Given,
E_a = 49 kJ / mol^-1
T = 313 K
A = 9 × 10² s^-1
R = 8.314 J K^-1 mol^-1
log k = – E_a / (2.303 RT) + log A
log k = – 49000 / (2.303 × 8.314 × 313) + log 9 × 10¹⁰
log k = 2.77843
k= antilog (2.77843)
k = 6 × 10² s^-1.

Half-life (t_1?2) is time required for a quantity to reduce to half of its initial value. The half-life of a zero-order reaction is directly proportional to its initial concentration. They are related as:
t_1/2 = [R]₀/2k.

Given, rate constant k= 0.00058 s^-1
Time t= 50 minutes = 50 × 60 = 3000 seconds
First-order integrated rate equation, k = \(\frac{2.303}{t}\)log\(\frac{100}{100-a}\)
log\(\frac{100}{100-a}\) = \(\frac{kt}{2.303}\)
log\(\frac{100}{100-a}\) = \(\frac{0.00058 \times 3000}{2.303}\) = 0.756
a=82.44 percent.

Let initial concentration of reactant = [A]
Given concentration of reactant = 2 x [A]
We know that rate of a second order reaction = k x [A]²
Therefore, rate for 2 x [A] = k x 4 x [A]²
Therefore, rate quadruples when concentration of reactant is doubled in a second order kinetic reaction.

The reaction C₂H₅COOC₂H₅ + H₂O → C₂H₅COOH + C₂H₅OH is a pseudo first order reaction and follows first order kinetics with rate law R=k[C₂H₅COOC₂H₅].
Given, rate of reaction R=0.39 s^-1 and rate constant k=0.2
So, 0.39 = 0.2 × [C₂H₅COOC₂H₅]
[C₂H₅COOC₂H₅]=1.95 M.

Given,
N₂O₅ → 2N0₂ + 1/2 O₂
Rate = k[N₂O₅] ¹
Rate =6.25 x 10^-4 x [1.25]
Rate = 7.75 x 10^-4.

Given,
k= 3.28 × 10^-4 s^-1
The general formula to find units for rate constant, k=(mol L^-1)^1-ns^-1 where n is order of reaction. The value of n must be 1 for (mol L^-1)^1-ns^-1 to become s^-1. Therefore, k=3.28 × 10^-4s^-1 represents a first order reaction.

The number of collisions per unit volume of reaction mixture is known as collision frequency. According to this ory, reactant molecules are assumed to be hard spheres and reaction is postulated to occur when molecules collide with each or.

The reaction for inversion of cane sugar is C₁₂H₂₂O₁₁ + H₂O ? glucose + fructose.
In this reaction water is present in excess and belongs to a pseudo first order reaction, even though molecularity is 2 order of reaction is 1 so rate law R=k[C₁₂H₂₂O₁₁].

The reaction involved in acid hydrolysis of CH₃COOC₂H₅ in aqueous solution = CH₃COOC₂H₅ + H₂O ? CH₃COOH + C₂H₅OH. The rate law for reaction is R=k[CH₃COOC₂H₅][H₂O], however, water is present in large excess and is assumed to be constant during course of reaction. So rate law becomes R=k [CH₃COOC₂H₅] and follows first order kinetics.