Chemical Kinetics Important Questions

Very Short Answer Questions (1 Marks Questions)

Ques. Define the term rate of reaction.

Ans. The ratio between the increased concentration of product and decrease in the concentration of reactant per unit time.

Ques. What is the activation of energy?

Ans. To form the activated complex from the reactant molecules the amount of minimum extra energy absorbed is said to be the activation of energy.

Ques. Define the term half-life period of reaction.

Ans. The amount of time used for the half-reaction to complete is known as the half-life of a reaction.

Ques. Expand the given rate of reaction in form of the formation of ammonia.

N₂(g) + 3H₂(g) → 2NH₃(g) (All India 2013)

Ans. \(\frac{-d\left[\mathrm{~N}_{2}\right]}{d t}=\frac{-1}{3} \frac{d\left[\mathrm{H}_{2}\right]}{d t}=+\frac{1}{2} \frac{d\left[\mathrm{NH}_{3}\right]}{d t}\)

Ques. What is rate constant (k)?

Ans. When the concentration of reaction is taken as unity and it is the rate of reaction.

Ques. Give Arrhenius equation with the relation between temperature and rate constant.

Ans. k = A e^{-Ea /RT}

Ques. Define the term ‘order of reaction’.

Ans. The order of the reaction is the sum of powers of any reactants in rate expression. For example, Rate = K [A]^x [B]^y, it can be written as the order of reaction = x + y.

Ques. For given reaction: A + B → P, the rate law r = k [A]^1/2 [B]². What will be the order of reaction? (CBSE All India 2013)
Ans. Order of reaction = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\) or 2.5

Ques. What are zero-order reactions?

Ans. When the rate of a chemical reaction is said to be proportional to the zero power to that of the reactant’s concentration, it is a zero-order reaction. Rate ∝ [A]⁰

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Short Answer Questions (2 Marks Questions)

Ques. Distinguish between rate expression and rate constant of a reaction. (CBSE 2011)
Ans. Rate expression can be said a mathematical expression that is used to determine the rate of a reaction in form of the molar concentration of reactants.

The rate constant is the rate of the reaction when the concentration of reactants is unity. In rate law, it is proportionality constant. The rate constant is independent of the initial concentrations of the reactants.

Ques. Define electrophoresis.

Ans. It is also known as the electrokinetic process which is used for the separation of charged particles by the use of electrical charge in a fluid. In life sciences, it is used for the process of the separation of DNA or protein molecules. It is also used in laboratories for the bifurcation of molecules based on terms like density, size and purity.

Ques. For the below reaction:
\(2 \mathrm{NH}_{3}(g) \stackrel{\mathrm{Pt}}{\longrightarrow} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\)

Rate = k
(i) Write the order and molecularity of the above reaction.
(ii) Also write the unit of k. (CBSE Sample paper 2018)

Ans. \(2 \mathrm{NH}_{3}(g) \stackrel{\mathrm{Pt}}{\longrightarrow} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\)

Rate = k
(i) For the above reaction, Order of reaction: Zero-order Molecularity = 2 (bimolecular)
(ii) Unit of k = mol L^-1 s^-1 or atom s^-1

Ques. (i) For a reaction A + B → P, the rate law is given as follows:
r = k[A]^1/2 [B]². Find the order of this given reaction?

(ii) Find the half-life of the first-order reaction which has a rate constant k = 5.5 × 10^-14 s^-1. (CBSE 2013)

Ans. Order of reaction 1
= 1/2 + 2 = 2.5

(ii) t_1/2 = \(\frac{0.693}{k}\) = \(\frac{0.693}{5.5 \times 10^{-14}}\)

= 1.26 × 10¹³ s

Ques. List the buffers used for DNA electrophoresis.

Ans. The buffers used for DBA electrophoresis are as below:

• Tris-acetate-EDTA (TAE) running buffer
• Tris-borate-EDTA (TBE)

Ques. (i) What will be the order of the reaction whose rate constant has given the same units as the rate of reaction?

(ii) For a reaction: A + H₂O → B; Rate ∝ [A], what is the order of this reaction. (Comptt. All India 2017)

Ans. (i) The reaction whose rate constant has the same units as the rate of reaction, then it will have a zero-order of reaction.

(ii) The reaction A + H₂O → B; Rate ∝ [A]

The order of this reaction will be a pseudo-first-order reaction because the rate of reaction given here depends only on the concentration of A only.

