NEET Chemical Kinetics Previous Year Questions with Solutions

For the chemical reaction 2A → B, the rate of reaction is defined with respect to the disappearance of reactant A divided by its stoichiometric coefficient: Rate = - (1/2) (d[A]/dt)

• • Order = 0.
• • Rate constant k = 1.0 mol L⁻¹ min⁻¹.
• • -d[A]/dt = 2k.

Separate variables and integrate from time t = 0 to time t: - ∫_{[A]₀}^{[A]_t} d[A] = 2k ∫_0^t dt

• • Linear relationship between concentration consumed and time.

75% completion means 75% of the initial reactant has been consumed: [A]₀ - [A]_t = 0.75 × [A]₀ = 0.75 × 2.0 M = 1.5 M Substitute this into the integrated rate law: 1.5 M = 2 × (1.0 mol L⁻¹ min⁻¹) × t

• • Time required = 0.75 min.
• • Matches Option (2).

The Arrhenius equation describes how the rate constant depends on absolute temperature: k = A e^{-E_a / RT} Taking natural logarithm on both sides: ln k = ln A - (E_a / R)(1/T)

• • y = mx + c with y = ln k, x = 1/T, slope m = -E_a/R, and intercept c = ln A.
• • From the graph, the y-intercept is clearly 6 ⇒ ln A = 6.

We need the temperature at which the rate constant becomes k = e² min⁻¹.

• • Substitute ln k = 2 and ln A = 6 into the linear equation:
• • 2 = 6 - \frac{E_a}{RT} \implies \frac{E_a}{RT} = 4

Convert E_a to joules: E_a = 6.64 kJ mol⁻¹ = 6640 J mol⁻¹. Substitute R = 8.3 J K⁻¹ mol⁻¹.

• • Temperature = 200 K.
• • Matches Option (3).

For a first-order reaction with rate constant k, the time t required for reactant concentration to decrease from [R]₀ to [R]_t is:

• • For 99.9% completion, [R]_t = [R]₀ - 0.999 [R]₀ = 0.001 [R]₀ = 10⁻³ [R]₀.
• • Therefore: [R]₀ / [R]_t = 1 / 10⁻³ = 10³ = 1000.

Substitute k = 2.303 log(2) / t_{1/2} = 0.693 / t_{1/2} into the integrated rate equation:

• • This is a fundamental theorem of first-order chemical kinetics.

Substitute t_{1/2} = 1 minute:

• • Matches Option (4) exactly.

The rate constant k has units of s⁻¹, which uniquely characterizes a first-order chemical reaction:

• • Given [R]₀ = 7.2 mol L⁻¹ and [R]_t = 0.9 mol L⁻¹.
• • Ratio [R]₀ / [R]_t = 7.2 / 0.9 = 8 = 2³.

Substitute the given value log 2 = 0.301 into the integrated rate law:

• • Rounding to three significant digits yields 69.3 s.

Confirm via consecutive half-lives:

• • t_{1/2} = 0.693 / 0.03 = 23.1 s.
• • Total time = 3 × 23.1 s = 69.3 s.
• • Matches Option (1) exactly.

The Arrhenius equation describes how the rate constant k of a reaction depends exponentially on absolute temperature T and activation energy Ea:

Writing the equation at two distinct absolute temperatures T1 and T2 with rate constants k1 and k2:

Subtracting the first equation from the second eliminates the unknown frequency factor A entirely:

Because R is known, experimental determination of the rate constant at two distinct temperatures permits exact calculation of Ea without prior knowledge of orientation factors or pre-exponential constant A.

Taking the natural logarithm of both sides of the Arrhenius rate expression:

Aligning variables to identify graphical features: - Dependent variable: y = ln k - Independent variable: x = 1 / T - Slope: m = -Ea / R (strictly negative) - Y-intercept: c = ln A (positive finite value when extrapolated to 1/T = 0)

As absolute temperature T increases, 1/T decreases, and the reaction rate constant k increases (ln k rises). Conversely, moving along the positive x-axis (increasing 1/T, meaning cooling), ln k falls linearly. The graph must therefore be a straight line sloping downward from left to right, exactly as illustrated in Plot (2).

