Chemical Kinetics MCQ Quiz - Objective Question with Answer for Chemical Kinetics - Download Free PDF

CONCEPT:

Determining Reaction Order from Experimental Data:

• For a rate law: Rate = k[X]m[Y]n
• Compare experiments where one reactant's concentration changes and the other remains constant.
• The rate change gives the exponent (order) for that reactant.

EXPLANATION:

•  Compare row 1 and row 2 (Y is constant):
  • [X] doubles from 0.1 → 0.2
  • Rate doubles from 1.5 × 10⁻³ → 3.0 × 10⁻³
  • ⇒ Order w.r.t X = 1
•  Compare row 2 and row 3 (X is constant):
  • [Y] doubles from 0.2 → 0.4
  • Rate doubles from 3.0 × 10⁻³ → 6.0 × 10⁻³
  • ⇒ Order w.r.t Y = 1

Therefore, the correct order of the reaction is 1 with respect to X and 1 with respect to Y.

EXPLANATION:

Arrhenius Equation and Its Graphical Form:

• The Arrhenius equation is:
  k = Ae^−E_a/RT
• Taking natural log on both sides gives:
  ln k = ln A − E_a / RT
• It can be rearranged as:
  ln k = −(E_a/R) × (1/T) + ln A

This is in the form of a straight-line equation:

y = mx + c, where

• y = ln k
• x = 1/T
• Slope m = −E_a/R (a negative value)
• Intercept c = ln A

• Since the slope is negative, the graph of ln k vs 1/T must be a straight line with a negative slope.
• Among the provided graphs, only the first option shows a line sloping downward as 1/T increases.

Therefore, the correct graph of ln k vs 1/T is the first graph.

CONCEPT:

Zero-Order Reaction Kinetics:

• In a zero-order reaction, the rate is independent of the concentration of the reactant.
• The concentration at any time is given by:

  [A] = [A]0 − kt

• Rearranging for time:

  t = ([A]0 − [A]) / k

EXPLANATION:

• Given:
  • Initial amount [A]0 = 10 µg
  • Final amount [A] = 5 µg
  • Rate constant k = 0.1 µg/year
• Using the formula:

  t = (10 − 5) / 0.1 = 5 / 0.1 = 50 years
  Therefore, the time required is 50 years.

CONCEPT:

Arrhenius Equation:

The Arrhenius equation is:

log(k2/k1) = (Ea / 2.303R) × (1/T1 − 1/T2)

EXPLANATION:

• Given:
  • k1 = 1.2 × 10−3 at T1 = 500 + 273 = 773 K
  • k2 = 3.6 × 10−1 at T2 = 700 + 273 = 973 K
  • R = 8.314 J K−1 mol−1
• Using the equation:

  log(3.6 × 10−1 / 1.2 × 10−3) = (Ea / 2.303 × 8.314) × (1/773 − 1/973)

  log(300) ≈ 2.477

  (1/773 − 1/973) ≈ 2.66 × 10−4

• 2.477 = (Ea / 19.147) × 2.66 × 10−4

  Ea = (2.477 × 19.147) / 2.66 × 10−4 ≈ 178000 J/mol

  Ea = 178 kJ/mol

 Therefore, the activation energy (Ea) is 178 kJ/mol.

EXPLANATION:

First-Order Reaction Kinetics

• In a first-order reaction, the time required for a certain percentage of the reactant to decompose depends only on the half-life.
• The relationship between time and completion percentage for a first-order reaction is:

  t75% = 2 × t1/2

  t87.5% = 3 × t1/2

  t99.9% = 10 × t1/2​

• To calculate time for 75% completion of the reaction:

  t75% = 2 × t1/2 = 2 × 30 = 60 min

Therefore, the time taken for 75% completion of the reaction is 60 minutes.

Concept:

• For a general reaction : aA + bB → cC + dD
  • Rate = k [A]x [B]y  [where x + y = order of reaction]
  • k = Rate/ [A]x [B]y  
  • k = concentration/time x 1/(concentration)n          [where n = x + y i.e. order of reaction]
• The sum of powers of the concentration the reactants in the rate law expression is known as the order of reaction.
• The SI unit of concentration is 'molL-1' and the SI unit of time is 's', then SI units of k for different reactions can be found out using the above formula.

