CBSE Class 12 Chemistry Chemical Kinetics MCQs Set 09

MCQs

Question. The unit of rate constant depends upon the
(a) molecularity of the reaction
(b) activation energy of the reaction
(c) order of the reaction
(d) temperature of the reaction
Answer: (c) order of the reaction
Explanation: For the reaction,
\( xA + yB \rightarrow \text{Product} \)
\( r = k[A]^x[B]^y \)
\( \frac{dx}{dt} = k[A]^x[B]^y \)
\( \text{mol L}^{-1}\text{s}^{-1} = k(\text{mol L}^{-1})^x(\text{mol L}^{-1})^y \)
\( k = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^x(\text{mol L}^{-1})^y} \)
\( = (\text{mol L}^{-1})^{1-(x+y)}\text{s}^{-1} \)
Where \( (x + y) = \text{order of the reaction} \)

Question. In the presence of a catalyst, heat evolved or absorbed during reaction
(a) increases
(b) decreases
(c) remains unchanged
(d) may increase or decrease
Answer: (c) remains unchanged
Explanation: There is no effect on heat evolved or absorbed during the reaction in the presence of a catalyst. It is because catalyst influence the rate of reaction and does not participate in the reaction.

Question. Which of the following statement is correct?
(a) The rate of a reaction decreases with passage of time as the concentration of reactants decreases.
(b) The rate of a reaction is same at any time during the reaction.
(c) The rate of a reaction is independent of temperature change.
(d) The rate of a reaction decreases with increase in concentration of reactant(s).
Answer: (a) The rate of a reaction decreases with passage of time as the concentration of reactants decreases.
Explanation: The rate of a reaction depends upon the concentration of reactants. As reactants are being consumed to give products and hence slowly it will be consumed when reaction is completed. That is why rate of reaction is expressed as decrease in concentration of reactants with the passage of time.

Question. Rate law can be determined from balanced chemical equation if
(a) reverse reaction is involved.
(b) it is an elementary reaction.
(c) it is a sequence of elementary reactions.
(d) any of the reactants is in excess.
Answer: (b) it is an elementary reaction.
Explanation: In case of elementary reaction the rate law can be determined from balanced chemical equation. As, order of the elementary reaction is the sum of the stoichiometric coefficients of the reactants of a reaction.

Question. Which of the following expressions is correct for the rate of reaction given below?
\( 5\text{Br}^-(aq) + \text{BrO}_3^-(aq) + 6\text{H}^+(aq) \rightarrow 3\text{Br}_2(aq) + 3\text{H}_2\text{O}(l) \)
(a) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = 5\frac{\Delta[\text{H}^+]}{\Delta t} \)
(b) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = \frac{6}{5}\frac{\Delta[\text{H}^+]}{\Delta t} \)
(c) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = \frac{5}{6}\frac{\Delta[\text{H}^+]}{\Delta t} \)
(d) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = 6\frac{\Delta[\text{H}^+]}{\Delta t} \)
Answer: (c) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = \frac{5}{6}\frac{\Delta[\text{H}^+]}{\Delta t} \)
Explanation: Rate law expression for the given chemical equation is written as :
\( 5\text{Br}^-(aq) + \text{BrO}_3^-(aq) + 6\text{H}^+(aq) \rightarrow 3\text{Br}_2(aq) + 3\text{H}_2\text{O}(l) \)
Rate law expression for the above equation can be presented as :
\( \frac{-1}{5}\frac{\Delta[\text{Br}^-]}{\Delta t} = \frac{-\Delta[\text{BrO}_3^-]}{\Delta t} = \frac{-1}{6}\frac{\Delta[\text{H}^+]}{\Delta t} = \frac{+1}{3}\frac{\Delta[\text{Br}_2]}{\Delta t} \)
\( \implies \)
\( \frac{\Delta[\text{Br}^-]}{\Delta t} = \frac{-5\Delta[\text{BrO}_3^-]}{\Delta t} = \frac{-5}{6}\frac{\Delta[\text{H}^+]}{\Delta t} \)
\( \frac{\Delta[\text{Br}^-]}{\Delta t} = \frac{5}{6}\frac{\Delta[\text{H}^+]}{\Delta t} \)

Question. In a chemical reaction X → Y, it is found that the rate of reaction doubles when the concentration of X is increased four times. The order of the reaction with respect to X is:
(a) 1
(b) 0
(c) 2
(d) 1/2
Answer: (d) 1/2
Explanation: X → Y
Rate(r) ∝ [X]\(^n\) [Where n = Order of reaction]
If the concentration X is increased by 4 times
X' = 4X
Then, Rate(r') ∝ [X']\(^n\)
\( \frac{r'}{r} = \frac{[4X]^n}{[X]^n} = 2 \)
r' is new rate, X' is a new concentration
\( [4]^n = 2 \)
\( \therefore n = \frac{1}{2} \)
Order of reaction = \( \frac{1}{2} \)

