CBSE Class 12 Chemistry Chemical Kinetics NCERT Solutions

The average rate of a reaction can be calculated using the formula:

\text{Average Rate} = -\frac{\Delta[R]}{\Delta t}

where \(\Delta[R]\) is the change in concentration of reactant R and \(\Delta t\) is the change in time.

Given:

\(\left[R\right]_1 = 0.03 \text{ M}\)

\(\left[R\right]_2 = 0.02 \text{ M}\)

\(\Delta t = 25 \text{ minutes}\)

First, calculate the rate in M min⁻¹:

\text{Average Rate} = -\frac{\left[R\right]_2 - \left[R\right]_1}{t_2 - t_1} = -\frac{0.02 \text{ M} - 0.03 \text{ M}}{25 \text{ min}}

= -\frac{-0.01 \text{ M}}{25 \text{ min}} = \frac{0.01}{25} \text{ M min}^{-1} = 4 \times 10^{-4} \text{ M min}^{-1}

Now, convert the rate to M s⁻¹ by dividing by 60 (since 1 minute = 60 seconds):

\text{Average Rate} = \frac{4 \times 10^{-4} \text{ M min}^{-1}}{60 \text{ s min}^{-1}}

= \frac{4}{60} \times 10^{-4} \text{ M s}^{-1} \approx 0.0667 \times 10^{-4} \text{ M s}^{-1} = 6.67 \times 10^{-6} \text{ M s}^{-1}

Thus, the average rate of the reaction is \(4 \times 10^{-4} \text{ M min}^{-1}\) or \(6.67 \times 10^{-6} \text{ M s}^{-1}\).

For a reaction where the stoichiometry is involved, the rate is expressed with respect to the change in concentration of reactants or products, divided by their stoichiometric coefficients. For the reaction \(2A \rightarrow Products\), the rate is given by:

\text{Rate} = -\frac{1}{2} \frac{\Delta[A]}{\Delta t}

Given:

\(\left[A\right]_1 = 0.5 \text{ mol L}^{-1}\)

\(\left[A\right]_2 = 0.4 \text{ mol L}^{-1}\)

\(\Delta t = 10 \text{ minutes}\)

Calculate the change in concentration:

\(\Delta[A] = \left[A\right]_2 - \left[A\right]_1 = 0.4 \text{ mol L}^{-1} - 0.5 \text{ mol L}^{-1} = -0.1 \text{ mol L}^{-1}\)

Now, substitute these values into the rate expression:

\text{Rate} = -\frac{1}{2} \frac{-0.1 \text{ mol L}^{-1}}{10 \text{ min}}

= -\frac{1}{2} (-0.01 \text{ mol L}^{-1} \text{ min}^{-1})

= 0.005 \text{ mol L}^{-1} \text{ min}^{-1}

This can also be expressed as \(5 \times 10^{-3} \text{ M min}^{-1}\).

The order of a reaction is determined by the sum of the exponents of the concentration terms in the rate law. The given rate law is:

r = k \left[ A \right]^{1/2} \left[ B \right]^2

The exponent for reactant A is 1/2, and the exponent for reactant B is 2.

Therefore, the overall order of the reaction is the sum of these exponents:

\text{Order} = \frac{1}{2} + 2

= 0.5 + 2 = 2.5

The order of the reaction is 2.5.

The problem states that the reaction \(X \rightarrow Y\) follows second-order kinetics. The rate law for a second-order reaction is generally expressed as:

\text{Rate} = k[X]^2

where \(k\) is the rate constant and \([X]\) is the concentration of reactant X.

Let the initial concentration of X be \([X]_1 = a\). The initial rate (\(Rate_1\)) is:

Rate_1 = k(a)^2

Now, the concentration of X is increased to three times its original value. So, the new concentration is \([X]_2 = 3a\).

The new rate (\(Rate_2\)) will be:

Rate_2 = k(3a)^2

Rate_2 = k(9a^2)

Rate_2 = 9 \times (ka^2)

Since \(Rate_1 = ka^2\), we can substitute this into the equation for \(Rate_2\):

Rate_2 = 9 \times Rate_1

Therefore, if the concentration of X is increased to three times, the rate of formation of Y will increase by a factor of 9.