Chemistry - Chemical Kinetics Question with Solution | TestHub

Time ?

Where rotation $=-2.032$

We know $\mathrm{K}=\frac{1}{\mathrm{t}}\ell \mathrm{n}\left(\frac{{\mathrm{Q}}_{\infty }-{\mathrm{Q}}_0}{{\mathrm{Q}}_{\infty }-{\mathrm{Q}}_{\mathrm{t}}}\right)\left\{\because \mathrm{K}=1.1\times 10^{-2}\min \right\}$ $1.1\times 10^{-2}=\frac{1}{t}\ln \left(\frac{-10.74-23.09}{-10.74+2.032}\right)$

$1.1\times 10^{-2}=\frac{1}{\mathrm{t}}\ell n\left(\frac{34.83}{18.708}\right)$

$1.1\times 10^{-2}=\frac{1}{t}\ell n4$

$\mathrm{t}=\frac{\ln 4}{1.1\times 10^{-2}}$

$\mathrm{t}=\frac{1.38}{1.1\times 10^{-2}}\{$ Given $\ln 4=1.38\}$

$\mathrm{t}=126\min$.