Question. The half-life period for a zero order reaction is equal to:
(a) \( \frac{0.693}{k} \)
(b) \( \frac{2k}{[R]_0} \)
(c) \( \frac{2.303}{k} \)
(d) \( \frac{[R]_0}{2k} \)
(where [R]\(_0\) is initial concentration of reactant and k is rate constant).
Answer: (d) \( \frac{[R]_0}{2k} \)
Explanation: Half life period of a zero order reaction \( = \frac{[R]_0}{2k} \)
Where [R]\(_0\) = Initial concentration of reactant
k = Rate constant

Question. For a zero order reaction, the slope in the plot of [R] vs. time is:
(a) \( \frac{-k}{2.303} \)
(b) −k
(c) \( \frac{+k}{2.303} \)
(d) +k
(where [R] is the final concentration of reactant)
Answer: (b) −k
Explanation: For zero order reaction, r = k[A]°
Graphically,
slope = −k

Question. A first order reaction is 50% completed in \( 1.26 \times 10^{14} \) s. How much time would it take for 100% completion ?
(a) \( 1.26 \times 10^{15} \) s
(b) \( 2.52 \times 10^{14} \) s
(c) \( 2.52 \times 10^{28} \) s
(d) Infinite
Answer: (d) Infinite
Explanation: In a chemical reaction, rate of a reaction is calculated in specified time such as half life of the reaction for 50% completion of reaction but it is impossible to calculate time for 100% completion of reaction. Thus, it is assumed that reaction will be 100% complete only after infinite time.

Question. Consider a first order gas phase decomposition reaction given below:
A(g) → B(g) + C(g)
The initial pressure of the system before decomposition of A was ‘p\(_i\)’. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘p\(_t\)’. The rate constant k for the reaction is given as:
(a) \( k = \frac{2.303}{t} \log \frac{p_i}{p_i - x} \)
(b) \( k = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t} \)
(c) \( k = \frac{2.303}{t} \log \frac{p_i}{2p_i + p_t} \)
(d) \( k = \frac{2.303}{t} \log \frac{p_i}{p_i + x} \)
Answer: (b) \( k = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t} \)
Explanation: Let us consider a first order gas phase decomposition reaction:
The initial pressure of the system before decomposition of A is ‘p\(_i\)’ After lapse of time ‘t’, total pressure of the system increased by x units and became ‘p\(_t\)’.
Hence, the pressure of A decreased by x atom.
A(g) → B(g) + C(g)
Initial pressure: \(\quad\) p\(_i\) atom \(\quad\) 0 \(\quad\) 0
Pressure after time t: (p\(_i\) − x) \(\quad\) x atm \(\quad\) x atm
p\(_t\) = (p\(_i\) − x) + x + x
= p\(_i\) + x atm
x = p\(_t\) − p\(_i\)
= p\(_i\) − (p\(_t\) + p\(_i\))
p\(_A\) = 2p\(_i\) − p\(_t\)
Rate constant, \( k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \)
\( k = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t} \)

Question. For the reaction A → B, the rate of reaction becomes three times when the concentration of A is increased by nine times. What is the order of reaction ?
(a) 1
(b) 2
(c) 1/2
(d) 0
Answer: (c) 1/2
Explanation: For a chemical reaction,
A → B
Rate of reaction r ∝ [A]\(^n\) …(i)
If concentration of A is increased by nine times, then rate of reaction becomes three times,
r' = 3r
A' = 9A
r' ∝ [A']\(^n\)
3r ∝ [9A]\(^n\) …(ii)
From equation (i) and (ii)
\( \frac{r}{3r} = \frac{[A]^n}{[9A]^n} \)
\( \frac{1}{3} = \left(\frac{1}{9}\right)^n \)
\( \frac{1}{3} = \left(\frac{1}{3}\right)^{2n} \)
\( 1 = 2n \)
\( n = 1/2 \)
\( \therefore \) Order of reaction = [1/2]

Question. The value of rate constant of a pseudo-first-order reaction:
(a) depends on the concentration of reactants present in small amount.
(b) depends on the concentration of reactants present in excess.
(c) is independent of the concentration of reactants.
(d) depends only on temperature.
Answer: (b) depends on the concentration of reactants present in excess.
Explanation: Rate constant of a pseudo-first order reaction depends on the concentration of reactants present in excess. In a chemical reaction,
A + B → C
Where rate of reaction R = \( k[A][B] \)
But if concentration of B is in excess, then concentration of B will remain constant. The chemical reaction will be termed as pseudo-first order reaction such that R = \( k'[A] \), where \( k' \) is new rate constant value.