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Long Answer Questions (3 Marks Questions)

Ques. If K = 2.5 × 10^-4 mol litres^-1 s^-1 is given. Find the rate of production of N₂ and H₂ rate for the decomposition of NH₃ on the platinum surface is zero order.

Ans. 2NH₃ → N₂ + 3H₂

For a zero-order reaction, 

Rate of reaction of \(\frac{d}{dt} [NH_3]\) = 2.5 × 10^-4 mol litres^-1 s^-1

\(-\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{2 \mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{dt}}\)

As per the rate law,

2.5 × 10^-4 mol L^-1 s^-1 = \(\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{dt}}\)

Therefore, the rate of production of N₂ is  2.5 × 10^-4 mol litres^-1 s^-1

Now, according to the rate law,

\(-\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{2 \mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{3 \mathrm{dt}}\)

\(\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}=-3 \times \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{2 \mathrm{dt}}\)

Now, the rate of production of H₂ is 3 times the rate of reaction = 

3 × 2.5 × 10^-4 mol L^-1 s^-1

= 7.5 × 10^-4 mol L^-1 s^-1

Ques. Give the difference between ‘order of reaction’ and ‘molecularity’ of reaction (CBSE Delhi 2014)

Ans. The difference between ‘order of reaction’ and ‘molecularity’ of reaction are as follows:

Ques. Give any two important criteria used in order for a reaction to occur. Which scientists gave collision theory? What do you call the collisions which lead to the formation of products?

Ans. The significant criteria include in the order of reaction are as follows:-

• The molecules must collide in reaction.
• The molecules must have the correct orientation or in the correct direction so that they must collide from the correct side.

The Collision Theory was given by scientists Max. Trautz and W. Lewis in 1918. The collisions which lead to the formation of products are called Effective collisions. 

Ques. The rate constant for a first-order reaction is given as 60 s^-1. Then find out how much time will it take to reduce the initial concentration of the reactant to its 1/16 ^th value?

Ans. t = \(\frac{2.303}{k}\) log (\(\frac{[A]_0}{[A]}\))

t = \(\frac{2.303}{60}\) log (\(\frac{a}{\frac{a}{16}}\))

t = 4.6 × 10^-2 seconds

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Very Long Answer Questions (5 Marks Questions)

Ques. During the process of a nuclear explosion, one of the products is given ⁹⁰Sr with a half-life of 28.1 years. If 1 µg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium if it is not lost metabolically. How much of it will be left/remain after 10 years and 60 years?

Ans. According to the question,

Here, k = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{28.1}\ y^{-1}\)

It is known that,

\(\begin{aligned} &t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\\ &\Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[R]}\\ &\Rightarrow 10=\frac{2.303}{\frac{0.693}{28.1}}(-\log [\mathrm{R}])\\ &\Rightarrow \log [R]=-\frac{10 \times 0.693}{2.303 \times 28.1}\\ &\Rightarrow[R] \ =\text{antilog }(-0.1071)\\ & \ \ \ \ \ \ \ \ \ \ \ \ =\operatorname{antilog}(\overline{1} .8929)\\ &\ \ \ \ \ \ \ \ \ \ \ \ =0.7814\ \mu \mathrm{g} \end{aligned}\)

So, 0.7814 μg of ⁹⁰Sr will remain after 10 years.

Further,

\(\begin{aligned} &t=\frac{2.303}{k} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \\ &\Rightarrow 60=\frac{2.303}{\frac{0.693}{28.1}} \log \frac{1}{[\mathrm{R}]} \\ &\Rightarrow \log [\mathrm{R}]=-\frac{60 \times 0.693}{2.303 \times 28.1} \\ &\Rightarrow[\mathrm{R}]=\text { antilog }(-0.6425) \\ &\quad \quad \quad=\operatorname{antilog}(\overline{1} .3575) \\ &\quad \quad \quad =0.2278\ \mu \mathrm{g} \end{aligned}\)

Therefore, it is calculated that 0.2278 μg of ⁹⁰Sr will remain after 60 Years.