• Auto catalyst: In certain reactions, one of the product acts as a catalyst. In the initial stages, the reaction is slow but as soon as the products come into existences the reaction rate increases. This type of phenomenon is known as auto-catalysis.

• In the permanganate titration of oxalic acid in the presence of H2​SO4​ (acid medium);
• There is a slow discharge of the colour of permanganate solution in the beginning but after some time, the discharge of the colour becomes faster.

Molecular equations

\(\rm 2KMnO_4 \ + \ 3H_2SO_4 \longrightarrow \ K_2SO_4 \ + \ 2MnSO_4 \ + \ 3H_2O \ + \ 5[O]\)

\(\begin{array}{*{20}{c}} {\rm COOH}\\ {| \ \ \,\,\,\,\,\,\,\,\,\,\,}\\ {\rm COOH} \end{array} .\rm 2H_2O \ + \ [O] \ \xrightarrow{60 - 70^\circ C} \ 2CO_2 \ + \ 3H_2O] \ \times \ 5\)

\(\rm 2KMnO_4 \ + \ 2H_2SO_4 + 5 \begin{array}{*{20}{c}} {\rm COOH}\\ {| \ \ \,\,\,\,\,\,\,\,\,\,\,}\\ {\rm COOH} \end{array} . \rm 2H_2O \xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ } \ K_2SO_4 \ + \ 2MnSO_4 \ + \ 18H_2O \ + \ 10CO_2\)

• This is due to the formation of MnSO4​ during the reaction which acts as a catalyst for the same reaction.

So, Mn⁺² is acting like auto catalyst for this reaction.

The correct answer is It increases the rate of reaction.

Key Points

• A catalyst is a substance that can be added to a reaction to increase the reaction rate without getting consumed in the process.
• Catalysts typically speed up a reaction by reducing the activation energy or changing the reaction mechanism.
• Enzymes are proteins that act as catalysts in biochemical reactions.
• Common types of catalysts include enzymes, acid-base catalysts, and heterogeneous (or surface) catalysts.
• A positive catalyst is a substance that increases the rate of reaction or the speed of the reaction without getting consumed in the process. Hence, Option 1 is correct.
• These catalysts can be recovered at the end of the reaction and reused as they do not get altered in any way during the reaction.
• Catalysts can be of 2 types:
• Homogenous catalysts- those catalysts which are in the same phase or state (mostly liquid or gas) as the reactants.
• Heterogeneous catalysts- those catalysts which are in a different phase than the reactants. Catalysts that are involved in biochemical reactions are called enzymes.

The correct answer is Zero.

• The order of reaction represents the number of reactants whose concentration directly affects the rate of reaction.
• In zero Order Reactions, the rate of reaction is independent of the concentration of the reactants.
• In the first-order reaction, the rate of reaction depends on the concentration of only one of the reactants.
• In second-order reactions, the rate of reactions can be obtained either from the concentration of one reactant squared or from the concentration of two separate reactants.

Additional Information

• The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters (normally rate coefficients and partial reaction orders).

CONCEPT:

• Half-life: The half-life of a chemical reaction can be defined as the time taken for the concentration of a given reactant to reach 50% of its initial concentration. It is denoted by the symbol ‘t1/2’ and is usually expressed in seconds.
• The half life period of first order reaction may be calculated as given below:

​N = N0 / 2n

Where, n = no. of half-lives

n  = Total time (t) / Half-life period (t½)

N0 = Initial concentration of the substance           

N = Amount of radioactive substance left after n- half-live

Calculations:

Given:  t½ = 20 min ; Let N0 = x ;

After completion 75% remaining amount of reactant

N = 25% of x = x / 4

To find: t =?

We know, N = N0 / 2n

 ⇒ x / 4 = x / 2n

 ⇒ 2n = 4   (·.· 22 = 4)

 ⇒  n = 2 

Also, n = Total time (t) / Half-life period (t½)

⇒ 2 = t / 20

⇒ t = 40 min

Hence, first order reaction takes 40 minutes to 75% completion.

Concept:

Nitrogen dioxide (NO₂) is a highly reactive gas which is also termed as nitrogen oxide and it has a high temperature of reddish-brown gas.

