As.ΔH = - 33 kJ mol^-1, the reaction is exothermic. The activation energy diagram will be as shown in fig.
ΔH = E_f - E_b= - 33 kJ
k_f = Ae^-E_f/RT
kb = Ae^-E_b/RT
\(K_c={k_f\over k_b}=e^{(E_b-E_f)/RT}\)
In \(K_c={E_b-E_f\over RT}\ or\ log\ K_c={E_b-E_f\over 2.303RT}={30000\ J\ mol^{-1}\over 2.303(8.314JK^{-1}mol^{-1})300K}=5.2227\)
K_c = Antilog 5·2227 = 1·67 x 10⁵
Substituting \(E_b={31\over 20}E_f,\)We get
\(E_f-{31\over 20}E_f=-33\ or\ {-{11\over 20}}E_f=-33\ or\ E_f={33\times20\over 11}=60kJ\ mol^{-1}\)
E_b = E_f + 33 = 60 + 33 = 93 kJ mol^-1

From the given data, we have

Graph of 10 k vs lIT. The plot obtained is as shown in the Fig.
Slope of the line = \({y_2-y_1\over x_2-x_1}=-20.62\times10^3K\)
From Arrhenius eqn., Slope = -\({E_a\over R}\)(for plot of In k of Iff)
E_a = - Slope x R
= 20·62 x 10³ K x (8·314 JK^-I mol^-1)
= 171.4 kJ mol^-1
Further, In k = In A -\({E_a\over RT}\) or In A = in k+\({E_a\over RT}\)
Substituting T = 633 K, k = 0·19 x 10^-4 s^-1,
i.e. In k = - 10·87, we get
In \(A=-10.87+{171400\over8.314\times633}=-10.87+32.57=21.70\)
or A = 2·65 x 10⁹M^-1 s^-1.

(i) As pressure of X is changing with time, it cannot be a zero order reaction. Let us now check it for 1st order.
At t = 100 min, \(k={2.303\over 100}log{P_0\over P_t}={2.303\over 100}log {800\over 400}=6.932\times 10^{-3}min^{-1}\)
At t = 200 min, \(k={2.303\over 200}log{800\over 200}={2.303\over 800}log4=6.932\times10^{-3}min^{-1}\)
As k comes out to be constant, hence it is a reaction of 1st order
(ii)\(t_{75./.}={2.303\over k}log{100\over 100-75}={2.303\over 6.932\times10-3min}log4=200min\)
(iii) 2x (g)⟶ 3 y (g) + 2z (g)
Initial Pressure             800mm      0             0
Pressure after time t     800 - 2p    3 p           2 p
When pressure of X is 700 mm, 800 - 2 p = 700 or p = 50 mm
Total pressure = (800 - 2 p) + 3 p + 2 p = 800 + 3 p = 800 + 3 x 50 = 950 mm.

Suppose initial pressure of N₂O₅ is P mm and decrease is pressure of N₂O₅ in time t is p mm.
Then N₂O₅ (g) = 2 N02 (g) + 0·5 02 (g)
Initial   P mm
After time t,   (P-p)  2p   0·5 p Total = P + 1·5p
P ∝ 600 mm and (P + 1·5p) ∝ 960 mm or 1·5p ∝ 360 mm or p o∝ 240 mm
Mole fraction of N₂O₅ decomposed\(={p\over P}={240\over 600}=0.4\)

\(t_1={2.303\over k_1}log{a\over 0.10a}=t_1={2.303\over k_2}log{a\over a-0.25a}\)
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.