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, \({Rate\ of\ 2nd\ reaction\over Rate\ of\ 1st\ reaction}={27R_2\over 8R_1}={27\over 8}\)

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.

\(k={0.693\over t_{1/2}}={0.693\over 2.1hr}=0.33hr^{-1}\)
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 =\({22.4\over 63}\times6.138\ L=2.2176L\)

(a) Rate expression is a way of expressing rate of reaction, e.g.
N₂(g) + 3H₂(g)⇾ 2NH₃(g)
\({-d[N_2]\over dt}={-{1\over 3}}{d[H_2]\over dt}=+{1\over 2}{d[NH_3]\over dt}\)
Rate constant is defined as equal torate of reaction when molar cone. of reactants is unity. Its unit depends upon order of reaction.
(b) (i)\(k={2.303\over t}log{[R]_0\over [R]}={2.303\over 20}log{0.400\over 0.289}={2.303\over 20}[10 0.400 - 10 0.289]\)
\(={2.303\over 20}=[\bar1.6021-\bar1.4609]={2.303\over 20}\times1.1412\)
\(={0.3521\over 20}=0.016285\ min^{-1}\)
\(k={2.303\over 40}log{0.400\over 0.209}=={2.303\over 40}[log0.400-log0.209={2.303\over 40}\times0.2820\)
\(={0.6494\over 40}=0.01623\ min^{-1}\)
\(k={2.303\over 60}[log 0.400 - log 0.151] ={2.303\over 60}[\bar1.6021- \bar1.1790] ={2.303\over 60}\times 0.4231\)
\(={0.9739\over 60}=0.01623 \ min^{-1}\)
\(k={0.01625+0.01623+001623\over 3}\)
= 0.016236 min^-1
(ii) Initial rate = k[N₂O₅]
= 0.016236 x 0.4 = 0.00649 mol L^-1 S^-1
(iii) \(k={2.303\over t}log{[R]_0\over [R]}\Rightarrow k={2.303\over t}log{0.4\over 0.35}\)
\(t={2.303\over 1.6236\times10^{-12}}(log40-log35)={2.303\over 1.625\times10^{12}}(1.6021-1.5441)\)
\(t={2.303\times10^2\times0.0580\over 1.6236}={13.3574\over 1.6236}=8.227min.\)

Calculation of activation energy of reaction (i)
T₁= 300 K, T₂ = 310 K, k₁ = k, k₂ = 2 k
\(log{k_2\over k_1}={E_A\over 2.303E}\left(T_2-T_2\over T_1T_2\right ),ie.,\ log2={E_a\over 2.303\times8.314}\times{10\over 300\times310}\ or\ E_a=53.60kJmol^{-1}\)
Calculation of rate constant of reaction (i) at 310 K
\(k={0.693\over t_{1/2}}={0.693\over 30\ min}=2.31\times10^{-2}min^{-1}\)
Rate constant of reaction (ii) at 310 K = 2 x 2·31x 10^-2 min^-1 = 4·62 x 10^-2 min^-1
Energy 0f ac tiva tion 0f reac tion (ii) =\({53.60kJ\ mol^{-1}\over 2}=26.80kJ\ mol^{-1}\)
Aim. To calculate k for reaction (ii) at 300 K
\(log{4.62\times10^{-2}\over k_{300}}={26.80\over 2.303\times8.314\times10^{-3}}\times{10\over 300\times310}=0.0151\)
or \({4.62\times10^{-2}\over k_{300k}}=Antilog\ or\ 0151 = 1.035\ or \ k_{300k}={4.62\times10^{-2}\over 1.035}=4.46\times10^{-2}min^{-1}\)