From the question, the reaction given is:

2N₂O₅ (g) → 4NO₂ (g) + O₂ (g)

Rate of reaction \(\Rightarrow \frac{{ - 1}}{2}\frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{1}{4}\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{d\left[ {{O_2}} \right]}}{{dt}}\)

Rate of reaction \(\Rightarrow \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{{ - 4}}{2}\frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}}\)

Rate of reaction \(\Rightarrow \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = - 2\frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}}\)

Calculation:

According to the question,

\(\frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = - \frac{{\left( {2.75 - 3} \right)}}{{30}} = \frac{{0.25}}{{30}}\;M\;{\rm{mi}}{{\rm{n}}^{ - 1}} = \frac{1}{{120}}\;M\;{\rm{mi}}{{\rm{n}}^{ - 1}}\)

Now,

\(\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = 2 \times \frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = 2 \times \frac{1}{{120}}\)

\(\therefore \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = \frac{1}{{60}}M\;{\rm{mi}}{{\rm{n}}^{ - 1}}\)

Concept:

Rate of a Reaction:

• The rate of a reaction is the velocity of a reaction.
• It is the amount of chemical change occurring with time.
• As the reaction continues, the amount or concentration of reactants decreases, and the concentration of products increases.

Thus it can be regarded as the rate of decrease of reactants or the rate of increase of products.

Rate Law:

• Rate law states that the rate of a reaction is proportional to the concentration of the reactants raised to the power of the order of the reaction.

In mathematical terms, we can say that

\( Rate = - (dc/dt) \); the rate of decrease of reactants

\(Rate= dx/dt\) ; the rate of increase of reactants

\(Rate = k[C]^n\) ; where 'k' = rate constant and 'n' = order of a reaction.

​Order of a reaction:

• The order of a reaction is the number of concentration terms on which the reaction rate depends.
• If multiple concentration terms are involved, then the order of a reaction is the sum of all powers.
• For example if -

\(- (dc/dt) = k[A]^x[B]^y ,\) then

\( order = x + y​\)

Calculation:

Given: 

The reaction is A + B→ C\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % GGUaGaaGimaiaaigdaaeaacaGGUaGaaGimaiaaisdaaaGaeyypa0Za % aiWaaeaadaWcaaqaaiaac6cacaaIXaaabaGaaiOlaiaaikdaaaaaca % GL7bGaayzFaaWaaiWaaeaadaWcaaqaaiaac6cacaaIYaaabaGaaiOl % aiaaikdaaaaacaGL7bGaayzFaaaaaa!4596! \frac{{.01}}{{.04}} = \left\{ {\frac{{.1}}{{.2}}} \right\}\left\{ {\frac{{.2}}{{.2}}} \right\}\)

Let the order of the reaction w.r.t A be 'n' and w.r.t B be 'm'.

\(\frac{{R_1}}{{R_2}} = \frac{{.01}}{{.04}} = \frac{{k[.1]^n[.2]^m}}{{k[.2]^n[.2]^m}}\)

\(\frac{{.01}}{{.04}} = \left\{ {\frac{{.1}}{{.2}}} \right\}^n \)

\(\left\{ {\frac{{.1}}{{.2}}} \right\}^2 = \left\{ {\frac{{.1}}{{.2}}} \right\}^n\)

Equating the powers as the bases are equal, we get 'n' = 2.

again, \(\frac{{R_2}}{{R_3}} = \frac{{.04}}{{.08}} = \frac{{k[.2]^n[.2]^m}}{{k[.2]^n[.8]^m}}\)

or,  \({.04\over.08}= \left\{ {\frac{{.1}}{{.4}}} \right\}^m\)

 \([{{.1\over .4}}]^{1\over2}= \left\{ {\frac{{.1}}{{.4}}} \right\}^m\)

Equating the powers as the bases are equal, we get 'm' = ½

Hence, the total order =  \(m + n = 2.5\)

Explanation:- 

R → P One mole of the reactant R produces one mole of the product P. If [R]₁ and [P]₁ are the concentrations of R and P respectively at time t₁ and [R]₂ and [P]₂ are their concentrations at time t₂ then,

Δt = t₂ – t₁, Δ[R] = [R]₂ – [R]₁, &

 Δ [P] = [P]₂ – [P]₁ 

• Consider a general reaction

a A + b B → c C + d D

Where a, b, c and d are the stoichiometric coefficients of reactants and products.

The rate expression for this reaction is

Rate ∝ [A]x [B]y , Rate = k [A]^x[B]^y 

⇒ dR/dt =  k [A]x[B]^y 

where k is a proportionality constant called rate constant.

