JEE Mains and Advanced

Numerical 1: At 298 K, the rate constant for a first-order reaction is 2.5 x 10^-3 s^-1. What is the half-life of the reaction? (a) 1.0 x 10^-3 s (b) 2.8 x 10^-3 s (c) 182.3 s (d) 277.8 s Answer: (c) 182.3 s

Explanation: The half-life of a first-order reaction is given by the equation:

$$t_{1/2} = \frac{0.693}{k}$$

Substituting the given rate constant, we get: $$t_{1/2} = \frac{0.693}{2.5 \times 10^{-3} \text{ s}^{-1}} = 277.2\text{ s}$$ Therefore, the half-life of the reaction is 277.2 s, which is closest to option (d).

Numerical 2: The rate constant for the decomposition of N2O5 at 25°C is 4.0 x 10^-5 s^-1. What is the rate of decomposition of N2O5 if the initial concentration is 0.020 M? (a) 8.0 x 10^-6 M s^-1 (b) 4.0 x 10^-5 M s^-1 (c) 2.0 x 10^-4 M s^-1 (d) 8.0 x 10^-8 M s^-1 Answer: (a) 8.0 x 10^-6 M s^-1

Explanation: The rate of a first-order reaction is given by the equation: $$ \text{Rate} = -\frac{d[\text{N}_2\text{O}_5]}{dt} = k[\text{N}_2\text{O}_5] $$

Substituting the given rate constant and initial concentration, we get: $$ \text{Rate} = -4.0 \times 10^{-5} \text{ s}^{-1} \times 0.020 \text{ M} = 8.0 \times 10^{-6} \text{ M s}^{-1} $$ Therefore, the rate of decomposition of N2O5 is 8.0 x 10^-6 M s^-1.

CBSE Board Exams

Numerical 1: The activation energy of a reaction is 100 kJ mol^-1. At what temperature will the reaction be twice as fast as it is at 25°C? Answer: 333 K

Explanation: We can use the Arrhenius equation to solve this problem:

$$ k = Ae^{-\frac{E_a}{RT}}$$

where:

• k is the rate constant
• A is the pre-exponential factor
• Ea is the activation energy
• R is the ideal gas constant
• T is the temperature

We want to find the temperature at which the reaction is twice as fast as it is at 25°C, so we can set the ratio of the rate constants at the two temperatures equal to 2:

$$\frac{k_2}{k_1} = 2$$

Substituting the Arrhenius equation into this expression, we get: $$\frac{Ae^{-\frac{E_a}{RT_2}}}{Ae^{-\frac{E_a}{RT_1}}} = 2$$

Simplifying, we get: $$\frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}} = 2$$ Taking the natural logarithm of both sides, we get: $$-\frac{E_a}{RT_1} + \frac{E_a}{RT_2} = \ln 2$$

Rearranging, we get: $$\frac{E_a}{RT_2} = \frac{E_a}{RT_1} - \ln 2$$

Substituting the given values, we get: $$\frac{100000 \text{ J mol}^{-1}}{8.314 \text{ J mol}^{-1} \text{ K}^{-1} T_2} = \frac{100000 \text{ J mol}^{-1}}{8.314 \text{ J mol}^{-1} \text{ K}^{-1} (298 \text{ K})} - \ln 2$$

Solving for T2, we get: $$T_2 = 333 \text{ K}$$ Therefore, the reaction will be twice as fast at 333 K compared to 25°C.

Numerical 2: A reaction is first order with respect to both A and B. The rate constant for the reaction is 2.0 x 10^-3 L mol^-1 s^-1. What is the rate of the reaction if the initial concentrations of A and B are both 0.10 M? Answer: 2.0 x 10^-5 M s^-1

Explanation: The rate law for a first-order reaction with respect to both A and B is: $$ \text{Rate} = k[\text{A}][\text{B}] $$

Substituting the given rate constant and initial concentrations, we get: $$ \text{Rate} = 2.0 \times 10^{-3} \text{ L mol}^{-1} \text{ s}^{-1} \times 0.10 \text{ M} \times 0.10 \text{ M} = 2.0 \times 10^{-5} \text{ M s}^{-1} $$ Therefore, the rate of the reaction is 2.0 x 10^-5 M s^-1.