NEET Previous Year Questions (2014-2025): Chemical Kinetics

Ans: (c)
Substitute the values into the formula:
t = - (1 / 0.03) × ln(0.9 / 7.2)

Take the natural logarithm (ln):
ln(0.125) = -2.079

Substitute into the equation:
t = - (1 / 0.03) × (-2.079)

Simplify:
t = (1 / 0.03) × 2.079
t = 69.3 s

Therefore, the time taken for the concentration to reduce from 7.2 mol L^-1 to 0.9 mol L^-1 is 69.3 seconds.

Ans: (c)

• In the given reaction:C(s) + 2H₂(g) → CH₄(g); ΔH = -74.8 kJ mol^-1
• ΔH is negative, indicating that the reaction is exothermic.
• In an energy diagram for this reaction:
  • The reactants (C and H₂) start at a higher energy level.
  • The products (CH₄) are at a lower energy level, showing that energy is released during the reaction.
  • The difference in energy between the reactants and products corresponds to the magnitude of ΔH (74.8 kJ mol^-1).

Therefore, the correct energy diagram will show the reactants at a higher energy level than the products, with an energy drop of 74.8 kJ mol^-1.

Ans: (b)

From the formula for half-life:

t₁/₂ = 0.693 / k

• k = 0.693 / t₁/₂
• k = 0.693 / 1
• k = 0.693 min⁻¹

The time required for 99.9% completion can be calculated using the formula:
t = (2.303 / k) x log(1 / (1 - fraction completed))

• t = (2.303 / 0.693) x log(1 / (1 - 0.999))
• t = (2.303 / 0.693) x log(1 / 0.001)
• t = (2.303 / 0.693) x log(1000)
• t = (2.303 / 0.693) x 3
• t ≈ 10 minutes

Therefore, the time required for 99.9% completion of the reaction is approximately 10 minutes.

Ans: (d)
To calculate the value of E_a
The equation used is

Hence, E_a can be calculated if the value of the rate constant k is known at two different temperatures, T₁ and T₂.

Ans: (d)
The Arrhenius equation is given as

In⁡ k vs 1 / T gives a straight line graph with slope =  and intercept = InA

Ans: (a)
To solve this problem, we will use the Arrhenius equation, which relates the rate constant (k) of a chemical reaction to the temperature (T) and the activation energy (E_a). The equation in its logarithmic form is:

where:

• K is the rate constant,
• A is the pre-exponential factor (frequency factor),
• E_a is the activation energy,
• R is the universal gas constant
• T is the temperature in Kelvin,
• $2.303$ is the factor to convert from natural log to common log.

According to the problem, the rate of the reaction quadruples (k₂$=$4k₁) when the temperature increases from $27$^∘C (which equals $27+273.15=300.15$K) to $57$^∘C (which equals $57+273.15=330.15$K). Using the logarithmic form of the Arrhenius equation, for two different temperatures we have:


So,

Therefore, the activation energy Ea is 38.07kJ/mol, which best matches Option A:
38.04kJ/mol

Ans: (a)
The given reaction is:
2HI_(g) → H₂_(g) + I₂_(g)
The stoichiometric coefficients indicate the ratio of the change in concentration of each substance involved in the reaction.

• For every 2 moles of HI that are consumed, 1 mole of H₂ and 1 mole of I₂ are produced.

This gives us the following relationship for the rate of change of concentration:

• Δ[HI] is the concentration change of HI (reactant).
• Δ[H₂] is the concentration change of H₂ (product).

From the stoichiometry of the balanced equation, we know:

• For every 2 moles of HI consumed, 1 mole of H₂ is produced. Therefore, the rate of consumption of HI is twice the rate of formation of H₂.

Thus, the correct rate expression is: -Δ[HI] / Δt = 2Δ[H₂] / Δt
This matches with option (a).

Ans: (b)
Effective collisions are those that lead to a chemical reaction. For a collision to be effective, the following conditions must be met:

A: Energy greater than threshold energy: The colliding particles must have energy greater than the activation energy threshold for the reaction to occur.

B: Breaking of old bond in reactant: During an effective collision, bonds in the reactant molecules must break to form products.

C: Formation of new bond in product: New bonds must form in the product molecules as a result of the effective collision.

E: Proper orientation: The molecules involved in the collision must be oriented in such a way that the reacting atoms or groups can interact effectively.

D: High activation energy is not necessary for an effective collision. It refers to the energy barrier the reactants need to overcome, but the effective collision occurs when the energy exceeds the threshold energy, not necessarily when the activation energy is high.
Thus, the correct answer is Option B: A, B, C, E only.

Ans: (a)

For a first-order reaction, the rate constant (k) is related to the time taken to decompose a certain percentage of reactant using the following equation:
ln ([A]₀/[A]) = kt
Where:

• [A]₀ is the initial concentration,
• [A] is the concentration after time t,
• k is the rate constant,
• t is the time.

For first-order reactions, the equation becomes:

Given that 40% of SO₂Cl₂ decomposes, this means that 60% remains after 560 seconds.
So, [A] = 0.60[A]₀ and [A]₀/[A] = 1/0.60 = 1.6667.
Now, using the formula:
log (1.6667) = 0.2219 (logarithmic calculation).

Using this in the first-order equation:
log (1.6667) = kt / 2.303
0.2219 = k * 560 / 2.303

Solving for k:
k = (0.2219 * 2.303) / 560
k ≈ 0.0009127 s⁻¹

Convert to min⁻¹:
k ≈ 0.0009127 × 60 ≈ 0.05476 min⁻¹ = 5.476 × 10⁻² min⁻¹

Thus, the rate constant is 5.476 × 10⁻² min⁻¹, so the correct answer is (a).

Ans: (c)
We can determine the order of the reaction with respect to A and B using the given data. The rate equation for a reaction is of the form:
Rate = k[A]^m[B]ⁿ
Where:

• m is the order with respect to A,
• n is the order with respect to B,
• k is the rate constant.

We can calculate the orders with respect to A and B by comparing two experiments where one concentration is constant and the other changes.
Step 1: Find the order with respect to A (m):

• Compare experiments 1 and 2, where the concentration of B is constant (0.1 M) and the concentration of A changes from 0.1 M to 0.2 M.
• Rate 1 = 2 × 10⁻³, Rate 2 = 4 × 10⁻³.
• The ratio of the rates is:
• (Rate 2 / Rate 1) = (4 × 10⁻³) / (2 × 10⁻³) = 2.
• The ratio of the concentrations of A is:
• ([A]₂ / [A]₁) = (0.2 / 0.1) = 2.
• Since the rate doubles when the concentration of A doubles, the order with respect to A is 1.

Step 2: Find the order with respect to B (n):

• Compare experiments 2 and 3, where the concentration of A is constant (0.2 M) and the concentration of B changes from 0.1 M to 0.2 M.
• Rate 2 = 4 × 10⁻³, Rate 3 = 1.6 × 10⁻².
• The ratio of the rates is:
• (Rate 3 / Rate 2) = (1.6 × 10⁻²) / (4 × 10⁻³) = 4.
• The ratio of the concentrations of B is:
• ([B]₃ / [B]₂) = (0.2 / 0.1) = 2.
• The rate quadruples when the concentration of B doubles, which suggests the order with respect to B is 2.

Thus, the order of the reaction is 1 with respect to A and 2 with respect to B, so the correct answer is (c) 1, 2.

Ans: (c)
According to the Arrhenius equation:

ln k = -E_a/RT + ln A

Where:

• k is the rate constant,
• T is the temperature,
• E_a is the activation energy,
• R is the universal gas constant,
• A is the pre-exponential factor.

This equation shows that a plot of ln k versus 1/T should give a straight line with a slope of -Ea/R.

• The plot will have a negative slope because E_a is positive (activation energy is always positive).
• Therefore, the plot should be a downward-sloping line, which corresponds to option (c).

Thus, (c) is the correct plot that represents the variation of ln k with 1/T according to the Arrhenius equation.

Ans: (c)
We can use the Arrhenius equation to calculate the activation energy (E_a) of the reaction:
ln(k₂ / k₁) = (E_a / R) * (1/T₁ - 1/T₂)
Where:

• k₁ = 0.04 s⁻¹ (rate constant at T₁ = 500 K),
• k₂ = 0.14 s⁻¹ (rate constant at T₂ = 700 K),
• R = 8.31 J K⁻¹ mol⁻¹ (gas constant),
• T₁ = 500 K,
• T₂ = 700 K.

Now, substitute the known values into the equation:
ln(0.14 / 0.04) = (E_a/ 8.31) * (1/500 - 1/700)
First, calculate the ratio:
ln(0.14 / 0.04) = ln(3.5) = 0.5441
Now calculate the temperature difference:
(1/500 - 1/700) = (700 - 500) / (500 * 700) = 200 / 350000 = 0.0005714 K⁻¹
Now, substitute into the equation:
0.5441 = (E_a / 8.31) * 0.0005714
Solve for E_a:
E_a = (0.5441 * 8.31) / 0.0005714
E_a ≈ 18219 J
Therefore, the activation energy E_a is approximately 18219 J, which corresponds to option (c).

Ans: C

Ans: (b)

Step 1: Assertion (A)

The statement says "A reaction can have zero activation energy." This is correct. Certain reactions, such as free-radical reactions, can proceed without any extra energy barrier.

Step 2: Reason (R)

The reason given is "The minimum extra amount of energy required by reactant molecules so that their energy becomes equal to the threshold value is called activation energy." This is the correct definition of activation energy.

Step 3: Relationship Between A and R

Although both the assertion and the reason are true, the reason does not actually justify the assertion. Knowing the definition of activation energy does not explain why in some cases, activation energy can be zero.

Therefore, Option (b) is the correct answer.