The relationship between rate constants k1, k2 at absolute temperatures T1, T2 is given by:

Given: - k2 / k1 = 4 ⟹ log10(4) = 0.6021 - T1 = 27°C = 300 K - T2 = 57°C = 330 K - R = 8.314 J K⁻¹ mol⁻¹ - 2.303 × R = 19.147 J K⁻¹ mol⁻¹

Cross-multiplying to isolate Ea:

Dividing by 1000 to convert Joules to Kilojoules:

According to the given rate law, the initial rate of reaction r₁ at initial reactant concentrations [A] and [B] is:

The initial concentration of A is tripled while [B] remains strictly constant:

Substituting the new concentrations into the rate equation gives the new initial rate r₂:

Comparing r₂ with r₁:

According to the Arrhenius equation, if activation energy were zero (Ea = 0), the rate constant would equal the frequency factor A and become temperature-independent:

For any chemical reaction involving breaking of chemical bonds and formation of new bonds, a finite energy barrier must be crossed. Therefore, a chemical reaction cannot have zero activation energy. Assertion A is FALSE:

Activation energy (Ea) is formally defined as the minimum excess energy that reactant molecules must absorb to acquire threshold energy (E_threshold) required for effective collisions. Reason R is TRUE:

For the general transformation $A \longrightarrow B$, the standard enthalpy of reaction is defined as the difference between the potential energy of the products and reactants: $$\Delta H = H_B - H_A$$ Given $\Delta H = -4.2\text{ kJ mol}^{-1} < 0$, heat is evolved to the surroundings, establishing that the process is exothermic.

Because $\Delta H < 0$: $$H_B - H_A < 0 \implies H_B < H_A$$ Thus, on the potential energy diagram against reaction coordinate, the baseline energy plateau for product $B$ must lie strictly below that of reactant $A$.

The forward activation enthalpy is $E_a = 9.6\text{ kJ mol}^{-1}$. Reactant $A$ must surmount an energy barrier of height $9.6\text{ kJ mol}^{-1}$ to reach the activated transition state ($A^{\ddagger}$) before descending to product $B$. Graph (2) correctly depicts this energy barrier with the product plateau $B$ lower than $A$.

The Arrhenius equation expresses the dependence of the rate constant $k$ on temperature $T$: $$k = A e^{-E_a / RT}$$ Taking the natural logarithm ($\ln$) of both sides:

Comparing this expression with the standard equation of a straight line $y = c + mx$, where $y = \ln k$ and $x = \frac{1}{T}$: - $y$-intercept $= \ln A$ - $\text{Slope } m = -\frac{E_a}{R}$

Given slope $= -5 \times 10^3\text{ K}$ and $R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$: $$-\frac{E_a}{R} = -5 \times 10^3\text{ K} \implies E_a = (5 \times 10^3\text{ K}) \times (8.314\text{ J K}^{-1}\text{mol}^{-1})$$

According to collision theory, the collision frequency ($$Z$$) is defined as the number of collisions per second per unit volume of the reacting mixture. For a bimolecular reaction $$A + B \to \text{Products}$$, the collision frequency is directly proportional to the product of molar concentrations: $$Z_{AB} \propto [A][B]$$.

When the concentration of reactants is increased, the number of reactant particles per unit volume increases. As a consequence, molecules collide more frequently with each other, leading to a direct increase in the collision frequency ($$Z$$).

Activation energy ($$E_a$$) is the minimum extra energy required by reactant molecules to reach the activated transition state. It is an intrinsic feature of the reaction mechanism and is completely independent of reactant concentration (changed only by a catalyst).

Heat of reaction ($$\Delta_r H$$) is the difference in enthalpy between products and reactants per mole. It is a thermodynamic property determined by chemical bond energies, not affected by initial concentration.

For a first order reaction $$R \to \text{Products}$$, the integrated rate equation in common logarithmic form is: $$k = \frac{2.303}{t} \log\left(\frac{[R]_0}{[R]}\right) \implies t = \frac{2.303}{k} \log\left(\frac{[R]_0}{[R]}\right)$$

Given: - Rate constant $$k = 4.606 \times 10^{-3}\text{ s}^{-1}$$ - Initial mass of reactant $$[R]_0 = 2.0\text{ g}$$ - Final mass of reactant remaining $$[R] = 0.2\text{ g}$$ (Notice that for a given volume, the ratio of masses is identical to the ratio of molar concentrations).

$$\frac{[R]_0}{[R]} = \frac{2.0}{0.2} = 10$$ $$\log(10) = 1$$

$$t = \frac{2.303}{4.606 \times 10^{-3}} \times 1$$ Notice that $$4.606 = 2 \times 2.303$$: $$t = \frac{1}{2 \times 10^{-3}} = \frac{1000}{2} = 500\text{ seconds}$$ This confirms Option (3).