Question. The slope in the plot of ln[R] vs. time gives:
(a) +k
(b) \( \frac{+k}{2.303} \)
(c) −k
(d) \( \frac{-k}{2.303} \)
(where [R] is the final concentration of reactant.)
Answer: (c) −k
Explanation: For a first order reaction, r = k[R]
ln[R] = −kt + C
slope = −k
 

Assertion and Reason Based MCQs

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Question. Assertion (A): Rate of reaction doubles when concentration of reactant is doubled if it is a first order reaction.
Reason (R): Rate constant also doubles with twice increase in the concentration of reactant.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
Explanation: For first order reaction
Rate\( _1 \) = \( k[A]_1 \)
[A]\(_2\) = 2[A]\(_1\)
Rate\( _2 \) = \( k[2A_1] \)
Rate\( _2 \) = \( 2 \times k[A_1] \)
Rate\( _2 \) = \( 2 \times \text{Rate}_1 \)
For a given reaction, rate constant is constant and independent of the concentration of the reactant.

Question. Assertion (A): The rate of reaction increases with the increase in temperature.
Reason (R): The reactant molecules collide less frequently on increasing temperature.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
Explanation: As the temperature of a reaction is increased, the rate of the reaction increases because the reactant molecules collide more frequently and with greater energy per collision. Such that effective collisions will increase the conversion of reacts into products in short period of time.

Question. Assertion (A): Dust particles suspended in the air inside grain elevators (unheated) can sometimes react explosively.
Reason (R): The dust particles have large surface area for the reaction.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Explanation: Dust particles of large surface area released from wheat, millet, barley are suspended in air. Then, in gram elevator, due to friction between particles with elevator in the presence of oxygen, creates spark in fuel (dust). Thus, explosion takes place due to increase in temperature of particles and elevator created by friction.

Question. Assertion (A): Elementary reactions have same value of order and molecularity.
Reason (R): Molecularity is the number of molecules that participate in the reaction, while order is an experimental quantity.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Explanation: An elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state. Molecularity is the number of molecules that participate in the reaction and it is calculated as every single step in a reaction whether it is elementary or complex while order is an experimental quantity Thus, in elementary reaction both molecularity and order are same.

Question. Assertion (A): Hydrolysis of an ester follows first order kinetics.
Reason (R): Concentration of water remains nearly constant during the course of the reaction.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
Explanation:
\( \text{CH}_3\text{COOC}_2\text{H}_5(\text{ester}) + \text{H}_2\text{O}(\text{excess}) \rightarrow \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \)
Hydrolysis of an ester follows first order kinetics as [H\(_2\)O] remains nearly constant during the course of the reaction. It is pseudo first order reaction.

Question. Assertion (A): For complex reactions molecularity and order are not same.
Reason (R): Order of reaction may be zero.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Explanation: For a complex reaction,
Order of overall reaction is experimental quantity that is equal to the molecularity of slowest step of elementary reaction.
As rate of overall reaction depends upon total number of molecules involved in slowest step of the reaction. Hence, for complex reaction, molecularity and order are not same. Order of reaction may be any integer, or rational number where as molecularity can be a whole numbers 1, 2, 3.

Question. Assertion (A): Order of the reaction can be zero or fractional.
Reason (R): We cannot determine order from balanced chemical equation.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Explanation: Order of a reaction may be zero or fractional. It can be determined through the rate law expression by sum of power of reactants as order of a reaction. It is an experimental quantity which cannot be analyzed by balanced chemical equation.

Question. Assertion (A): For a first order reaction, half-life period is independent of initial concentration of the reacting species.
Reason (R): The half-life of a reaction is the time in which the reactant concentration is reduced to one half of its initial concentration.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Explanation: For a first order reaction, \( t_{1/2} = \ln2/k \)
where half-life period is independent of initial concentration of the reacting species and is calculated from the rate constant.

