Zero order.

A reaction takes place because the molecules collide. The chances for a large lnumber of molecules or ions to collide simultaneously are less. Hence, the reactions of higher order are less.

Complex reaction proceeds through several elementary reactions.  Molecularity of each elementary reaction may be different, therefore, molecularity of complex reaction can't be determined.  Order of complex reaction is determined by slowest step in mechanism (involving elementary reactions).

Activated complex is intermediate compound between reactants and products as shown above. It is highly unstable as it has highest energy. It readily charges into product. Those molecules which can form activated complex can lead to formation of products, e.g

Log K = log A - Ea/2.303RT
Log K’ = log A - Ea/2.303RT
Log K = log A - Ea/2.303RT
Log K’/K = Ea-Ea/2.303 RT = 9.8037
Or K’/K = 6.36 x 109
UR catalysted /UR uncatalysed = K’/K = 6.4 x 10 9

2HI ⇌ H₂ (g) + I₂ (g)
Initial cone.           2x         0          0
Equilibrium cone. 2x-0.4x  0.2x     0.2x
At equilibrium [HI] = 0.96 mol L^-1
2x - 0.4x = 0.96
1.6x = 0.96
x= 0.60 mol L^-1
[H₂] at equilibrium 0.2x = 0.2 x 0.60 = 0.12 mol L^-1
[I₂]at equilibrium O.2x = 0.2 x 0.60 = 0.12 mol L^-1

\({ T }_{ 1 }=295K, \ { T }_{ 2 }=305K, \ { k }_{ 2 }=2{ k }_{ 1 } \ (given)\)
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
\(\log { 2 } = \ \frac { { E }_{ a } }{ 2.303\times 8.314 } \left( \frac { 305-295 }{ 305\times 295 } \right) \)
\({ E }_{ a } \ = \ \frac { 19.147\times 305\times 295\times 0.3010 }{ 10 } \)
\( { E }_{ a } \ = \ 51854.8 \ J \ { mol }^{ -1 }\)
\(=51.85 \ KJ \ { mol }^{ -1 }\)

\(k={2.303\over t}log{[R]_0\over [R]}\)
\(={2.303\over 100}log{[R]_0\over {40\over 100}[R]_0}\)
[60% is complete, 40% is left]
\(k={2.303\over 100}(log5-log2)\)
\(k={2.303\over 100}(0.6990-0.3010)\)
\(k={2.303\over 100}\times0.3980\ min^{-1}\)
\(t_{90./.}={2.303\over k}log{[R]_0\over [R]}\)
[90% is complete, 10% is left]
\(t_{90./.}={2.303\over k}log{[R]_0\over {10\over 100}[R]_0}\)
\(={2.303\times100\over 2.303\times0.3980}log10\)
\(={100\over 0.3980}=251.26min\)

(c) : 60 days = 3 half-lives, i.e., n = 3 ; [A] = \(\frac { [A]_{ 0 } }{ { 2 }^{ n } } =\frac { 100 }{ { 2 }^{ 3 } } =\frac { 100 }{ 8 } =12.5g\)

(c) : 75% of reaction is completed in two half-lived  i.e., 2 \(\times\) t ₁₂ = 32 min or t _1/2 = 16 min

(c) : On the basis of given units of k, the reactions of 3 rd order.

For the first order reaction
\(t_1=30sec\)
\(t_2=30sec\)
\(k_1=\frac{2.303}{t_1}\log \frac{a}{(a-x)}\)
\(\Rightarrow\ \ \ \ k_1=\frac{2.303}{30}\log \frac{0.60}{0.30}\)
\(=\frac{2.303}{30}\log 2\)
\(\Rightarrow\ \ \ \ k_1=\frac{2.303}{30}\times 0.3010\)
\(=0.0231 \ s^{-1}\)
and \(k_2=\frac{2.303}{t_2}\log \frac{a}{(a-x)}\)
\(=\frac{2.303}{60}\log \frac{0.60}{0.15}\)
\(=\frac{2.303}{60}\log 2^2\)
\(\Rightarrow\ \ k_2=\frac{2.303\times 2 \times 0.3010}{30}\)
\(=0.0231\ s^{-1}\)
\(\because\ k_1=k_2\)
Hence, the reaction is pseudo first order reaction.
(ii) \(Rate=\frac{\Delta x}{\Delta t}\)
\(=\frac{0.30-0.15}{60-30}=\frac{0.15}{30}\)
\(=0.005 \ mol\ L^{-1} \ s^{-1}\)

\(t_1={2.303\over k_1}log{a\over 0.10a}=t_1={2.303\over k_2}log{a\over a-0.25a}\)
As t₁ = t₂ \({2.303\over k_1}log{a\over 0.90a}={2.303\over k_2}log{a\over 0.7a};\ {k_2\over k_1}={log(100/75)\over log(100/90)}=2.73\)
But \(log{k_2\over k_1}={E_a\over 2.303R}\left(T_2-T_1\over T_1T_2\right)\)
Putting k₂/k₁) = 2.73, R = 8.314 J K^-1 mol^-1, T₁ = 298 K, T₂ = 308 K, we get E_a = 76.6 kJ mol^-1
Further, k - Ae^{-E_a / RT} or log k = log A -\({E_a\over 2.303RT}\)
Putting A = 3.56 x 10⁹s^-1, R = 8.314 x 10^-3 kJ K^-1 mol^-1, E_a = 76.6 kJ mol^-1, T = 318 K, we get
k = 9.3 x 10^-4 S^-1.