Ea = 200 kJ mol^-1=200 \(\times\) 10³J mol^-1
A = 1.6 \(\times\) 10¹³s^-1; T = 600 K; R = 8.314 JK mol^-1
\(k=A{ e }^{ -\left( \frac { Ea }{ RT } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( \frac { 200\times 10^3}{ 8.314 \times 600 } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( 40.09 \right) }\)
\(k=1.6\times { 10 }^{ 13 }\times 3.8\times { 10 }^{ -18 }{ s }^{ -1 }\)
\(k=6.08\times { 10 }^{ -5 }{ s }^{ -1 }\)

For the first order reaction
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{A}_{\mathrm{o}}}{\mathrm{A}}\) (or)
\(\mathrm{k} =\frac{2.303}{t} \log \frac{P_o}{P_t} \)
Here, P_o=0.08 atm (P_t)=0.04 atm; t = 60 s
\(\mathrm{k}=\frac{2.303}{60} \log \frac{[0.08]}{[0.04]}\)
\(=0.0383 \times \log 2  \)
\(=0.0383 \times 0.3010  \)
\(=1.152 \times 10^{-2} \mathrm{~s}^{-1}\)

Let [A₀] = 100;
When t = t_99.9%; [A] = (100-99.9) = 0.1
\(k=\frac { 2.303 }{ t } \log\left( \frac { [{ A }_{ 0 }] }{ [A] } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log\left( \frac { 100 }{ 0.1 } \right) \)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } \log1000\)
\({ t }_{ 99.9\% }=\frac { 2.303 }{ K } (3)\)
\({ t }_{ 99.9\% }=\frac { 6.909 }{ K } \)
\({ t }_{ 99.9\% }=10\times \frac { 0.69 }{ K } \)
\({ t }_{ 99.9\% }={ 10 } t_{ 1/2 }\)

For a first order reaction
\(\\ \\ k=\frac { 2.303 }{ t } log\left( \frac { [{ A }_{ 0 }] }{ [A] } \right) \\ \)   ..(1)
Let[A₀] =100M
When 
t = t_90%; [A] = 10M (given that t_90% = 8hours)
t = t_80%; [A ] = 20M
\(k=\frac { 2.303 }{ { t }_{ 80\% } } \log\left( \frac { 100 }{ 20 } \right) \)
\({ t }_{ 80\% }=\frac { 2.303 }{ K } \log(5)\) ....(2)
Find the value of k using the given data
\(k=\frac { 2.303 }{ { t }_{ 90\% } } \log\left( \frac { 100 }{ 10 } \right) \)
\(k=\frac { 2.303 }{ 8 } \log10\)
\(k=\frac { 2.303 }{ 8 } ...(3)\)
Substitute the value of k in equation (2)
\({ t }_{ 80\% }\frac { 2.303 }{ 2.303/8hours } \log(5)\)
t_80%= 8 hours x 0.6989
t_80%= 5.59 hours

Pseudo first order reaction:
A second order reaction can be altered to a first order reaction by taking one of the reactants in large excess, such reaction is called pseudo first order reaction.
Example: Acid Hydrolysis of an ester,
\(C{ H }_{ 3 }COOC{ H }_{ 3(aq) }+{ H }_{ 2 }{ O }_{( l) }\overset { { H }^{ + } }{ \longrightarrow } C{ H }_{ 3 }COO{ H }_{ (aq) }+C{ H }_{ 3 }O{ H }_{ (aq) }\)
Rate = \(k\left[ { CH }_{ 3 }COOC{ H }_{ 3 } \right] \left[ { H }_{ 2 }O \right] \)
If the reaction is carried out with a large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e., the concentration of water remains almost constant.
Now, we can define k [H₂O] = k'; Therefore the above rate equation becomes
Rate = \(k'\left[ { CH }_{ 3 }COOC{ H }_{ 3 } \right] \)
Thus it follows first order kinetics.