2 (g) + O_2 (g)$
$N_2O_5$ ની શરૂઆતની સાંદ્રતા $3.00\, mol\, L^{-1}$ છે. અને $30$ મિનિટ બાદ તે $2.75\, mol\, L^{-1}$ છે. તો $NO_2$ તા સર્જનનો દર ................ $mol\, L^{-1}\, min^{-1}$ જણાવો.