Case-based MCQs

I. Read the passage given below and answer the following questions:
The rate of a reaction, which may also be called its velocity or speed, can be defined with relation to the concentration of any of the reacting substances, or to that of any product of the reaction. If the species chosen is a reactant which has a concentration \( c \) at time \( t \) the rate is - \( dc/dt \), while the rate with reference to a product having a concentration \( x \) at time \( t \) is \( dx/dt \). Any concentration units may be used for expressing the rate; thus, if moles per liter are employed for concentration and seconds for the time, the units for the rate are moles litre\(^{-1}\)sec\(^{-1}\). For gas reactions pressure units are sometimes used in place of concentrations, so that legitimate units for the rate would be (mm of Hg) sec\(^{-1}\) and atm. sec\(^{-1}\).
The order of a reaction concerns the dependence of the rate upon the concentrations of reacting substances; thus, if the rate is found experimentally to be proportional to the \( \alpha \)\(^\text{th}\) power of the concentration of one of the reactants A, to the \( \beta \)\(^\text{th}\) power of the concentration of a second reactant B, and so forth, via.,
rate = \( k C_A^\alpha C_B^\beta \)
the overall order of the reaction is simply
\( n = \alpha + \beta \)
Such a reaction is said to be of the \( \alpha \)\(^\text{th}\) order with respect to the substance A, the \( \beta \)\(^\text{th}\) order with respect to B.

In the following questions, a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices on the basis of the above passage.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

Question. Assertion: Rate of reaction is a measure of change in concentration of reactant with respect to time.
Reason: Rate of reaction is a measure of change in concentration of product with respect to time.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Explanation: Rate of a reaction is the rate of change of concentration of reactant or product with time.

Question. Assertion: For a chemical reaction, P + 2Q → Products, Rate = \( k[P]^{1/2}[Q]^1 \) so the order of reaction is 1.5
Reason: Order of reaction is the sum of stoichiometric coefficients of the reactants.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
Answer: (c) Assertion is correct statement but reason is wrong statement.
Explanation: For the given chemical reaction, order is the sum of powers of the reactants which is calculated experimentally i.e., \( 1/2 + 1 = 1.5 \)
In an elementary reaction, order can be the sum of stoichiometric coefficients of the reactants but the given reaction is a complex reaction.

Question. Assertion: The unit of \( k \) is independent of order of reaction.
Reason: For zero order reaction, the unit of \( k \) is mol L\(^{-1}\)s\(^{-1}\).
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
Answer: (d) Assertion is wrong statement but reason is correct statement.
Explanation: For a general reaction
\( aA + bB \rightarrow cC + dD \)
Rate = \( k[A]^x \times [B]^y \)
Where \( x + y = n \) = order of the reaction
Unit of \( k = [\text{mol L}^{-1}]^{1-n}\text{ s}^{-1} \)
Thus, \( k \) is dependent on the concentration of the reactants so, unit of \( k \) for different orders have different unit. For zero order, \( n = 0 \), thus unit of \( k \) is mol L\(^{-1}\)s\(^{-1}\).

Question. Assertion: Reactions can occur at different speeds.
Reason: Rate of reaction is also called speed of reaction.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Explanation: Rate of reaction is the speed with which a reaction takes place which can occur at different speeds. The frequency of collisions such that, more often molecules collide with each other, the faster the reaction proceeds and the energy of collisions where more forcefully molecules collide with each other, the more likely they are to react and the faster the reaction proceeds.

II. Read the passage given below and answer the following questions:
The rate of the reaction is proportional to the concentration of the reactant. Hydrogenation of ethene results in the formation of ethane. The rate constant, k for the reaction was found to be \( 2.5 \times 10^{-15}\text{s}^{-1} \). The concentration of the reactant reduces to one-third of the initial concentration in 5 minutes.
The following questions are multiple choice questions. Choose the most appropriate answer:

Question. Find the order of reaction.
(a) Zero order
(b) First order
(c) Second order
(d) Fractional order
Answer: (b) First order
Explanation: Since the rate of the reaction is proportional to the concentration for the reactant i.e., ethene so, it is first order reaction.

Question. The rate law equation is:
(a) Rate = \( k[\text{C}_2\text{H}_6] \)
(b) Rate = \( k[\text{C}_2\text{H}_4]^2 \)
(c) Rate = \( k[\text{C}_2\text{H}_4] \)
(d) Rate = \( k[\text{C}_2\text{H}_4]^2 \)
Answer: (c) Rate = \( k[\text{C}_2\text{H}_4] \)
Explanation: \( \text{C}_2\text{H}_4 + \text{H}_2 \xrightarrow{\text{Catalyst}} \text{C}_2\text{H}_6 \)
Ethene → Ethane
Here, hydrogen is taken in excess so the reaction is first order reaction.
Rate law equation, Rate = \( k[\text{C}_2\text{H}_4] \)

