Let \(\left[A_{0}\right]=100 \%\), t = 50 minutes
Then [A]=100 - 40 = 60 %
(1) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{\mathrm{o}}\right]}{[\mathrm{A}]}\)
\(=\frac{2.303}{50} \log \left(\frac{100}{60}\right) \)
\(=\frac{2.303}{50} \log 1.667  \)
\(=\frac{2.303}{50} \times 0.2219 \)
\(\mathrm{k}=0.010216 \mathrm{~min}^{-1}  \)
\(\mathrm{k}=1 \times 10^{-2} \mathrm{~min}^{-1}\)
(2)  \( t =\frac{2.303}{0.010216} \log \left(\frac{100}{20}\right) \)
\(t =\frac{2.303}{{0.010216}}\times 0.6990\)
= 225.43 \(\times \) 0.6990
t = 157.58 min.
The time at which the reaction will be 80% complete is 157.58 min.

\(\mathrm{k}=0.693 / \mathrm{t}_{1 / 2}\)
\(\mathrm{k}=\frac{0.693}{8.0}=0.08 \mathrm{~min}^{-1}\)
For the first order reaction:
\(t =\frac{2.303}{k} \log \frac{\left[A_{0}\right]}{[A]}  \)
\(t =\frac{2.303}{0.087} \log \left(\frac{100}{1}\right)=26.47 \log 10^{2}  \)
\(t =2 \times 26.47 \log 10  \)
\(t =52.94 \mathrm{~min}\)

(i) The chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product.
(ii) The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.
Example:
Let us compare the following two reactions that you carried out in volumetric analysis.
1) Redox reaction between ferrous Ammonium Sulphate (FAS) and KMnO₄.
2) Redox reaction between oxalic acid and KMnO₄.
(i) The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe²⁺. In fact heating is required for the reaction between KMnO₄ and Oxalate ion and is carried out at around 60^oC.
(ii) The physical state of the reactant also plays an important role to influence the rate of reactions.
(iii) Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants.
Ex : Na_(s) + I₂(vap)   [Faster]
Na_(s) + I₂(s)   [Slower]
KI_(aq) + Pb(NO3)_2(aq) → PbI₂ (yellow) [Faster]
KI_(s) + Pb(NO3)_2(s) → PbI₂ (yellow)  [Slower]

\(\log K=\log A-\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ T } \right) \)
y = c + mx
\(m=-\frac { { E }_{ a } }{ 2.303R } \)
E_a = -2.303 Rm
E_a = -2.303 x 8.314 x (-4000)
E_a = 76,589J  mol^-1
E_a = 76.589 KJ mol^-1

(i) Let A = 100M, [A₀] - [A] = 20M,
For the zero order reaction
\(k=\left( \frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \right) \)
(i) 20% completion  \(k=\left( \frac { 20M }{ 20min } \right) \) = 1 mol L^-1 min^-1 
(ii) 80% completion
\(\mathrm{K}=1 \mathrm{~mol} \mathrm{~L}{ }^{-1} \mathrm{~s}^{-1} ;\left[\mathrm{A}_{0}\right]=100 \mathrm{M} ;\left[\mathrm{A}_{0}\right]-[\mathrm{A}]=80 \mathrm{M} ; \mathrm{t}=?\)
\(\therefore t=\left(\frac{\left[A_{O}\right]-[A]}{K}\right)=\frac{80}{1}=80 \mathrm{mins}\)