Chemistry 3 Marks Question

$\log\frac{\text{k}_{2}}{\text{k}_{1}}=\frac{\text{E}_\text{a}}{\text{2.303}\times\text{8.314 JK}^{-1}\text{mol}^{-1}}\bigg(\frac{\text{350-300}}{\text{350}\times\text{300 K}}\bigg)$

$\log\frac{\frac{0.693}{5}}{\frac{0.693}{20}}=\frac{\text{Ea}}{\text{2.303}\times\text{8.314}}\times\frac{\text{50}}{\text{350}\times\text{300}}$

$\log{4}=\frac{\text{E}_\text{a}}{\text{2.303}\times\text{8.314}}\times\frac{\text{50}}{\text{350}\times{300}}\text{ J mol}^{-1}$

$\text{E}_\text{a}=\text{2.303}\times\text{8.314}\times\text{300}\times\text{7}\times\text{0.6021 J}\text{mol}^{-1}$

$\text{= 24210 J mol}^{-1}=\text{24.210 kJ mol}^{-1}.$

$20\text{ }\text{min}=\frac{2.303}{k}\text{ }\log\frac{100}{75}\text{ }\text{ }\text{ }\text{ }\dots(1)$

$t=\frac{2.303}{k}\text{ }\log\frac{100}{25}\text{ }\text{ }\text{ }\text{ }\dots(2)$

Divide (1) equation by (2)

$\frac{20}{t}=\frac{\frac{2.303}{k}\text{ }\log\frac{100}{75}}{\frac{2.303}{k}\text{ }\log\frac{100}{25}}$

$=\frac{\log4/3}{\log4}$

$20/\text{t}=0.1250/0.6021$

$\text{t}=96.3\text{ }\text{min}$

1. $\text{k}=\frac{2.303}{\text{t}}\text{ }\log\frac{\text{[A]}o}{[\text{A}]}$

$\text{k}=\frac{2.303}{300}\text{ }\log\frac{1.6\times10^{-2}}{0.8\times10^{-2}}$

$\text{k}=\frac{2.303}{300}\text{ }\log2=2.31\times10^{-3}\text{S}^{-1}$

$\text{At}\text{ }600\text{s},\text{ }\text{k}=\frac{2.303}{\text{t}}\text{ }\log\frac{\text{[A]}o}{\text{[A]}}$

$=\frac{2.303}{600}\text{ }\log\ \frac{1.6\times10^{-2}}{0.4\times10^{-2}}$

$\text{k}=2.31\times10^{-3}\text{S}^{-1}$

k is constant when using first order equation, therefore, it follows first order kinetics.

OR

In equal time interval, half of the reactant gets converted into product and the rate of reaction is independent of concentration of reactant, so it is a first order reaction.

2. t_1/2 = 0.693/k

= 0.693/2.31$\times$10^-3

= 300 s

1. $\text{k}=\frac{2.303}{t}\log\frac{[A_{o}]}{[A]}$

$\text{k}=\frac{2.303}{\text{20 min}}\log\frac{0.400}{0.289}$

$\text{k} = 0.0163\text{ min}^{–1}$

2. $\text{k}=\frac{2.303}{\text{t}}\log\frac{[A_o]}{[A]}$

$\text{0.0163}=\frac{2.303}{\text{100}}\log\frac{0.400}{[A]}$

$[\text{A}] = 0.078\text{M}$

3. $\text{Initial rate R = k}[\text{N}_2\text{O}_5]$

$= 0.0163\text{ min}^{–1} \times (0.400\text{ M})$

$= 0.00652\text{M min}^{–1}.$

let rate constant = k₁ at 323 K.

let rate constant = k₂ at 373 K

$\log\frac{\text{k}_{2}}{\text{k}_{1}}=\frac{\text{Ea}}{\text{2.303R}}\bigg(\frac{\text{T}_{2}-\text{T}_{1}}{\text{T}_{1}\text{T}_{2}}\bigg)$

when T is 373K, k₂=3k₁

$\log\frac{\text{3k}_{1}}{\text{k}_{1}}=\frac{\text{Ea}}{\text{2.303}\times\text{8.314}}\bigg(\frac{\text{373}\text{-}\text{323}}{\text{323}\times\text{373}}\bigg)$

$\text{0.4771}=\frac{\text{Ea}}{\text{2.303}\times\text{8.314}}\times\frac{\text{50}}{\text{323}\times\text{373 J mol}^{-1}}$

$\text{Ea}=\text{22011.76 J mol}^{-1}=\text{22.012kJ mol}^{-1}.$

1. For a reaction to occur, there must be collision between the reacting species.
2. Only a certain fraction of total collisions are effective in forming the products.
3. For effective collisions, the molecule must possess the sufficient energy (equal or greater than threshold energy) as well as proper orientation.