Question. The half-life for the reaction is
(a) \( 2.772 \times 10^{-24}\text{ s} \)
(b) \( 2.772 \times 10^{-12}\text{ s} \)
(c) \( 1.386 \times 10^{-24}\text{ s} \)
(d) \( 1.386 \times 10^{-12}\text{ s} \)
Answer: (a) \( 2.772 \times 10^{-24}\text{ s} \)
Explanation: For first order reaction,
\( t_{1/2} = \frac{0.693}{k} \)
\( = \frac{0.693}{2.5 \times 10^{-15}\text{ s}^{-1}} \)
\( = 2.772 \times 10^{-24}\text{ s} \)

Question. The rate constant of the reaction after 5 minutes is
(a) \( 0.4290\text{ min}^{-1} \)
(b) \( 0.1297\text{ min}^{-1} \)
(c) \( 0.2197\text{ min}^{-1} \)
(d) \( 0.6591\text{ min}^{-1} \)
Answer: (c) \( 0.2197\text{ min}^{-1} \)
Explanation: t = 5 min
\( \frac{[R]_0}{[R]} = 3 \)
For first order reaction,
\( k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \)
\( = \frac{2.303}{5} \log 3 \)
\( = \frac{2.303}{5} \times 0.4771 \)
\( = 0.2197\text{ min}^{-1} \)

Question. OR
The slope of the curve in the reaction is
(a) \( k \)
(b) \( -k \)
(c) \( 2k \)
(d) \( -2k \)
Answer: (b) \( -k \)
Explanation: Slope = −k

III. Read the passage given below and answer the following questions:
Concentration dependence of rate is called differential rate equation. Integrated differential equations give relation between directly measured experimental data i.e. concentration at different times and rate constant. The integrated rate equations are different for the reactions of different reaction orders. The first-order reaction has a rate constant \( 1.15 \times 10^{-3}\text{s}^{-1} \).
The following questions are multiple choice questions. Choose the most appropriate answer:

Question. How long will 5 g of this reactant take to reduce to 3 g?
(a) 222.189 s
(b) 444.379 s
(c) 111.095 s
(d) 888.789 s
Answer: (b) 444.379 s
Explanation: Initial amount= 5 g
Final concentration = 3 g
Rate constant= \( 1.15 \times 10^{-3}\text{ s}^{-1} \)
We know that for a First order reaction
\( t = \frac{2.303}{k} \log \frac{[R_\text{initial}]}{[R_\text{final}]} \)
\( = \frac{2.303}{1.15 \times 10^{-3}} \log \frac{5}{3} \)
\( = \frac{2.303}{1.15 \times 10^{-3}} \times 0.2219 \)
\( = 444.379\text{ s} \)

Question. When the rate constant has same units as the rate of reaction, the order of the reaction is:
(a) Zero order
(b) First order
(c) Second order
(d) Fractional order
Answer: (a) Zero order
Explanation: In general, mathematical expression for rate constant is as follows,
Unit of \( k = [\text{mol L}^{-1}]^{1-n}\text{ s}^{-1} \)
Thus, \( n = 0 \), \( k = \text{mol L}^{-1}\text{s}^{-1} \) the rate constant has same units as the rate of reaction.
\( r = k[A]^0 \)
\( r = k \)
Unit of rate = mol L\(^{-1}\)s\(^{-1}\)
Unit of \( k \) = mol L\(^{-1}\)s\(^{-1}\)

Question. Under which condition a bimolecular reaction is kinetically first order reaction?
(a) When two reactants are involved.
(b) When one of the reactants is in excess.
(c) When one of the reactants does not involve in reaction.
(d) None of the options
Answer: (b) When one of the reactants is in excess.
Explanation: For a reaction, A + B → P, if B is in excess, then order of a reaction will be 1.
Rate = \( k[A][B] \)
If B is in excess then, R = \( k'[A] \).
Thus, a bimolecular reaction is kinetically first order reaction.

Question. For a reaction, A + H\(_2\)O → B
Rate ∝ [A]
The order of the reaction is:
(a) Zero order
(b) Fractional order
(c) Pseudo first order
(d) Second order
Answer: (c) Pseudo first order
Explanation: Slope = A + H\(_2\)O → B
r ∝ [A] \( (\because \text{[H}_2\text{O]} = \text{excess}) \)
It is called pseudo first order reaction.

Question. OR
The integrated rate equation for rate constant of a first order reaction is:
(a) \( k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \)
(b) \( k = \frac{1}{t} \left[ \frac{1}{[R]} - \frac{1}{[R]_0} \right] \)
(c) \( k = \frac{1}{2t} \left[ \frac{1}{[R]^2} - \frac{1}{[R]_0^2} \right] \)
(d) None of the options
Answer: (a) \( k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \)
Explanation: A → products
For first order reaction, rate depends on the concentration directly, such that \( r = k[R] \).
Then, Integrated rate equation is,
\( k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \)