At pH = 5, as half-life period is found to be independent of initial concentration of sucrose, this means with respect to sucrose, it is a reaction of first order, i.e., Rate = k [Sucrose].
If n is the order with respect to H⁺ion, t_1/2 ∝ [H⁺]^I-n,
i.e., 500 ∝ (10^-5)^I-n [PH = 5 means [H⁺] = 10^-5 M]                 .......(I)
and 50 ∝ (l0^-6)I-n [pH = 6 means [H⁺] = 10^-6 M]                    .......(ii)
Dividing (i) by (ii), 10 = (lo)l-n i.e. 1 - n = 1 or n = 0, i.e., order with respect to H+ion = O.Hence, overall rate law is Rate = k [Sucrose] [H⁺]^o.

If R_I is the rate of first reaction at 25°C, then as its temperature coefficient is 2, its rate at 35°C will be = 2 R₁, at 45°C = 2 x 2 R₁ = 4 R₁ and at 55°C = 2 x 4 R₁ = 8 R₁
If R₂.is the rate of the second reaction at 25°C, then as its temperature coefficient is 3, its rate at 35°C will be = 3 R₂, at 45°C = 3 x 3 R₂ = 9 R₂ and at 55°C = 3 x 9 R₂ = 27 R₂
Also, we are given R₁ = R₂, i.e., at 25°C, the rates are equal.
At 55⁰C, \({Rate\ of\ 2nd\ reaction\over Rate\ of\ 1st\ reaction}={27R_2\over 8R_1}={27\over 8}\)

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.

\(k={0.693\over t_{1/2}}={0.693\over 2.1hr}=0.33hr^{-1}\)
x = 99% of a = 0·99 a
\(t={2.303\over k}log{a\over a-x}={2.303\over 0.33hr^{-1}}log{a\over a-0.99a}={2.303\over 0.33}log 10^2=13.69hours\)
(ii) Amount decomposed = 99% of 6.2 g =\({99\over 100}\times 6.2g=6.138g\)
1 mol NH₂NO₂ (63g) produce N₂O at STP = 22·4 L
6.138 g will produce N₂O at STP =\({22.4\over 63}\times6.138\ L=2.2176L\)