On the basis above conclusions, the rate of reaction is given by

Rate = f × 2 (where f is the effective collisions and is total number of collisions per unit volume per second)

Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant k was proposed by Arrhenius. The equation, called Arrhenius equation is usually written in the form,

$\text{K}=\text{Ae}^{\text{-E}_\text{a}/\text{RT}}\ ....(\text{i})$

where A is a constant called frequency factor (because it gives the frequency of binary collisions of the reacting molecules per second per litre, E₀ is the energy of activation, Risa gas constant and Tis the absolute temperature. The factor e^-E_a/RT gives the fraction of molecules having energy equal to or greater than the activation energy, Ea.

The energy of activation (E_a) is an important quantity and it is characteristic of the reaction. Using the above equation, its value can be calculated.

Taking logarithm or both sides of equation (i), we get,

$\text{In k}=\text{In A}-\frac{\text{E}_\text{a}}{\text{RT}_1}$

If the value of the rate constant at temper-atures T₁ and T₂ are k₁ and k₂ respectively, then we have

$\text{In k}_1=\text{In A}-\frac{\text{E}_\text{a}}{\text{RT}_1}\ ....(\text{ii} )$

$\text{In k}_2=\text{In A}-\frac{\text{E}_\text{a}}{\text{RT}_2}\ ....(\text{iii} )$

Subtracting eqn. (i) from eqn. (ii), we get

$\text{In k}_2=\text{In k}_1=\frac{\text{-E}_\text{a}}{\text{RT}_2}+\frac{\text{E}_\text{a}}{\text{RT}_1}$

$=\frac{\text{E}_\text{a}}{\text{RT}_1}+\frac{\text{E}_\text{a}}{\text{RT}_2}$

$\text{or}\ \ \ \text{In}\frac{\text{k}_2}{\text{k}_1}=\frac{\text{E}_\text{a}}{\text{R}}+\bigg(\frac{1}{\text{T}_1}-\frac{1}{\text{T}_2}\bigg)$

$=\frac{\text{E}_\text{a}}{\text{R}}\bigg(\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\bigg)$

$\text{or}\ \ \ \text{log}\frac{\text{k}_2}{\text{k}_1}=\frac{\text{E}_\text{a}}{\frac{2.303}{\text{R}}}\bigg(\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\bigg)$

Thus knowing the values of the constant k₁ and k₂ at two different temperature T₁ and T₂, the value of E_a can be calculated.

Explanation:

1. Rate of conversion of diamond is imperceptible because it requires high activation energy.
2. Instantaneous rate of a reaction is rate of a reaction at a particular moment of time.
3. Average rate is obtained by dividing the change in concentration of any one of the reactant of product by the time taken for the change i.e. $\frac{\Delta\text{x}}{\Delta\text{t}}.$

1. Differential rate equation of reaction is

$\frac{\text{dx}}{\text{dt}}=\text{k}[\text{A}]^1[\text{B}]^2=\text{k}[\text{A}]\ [\text{B}]^2$

2. When cone. of B is tripled, it means cone. of B becomes (3 × B)

$\therefore$ New rate of reaction,

$\frac{\text{dx'}}{\text{dt}}=\text{k}[\text{A}]\ [\text{3B}]^2=9\text{k}[\text{A}]\ [\text{B}]^2=\bigg(\frac{\text{dx}}{\text{dt}}\bigg)$

i.e., rate of reaction will become 9 times.

3. When cone. of A is doubled and that of B is also doubled, then cone. of A becomes [2A] and that of B becomes [2B] rate of reaction,

$\frac{\text{dx''}}{\text{dt}}=\text{k}[\text{2A}]\ [\text{3B}]^2=8\text{k}[\text{A}]\ [\text{B}]^2$

i.e., the rate of reaction will become 8 times the rate as in (1).

1. $\text{Half life, t}_\frac{1}{2}=\frac{0.693}{\text{k}}$

$=\frac{0.693}{200\ \text{min}^{-1}}$

= 3.4 × 10^-3 s(approximately)

2. $\text{Half life, t}_\frac{1}{2}=\frac{0.693}{\text{k}}$

$=\frac{0.693}{2\ \text{min}^{-1}}$

= 0.35 min (approximately)

3. $\text{Half life, t}_\frac{1}{2}=\frac{0.693}{\text{k}}$

$=\frac{0.693}{4\ \text{years}^{-1}}$

= 0.173 years (approximately)

$\text{k}=\frac{2.303}{\text{t}}\log\frac{|\text{R|}_0}{\text{|R|}} $

$\text{k}=\frac{2.303}{24}\log\frac{1}{0.125}$

$\text{k}=\frac{2.303}{24}\log8$

$\text{k}=\frac{2.303}{24}\times0.9031$

$\text{k}=0.866 \text{ h}^{-1}$

$\text{t}_{\frac{1}{2}}=\frac{0.693}{\text{k}}$

$\text{t}_{\frac{1}{2}}=\frac{0.693}{0.0866\text{ h}^{-1}}\text{ or t}_{\frac{1}{2}}=8\text{ h}.$

 $\text{p}_\text{t}=\text{PN}_2\text{O}_5+\text{PN}_2\text{O}_4+\text{PO}_2$

$\text{PN}_2\text{O}_5=0.5-2\text{x}=0.5-2(\text{p}_\text{t}-0.5)=1.5-2\text{p}_\text{t}$