Ans: (d)
To understand this, let's first define what a first-order reaction means. The overall order of a reaction is the sum of the powers of the concentration terms in the rate law equation. In an overall first-order reaction, the sum of the powers should add up to 1.
Now, let's analyze the given options:

• (a) Rate = k[A]⁰[B]²: This would result in an overall order of 2, because the order of [A] is 0 and the order of [B] is 2. So, this is not a first-order reaction.
• (b) Rate = k[A][B]: This has a total order of 1 (1 for [A] and 1 for [B]), which is a second-order reaction in total, not first-order.
• (c) Rate = k[A]¹/₂[B]²: This is a mixed-order reaction with fractional exponents. The overall order is 1/2 + 2 = 2.5, which is not first order.
• (d) Rate = k[A]⁻¹/₂[B]³/₂: This gives an overall order of -1/2 + 3/2 = 1, which is first-order. Therefore, this is the correct rate law for a first-order reaction.

Thus, the correct answer is (d) Rate = k[A]⁻¹/₂[B]³/₂.

Ans: (c)
Given the reaction:
3A → 2B

The stoichiometry of the reaction shows that for every 3 moles of A that are consumed, 2 moles of B are produced.
Now, we can write the relationships between the rates of disappearance of A and appearance of B.

• The average rate of disappearance of A is given by:
  Rate of disappearance of A = -(Δ[A] / Δt)
• The average rate of appearance of B is given by:
  Rate of appearance of B = Δ[B] / Δt

From the stoichiometric ratio, we know that the change in the concentration of B is related to the change in the concentration of A by the ratio 2/3. This means for every 3 moles of A that disappear, 2 moles of B appear.
Therefore, the relation between the rate of disappearance of A and the rate of appearance of B is:
Δ[B] / Δt = (2/3) * (-Δ[A] / Δt)
Thus, the correct relation is: (c) - (2 Δ[A]) / (3 Δt).

Ans: (a)

• For zero order reaction
  r = k[A]0
  r = k (constant)
  hence, 'y' as 'rate' and 'x' as concentration will give desired graph.
• For first order reaction
  

hence, 'y' as 't_1/2' and 'x' as concentration will given desired graph.

Ans: (d)
A → Product
t = 0   0.1 M
t = 5 min    0.001 M

Ans: (b)

Ans: (d)
The enthalpy of reaction is negative, - 4.2 kJ mol^-1 . The reaction is an exothermic reaction i.e. the energy of product B, is less than the energy of reactant A. So, the potential energy profile for the reaction is

Ans: (c)
Arrhenius equation

Ans: (c)
The correct answer to this question is collision frequency. Here's an explanation for each option to understand why:

• Option A - Heat of Reaction: The heat of reaction, also known as the enthalpy change (ΔH), is determined by the difference in the energy levels of the reactants and products. It is a characteristic feature of a particular chemical reaction and is not typically affected by the concentration of reactants except in cases of extremely high or non-ideal concentrations where volume changes might somewhat influence pressure and therefore the heat of reaction in gases. Under normal conditions, however, changing the concentration of reactants does not alter the heat of reaction.
• Option B - Threshold Energy: Threshold energy is the minimum energy that reactant molecules must possess in order to undergo a successful collision, leading to a chemical reaction. This value is inherent to the specific reaction and is related to the strength of the bonds in the reactants as well as the energy required to form the activated complex during the reaction. The threshold energy does not change simply because the concentration of reactants is altered.
• Option C - Collision Frequency: Collision frequency corresponds to how often reacting particles collide in a given time interval. When the concentration of reactants is increased, there are more reactant particles per unit volume. This statistically leads to an increased number of collisions per unit time, thereby raising the collision frequency. This is in accordance with the collision theory of chemical kinetics.
• Option D - Activation Energy: Activation energy is the minimum energy barrier that must be overcome for reactants to be converted into products. It is a property of a particular reaction and is related to the nature of the reactants and the reaction pathway. Altering the concentration of reactants does not change the activation energy for that reaction.

Thus, the only factor among the provided options that changes with an increase in the concentration of reactants is indeed the collision frequency.

Ans: (a)

Ans: (c)
The first-order rate constant is given as,

99% completed reaction,

Ans: (c)

The rate of reaction is given as

Ans: (b)
For the first order reaction, t_1/2 = $\frac{0.693}{}$/ k
which is independent of initial concentration [A]₀.
For second order reaction, t_1/2 =
which depends on initial concentration [A]₀.

Ans: (b)
Half life of zero order

Ans: (c)

Ans: (a)

Ans: (b)

Ans: (b)
A catalyst provides an alternate path to the reaction which has lower activation energy.

Ans: (a)
At low pressure, rate is proportional to the surface coverage and is of first order while at high pressure it follows zero order kinetic due to complete coverage of surface area.

Ans: (c)
For zero order reaction unit of rate constant is mole per second.
For zero order
x = kt
 x = 0.6 × 10^-3 × 20 × 60 = 0.72 M 

Ans: (d)

According to Arrhenius equation,

Taking natural log on both the sides we get,

Comparing (1) with standard form of equation of line
y = mx + C
We get Slope,
Hence, if ln k is plotted against 1/T, slope of the line will be .

Ans: (c)
Half-life period of a first order reaction is independent of initial concentration,