Ques. For the below reaction:
2NO(g) + Cl₂(g) → 2NOCl(g)
The following data were collected and the following measurements were taken at 263 K

(i) Write the proper expression for rate law.
(ii) Find the value of rate constant and specify its units.
(iii) What will be the initial rate in experiment 4 for disappearance of Cl₂? (CBSE 2012)

Ans. (i) Rate = k[NO]²[Cl₂]

(ii) As given in experiment 1,
[NO] = 0.15 M; [Cl₂] = 0.15 M
Rate = 0.6 M/min.
Rate = k[NO]²[Cl₂]
0.6 M/min = k(0.15M)² (0.15 M)
k = 177.8 M^-2 min^-1
or
= 177.8 mol^-2 L² min^-1
Therefore, units of k are M^-2 min^-1 or mol^-2 L² min^-1.

(iii) As per experiment 4, [NO] = 0.25 M; [Cl₂] = 0.25 M and k = 177.8 M^-2 min^-1
Rate = k[NO]² [Cl₂]
= 177.8 M^-2 min^-1 × (0.25 M)² × 0.25 M
= 2.78 M min^-1.

Ques. When the temperature changes from 293 K to 313 K here the rate of a reaction becomes four times. So, find the energy of activation (E_a) of the reaction and assuming that it does not change with temperature. Take [R = 8.314 JK^-1 mol^-1, log 4 = 0.6021] (All India 2013)

Ans. We know,

Given: T₁ = 293 K , T₂ = 213 K

Let K₁ = K (consider), K₂ = 4K,

R = 8.314, E_a = ?

As, 

\(\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303\ R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)\)

By substituting values, we get

\(\log \frac{4 k_{2}}{k_{1}}=\frac{E_{a}}{2.303 \times\left(8.314 \mathrm{Jk}^{-1} m o l^{-1}\right)}\left(\frac{1}{293 k}-\frac{1}{313 k}\right)\)

log 4 = \(\frac{E_{a}}{2.303 \times 8.314}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)\)

E_a = 0.6021 × 2.303 × 8.314 × 293 × \(\frac{313}{20}\)

E_a = 52.863 KJ mol^-1

Ques. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with t_1/2 = 3hours. Calculate what will be the fraction of sample of sucrose remains after 8 hours.

Ans. In first order reaction:

k = \(\frac{2.303}{t}\) log (\(\frac{[R]_0}{[R]}\))

t_1/2 = 3 hours

k =  \(\frac{0.693}{t_{1/2}}\)

so, k =  \(\frac{0.693}{3}\) = 0.231 h^-1

0.231h^-1 = \(\frac{2.303}{8h}\) log (\(\frac{[R]_0}{[R]}\))       [k = \(\frac{2.303}{t}\) log (\(\frac{[R]_0}{[R]}\))]

log (\(\frac{[R]_0}{[R]}\)) = \(\frac{0.231 h^{-1} \times 8h}{2.303}\)

\(\frac{[R]_0}{[R]}\) = antilog(0.8024)

\(\frac{[R]_0}{[R]}\) = 6.3445

\(\frac{[R]}{[R]_0}\) = 0.1576 (approx)

= 0.158

Hence, the fraction of sucrose will be 0.158 after 8hours.

Ques. For 25% decomposition,first order reaction takes 20 minutes. Find the time when 75% of the reaction will be completed.
(take: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) (All India 2016)

Ans. Here t = 20 min , A₀ = 100%, A = 100-25 = 5%,

Then k = ?

k = \(\frac{2.303}{t}\) log (\(\frac{[A]_0}{[A]}\))

= \(\frac{2.303}{20}\) log (\(\frac{100}{75}\))

= \(\frac{2.303}{20}\) log 1.33

= \(\frac{2.303}{20}\) × 0.1248

 ∴ k = 0.0143 min^-1

So for 75% completion of reaction,

t = \(\frac{2.303}{k}\) log (\(\frac{[A]_0}{[A]}\))

20 min = \(\frac{2.303}{k}\) log (\(\frac{100}{75}\)) … (1)

t = \(\frac{2.303}{k}\) log (\(\frac{100}{75}\)) … (2)

dividing equation (1) by (2)

\(\begin{aligned} \frac{20}{t} &=\frac{\frac{2.303}{k} \log \left(\frac{100}{75}\right)}{\frac{2.303}{k} \log \left(\frac{100}{25}\right)} \\ &=\frac{\log 4 / 3}{\log 4} \end{aligned}\)

20 / t = 0.1250 / 0.6021

t = 96.3 min

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