Chemistry

Chemical Kinetics

256 Questions

Chemical kinetics involves the study of chemical reaction rates and the factors affecting them, such as temperature and concentration. This topic covers rate laws, half-life, and zero, first, and second order reactions. It is a crucial part of the chemistry syllabus for various competitive examinations.

Reaction rate parametersFirst order kineticsZero and second orderHalf-life of reactionRate constant units

Chemical Kinetics Questions

Multiple choice

The first-order reaction rate constant is =

  1. 2.303/t

  2. 2.303/t log ([A0]/ [A] )

  3. 2.303/t log

  4. 2.303 / k log ([A0]/ [A] )

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The integrated rate law for a first-order reaction is k = (2.303/t) * log([A0]/[A]).

Multiple choice some important compounds of magnesium and calcium the s-block elements chemistry

Reactant of the superphosphate production are stirred for ___________ minutes and then into one of the dens through the valve.

  1. 5-10 mins

  2. 10-15 mins

  3. 2-5 mins

  4. 20-30 mins
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A well powdered phosphate rock or bone ash mixed known quantity of concentrate sulphuric acid is introduced into a cast iron chamber provided with mechanical stirrer and two valves at bottom. Each valve opens in a big chamber called den. The mixture is stirred for 2 to 5 minutes and then  dumped mechanically into one of the dens through the valve. The mass is kept in the den for about 24 hours.

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

The rate of disappearance of A at two temperatures is given by $A\rightleftharpoons B$
i. $\frac {-d[A]}{dt}=2\times 10^{-2}[A]-4\times 10^{-3}[B]$ at 300 K
ii. $\frac {-d[A]}{dt}=4\times 10^{-2}[A]-16\times 10^{-4}[B]$ at 300 K
From the given values of heat of reaction which are incorrect

  1. $3.86 kcal$
  2. $6.93 kcal$
  3. $1.68 kcal$
  4. $1.68\times 10^{-2} kcal$
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
Explanation

$K _1=\frac {K _f}{K _b}=\frac {2\times 10^{-2}}{4\times 10^{-3}}=5$ at 300 K
$K _2=\frac {K _f}{K _b}=\frac {4\times 10^{-2}}{16\times 10^{-4}}=25$ at 400 K
$\therefore 2.303 log\frac {25}{55}=\frac {\Delta H}{2}\times \left [\frac {400-300}{400\times 300}\right ]$
or $\Delta H=3.85 kcal$

Hence, option A is correct and others are incorrect

Multiple choice

The rate of a chemical reaction is typically expressed in terms of:

  1. Concentration change over time

  2. Temperature change over time

  3. Pressure change over time

  4. Volume change over time
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The rate of a chemical reaction is usually measured by monitoring the change in concentration of reactants or products over time.

Multiple choice

A zero-order reaction is characterized by:

  1. A rate that is independent of reactant concentration

  2. A rate that is proportional to reactant concentration

  3. A rate that is inversely proportional to reactant concentration

  4. A rate that is proportional to the square of reactant concentration
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a zero-order reaction, the rate is constant and does not depend on the concentration of the reactants.

Multiple choice

The rate law for a first-order reaction is given by:

  1. Rate = k[A]^0

  2. Rate = k[A]^1

  3. Rate = k[A]^2

  4. Rate = k[A]^3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a first-order reaction, the rate is directly proportional to the concentration of the reactant raised to the power of 1.

Multiple choice

The half-life of a reaction is the time it takes for:

  1. The concentration of reactants to decrease by half

  2. The concentration of products to increase by half

  3. The rate of the reaction to decrease by half

  4. The temperature of the reaction to increase by half
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The half-life of a reaction is the time required for the concentration of the reactants to decrease to half of its initial value.

Multiple choice

The Arrhenius equation relates the rate constant (k) of a reaction to:

  1. Temperature (T)

  2. Activation energy (Ea)

  3. Concentration of reactants

  4. Order of the reaction
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

The Arrhenius equation shows the exponential relationship between the rate constant and the temperature and activation energy of the reaction.

Multiple choice

The rate-determining step of a reaction is:

  1. The slowest step in the reaction mechanism

  2. The step with the highest activation energy

  3. The step that produces the final product

  4. The step that consumes the reactants
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The rate-determining step is the slowest step in the reaction mechanism and determines the overall rate of the reaction.

Multiple choice

A population of bacteria grows according to the differential equation (\frac{dN}{dt} = kN), where (N) is the population size and (k) is a constant. If the initial population size is (N_0), what is the population size at time (t)?

  1. \(N(t) = N_0e^{kt}\)
  2. \(N(t) = N_0e^{-kt}\)
  3. \(N(t) = N_0 + kt\)
  4. \(N(t) = N_0 - kt\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The solution to the differential equation is (N(t) = N_0e^{kt}).

Multiple choice

In a chemical reaction, the rate of change of the concentration of a reactant (A) is given by the differential equation (\frac{d[A]}{dt} = -k[A]^2), where (k) is a constant. What is the order of the reaction?

  1. First order

  2. Second order

  3. Third order

  4. Fourth order
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The order of the reaction is determined by the exponent of the concentration term in the differential equation. In this case, the exponent is 2, so the reaction is second order.

Multiple choice

A certain drug is administered to a patient at a constant rate of $r$ milligrams per hour. The drug is eliminated from the body at a rate proportional to the amount of drug in the body. If the initial amount of drug in the body is $Q_0$ milligrams, what is the amount of drug in the body at time $t$?

  1. $Q(t) = Q_0 + rt$
  2. $Q(t) = Q_0 e^{-kt}$
  3. $Q(t) = \frac{Q_0}{1 + e^{-kt}}$
  4. $Q(t) = \frac{Q_0}{1 - e^{-kt}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The amount of drug in the body at time $t$ is given by the differential equation $\frac{dQ}{dt} = r - kQ$, where $k$ is the elimination rate constant. The general solution to this differential equation is $Q(t) = \frac{Q_0}{1 - e^{-kt}}$.

Multiple choice

A population of bacteria grows at a rate proportional to the number of bacteria present. If the initial population is $P_0$ and the population doubles in $T$ hours, what is the population at time $t$?

  1. $P(t) = P_0 e^{kt}$
  2. $P(t) = P_0 (1 + kt)$
  3. $P(t) = P_0 (1 - kt)$
  4. $P(t) = P_0 2^{t/T}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The population of bacteria at time $t$ is given by the differential equation $\frac{dP}{dt} = kP$, where $k$ is the growth rate constant. The general solution to this differential equation is $P(t) = P_0 e^{kt}$. Since the population doubles in $T$ hours, we have $P(T) = 2P_0$. Substituting this into the general solution, we get $2P_0 = P_0 e^{kT}$, which implies that $k = \frac{\ln 2}{T}$. Therefore, the population at time $t$ is $P(t) = P_0 e^{\frac{\ln 2}{T} t} = P_0 2^{t/T}$.