Thus, rate law is the expression in which reaction rate is given in terms of the molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation. 

For a general reaction

→ aA + bB → cC + dD Rate = k [A]x [B]y

Where x + y = n = order of the reaction k = x Rate [A] [B]y n concentration 1 = × where [A] [B] time concentration

Taking SI units of concentration, mol L–1 and time, s, the units of k for different reaction order are listed.

Since the unit of Rate Constant is given as k = 3 × 10-4 L mol-1s-1 the unit matches with the second Order

Concept:

Rate of the Reaction

The rate of the reaction is known as the speed at which the reactants are converted into products in a chemical reaction. 

Consider the reaction 

aA + bB → AB

Then the rate of the reaction is given by,

Rate = k [A]^a[B]^b

k-rate constant.

a- order of reaction with respect to A.

b- order of reaction with respect to B. 

[A] - concentration of A. & [B] - concentration of B.

Explanation:

For the given reaction-

2 NO(g) + O2(g) → 2 NO2(g) 

Rate (1) = k [NO]²[O₂]

Here the volume is reduced to half since volume is related to concentration by the equation: → C = n/V

→ C = n/V

So, when the volume is reduced to 1/2, then concentration becomes doubled, hence the rate of the reaction becomes 

Rate (2) = k [2NO]2[2O2]

So,

\(\begin{align} \frac{{rate\,\left( 1 \right)}}{{rate\,\left( 2 \right)}}\,\, &= \,\frac{{k{{\left[ {NO} \right]}^2}\left[ {{O_2}} \right]}}{{k{{\left[ {2NO} \right]}^2}\left[ {2{O_2}} \right]}}= \frac{1}{8} \end{align}\)

i.e. \(\frac{{rate\,\left( 1 \right)}}{{rate\,\left( 2 \right)}}\,\, = \,\frac{1}{8}\)

rate (2) = 8 × rate (1)

So, here when the volume is suddenly reduced to half its value by increasing the pressure on it, the rate of the reaction increase to eight times its initial value. 

Concept:

Rate of a Reaction:

• The rate of a reaction is the velocity of a reaction.
• It is the amount of chemical change occurring with time.
• As the reaction continues, the amount or concentration of reactants decreases, and the concentration of products increases.

Thus it can be regarded as the rate of decrease of reactants or the rate of increase of products.

Rate Law:

• Rate law states that the rate of a reaction is proportional to the concentration of the reactants raised to the power of the order of the reaction.

In mathematical terms, we can say that

Rate = - dc/dt ; the rate of decrease of reactants

Rate = - dx/dt; the rate of increase of reactants

→ Rate = k [C]; 

where 'k' = rate constant and 'n' = order of a reaction.

Calculation:
Given:

• The reaction is 2NO + Br2 → 2NOBr.
• The steps of the reaction are:

\(\rm NO \ + \ Br_2 \ \xrightarrow{Fast} \ NOBr_2\)  -- (I)

\(\rm NOBr_2 \ + \ NO \ \xrightarrow{Slow} \ 2NOBr\) --- (II)

The rate-determining step is \(\rm NOBr_2 \ + \ NO \ \xrightarrow{Slow} \ 2NOBr\) the because it is the slowest step.

According to rate law, the rate = k [NOBr₂][NO] - (a)

But the NOBr₂ is an intermediate and thus its concentration should be replaced.

For Reaction (I)

→ We know Equilibrium Constant K_c is 

→ \(K_c = \frac{[NoBr_2]}{[NO][Br_2]}\)  → [NOBr₂] = K_c × [NO] × [Br₂] --- (b)

Substituting (b) in Equation (a) we get 

Rate = k × Kc × [NO] × [Br2] × [NO]

∴ Rate = k' × [NO]² × [Br2]

Hence, for the reaction 2NO + Br2 → 2NOBr, the rate law is k [Br2][NO]².

Additional Information The rate-determining step of a reaction:

• There are multiple steps involved in a reaction involving intermediates.
• Each and every step has a different rate and it affects the overall rate of the reaction.
• Faster steps have a low effect on the reaction rate but when a step rate is slow, it slows down the whole reaction.
• Hence, the slowest step of the reaction becomes the rate-determining step.
• This is also known as the Bottleneck principle, as the rate of flow of water through the bottle depends on the size of the neck of the bottle.