CBSE Class 12 Chemistry Chapter 7: Chemical Kinetics NCERT Solutions

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

$$\backslash text\{Average\; Rate\}\; =\; -\backslash frac\{\backslash Delta[\backslash text\{Reactant\}]\}\{\backslash Delta\; t\}$$

Where $\backslash Delta[\backslash text\{Reactant\}]$ is the change in concentration of the reactant and $\backslash Delta\; t$ is the change in time.

Given:

$[R]\_1\; =\; 0.03\; \backslash text\{\; M\}$

$[R]\_2\; =\; 0.02\; \backslash text\{\; M\}$

$t\_1\; =\; 0\; \backslash text\{\; min\}$

$t\_2\; =\; 25\; \backslash text\{\; min\}$

Change in concentration, $\backslash Delta[R]\; =\; [R]\_2\; -\; [R]\_1\; =\; 0.02\; \backslash text\{\; M\}\; -\; 0.03\; \backslash text\{\; M\}\; =\; -0.01\; \backslash text\{\; M\}$

Change in time, $\backslash Delta\; t\; =\; t\_2\; -\; t\_1\; =\; 25\; \backslash text\{\; min\}\; -\; 0\; \backslash text\{\; min\}\; =\; 25\; \backslash text\{\; min\}$

Calculating the average rate in M min⁻¹:

$$\backslash text\{Average\; Rate\}\; =\; -\backslash frac\{-0.01\; \backslash text\{\; M\}\}\{25\; \backslash text\{\; min\}\}\; =\; \backslash frac\{0.01\}\{25\}\; \backslash text\{\; M\; min\}^\{-1\}\; =\; 4\; \backslash times\; 10^\{-4\}\; \backslash text\{\; M\; min\}^\{-1\}$$

Now, we convert the rate to M s⁻¹:

Since 1 minute = 60 seconds, $25\; \backslash text\{\; min\}\; =\; 25\; \backslash times\; 60\; \backslash text\{\; s\}\; =\; 1500\; \backslash text\{\; s\}$

$$\backslash text\{Average\; Rate\}\; =\; -\backslash frac\{-0.01\; \backslash text\{\; M\}\}\{1500\; \backslash text\{\; s\}\}\; =\; \backslash frac\{0.01\}\{1500\}\; \backslash text\{\; M\; s\}^\{-1\}\; =\; 6.67\; \backslash times\; 10^\{-6\}\; \backslash text\{\; M\; s\}^\{-1\}$$

Answer: The average rate of reaction is $4\; \backslash times\; 10^\{-4\}\; \backslash text\{\; M\; min\}^\{-1\}$ or $6.67\; \backslash times\; 10^\{-6\}\; \backslash text\{\; M\; s\}^\{-1\}$.

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.

The given reaction is $2A\; \backslash rightarrow\; Products$.

The rate of reaction can be expressed as:

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

Given:

$[A]\_1\; =\; 0.5\; \backslash text\{\; mol\; L\}^\{-1\}$

$[A]\_2\; =\; 0.4\; \backslash text\{\; mol\; L\}^\{-1\}$

$\backslash Delta\; t\; =\; 10\; \backslash text\{\; minutes\}$

Change in concentration, $\backslash Delta[A]\; =\; [A]\_2\; -\; [A]\_1\; =\; 0.4\; \backslash text\{\; mol\; L\}^\{-1\}\; -\; 0.5\; \backslash text\{\; mol\; L\}^\{-1\}\; =\; -0.1\; \backslash text\{\; mol\; L\}^\{-1\}$

Now, substitute these values into the rate expression:

$$\backslash text\{Rate\}\; =\; -\backslash frac\{1\}\{2\}\; \backslash frac\{-0.1\; \backslash text\{\; mol\; L\}^\{-1\}\}\{10\; \backslash text\{\; min\}\}\; =\; -\backslash frac\{1\}\{2\}\; \backslash times\; (-0.01\; \backslash text\{\; mol\; L\}^\{-1\}\; \backslash text\{\; min\}^\{-1\})$$

$$\backslash text\{Rate\}\; =\; 0.005\; \backslash text\{\; mol\; L\}^\{-1\}\; \backslash text\{\; min\}^\{-1\}$$

This can also be written as:

$\backslash text\{Rate\}\; =\; 5\; \backslash times\; 10^\{-3\}\; \backslash text\{\; M\; min\}^\{-1\}$

Answer: The rate of the reaction during this interval is $0.005\; \backslash text\{\; mol\; L\}^\{-1\}\; \backslash text\{\; min\}^\{-1\}$ or $5\; \backslash times\; 10^\{-3\}\; \backslash text\{\; M\; min\}^\{-1\}$.

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

The given rate law is $r\; =\; k\; \backslash left[\; A\; \backslash right]^\{1/2\}\; \backslash left[\; B\; \backslash right]^2$.

The exponent for reactant A is $1/2$.

The exponent for reactant B is $2$.

The overall order of the reaction is the sum of these exponents:

$$\backslash text\{Order\}\; =\; \backslash frac\{1\}\{2\}\; +\; 2\; =\; 0.5\; +\; 2\; =\; 2.5$$

Answer: The order of the reaction is 2.5.

The reaction is given as $X\; \backslash rightarrow\; Y$.

The problem states that this reaction follows second order kinetics.

The rate law for a second order reaction with respect to reactant X is:

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

Let the initial concentration of X be $[X]\_1$. The initial rate ($Rate\_1$) is:

$$Rate\_1\; =\; k[X]\_1^2$$

Now, the concentration of X is increased to three times its original value. Let the new concentration be $[X]\_2\; =\; 3[X]\_1$.

The new rate ($Rate\_2$) will be:

$$Rate\_2\; =\; k[X]\_2^2\; =\; k(3[X]\_1)^2$$

Expanding the term:

$$Rate\_2\; =\; k(9[X]\_1^2)\; =\; 9\; \backslash times\; (k[X]\_1^2)$$

Since $Rate\_1\; =\; k[X]\_1^2$, we can substitute this into the equation for $Rate\_2$:

$$Rate\_2\; =\; 9\; \backslash times\; Rate\_1$$

This shows that the new rate is 9 times the original rate.

Answer: If the concentration of X is increased to three times, the rate of formation of Y will increase by nine times.