CBSE Class 11 Chemistry Chapter 7 Equilibrium NCERT Solutions

1. When the volume of the container is suddenly increased, the concentration of the vapour molecules decreases because the same amount of vapour is now spread over a larger volume. This leads to an initial decrease in the vapour pressure. The system will then work to re-establish equilibrium by increasing the rate of evaporation.
2. Since the temperature is kept constant, the rate of evaporation, which depends only on temperature, remains unchanged initially. However, as the volume increases, the density of the vapour decreases, leading to fewer collisions between vapour molecules and the liquid surface. Consequently, the rate of condensation decreases initially.
3. When equilibrium is restored, the rate of evaporation will again become equal to the rate of condensation. Since the temperature remains constant, the system will reach a new equilibrium state. The vapour pressure of a liquid at a given temperature is independent of the volume of the container. Therefore, the final vapour pressure will be the same as the original vapour pressure.

The equilibrium constant $(K\_c)$ for the given reaction is expressed in terms of the concentrations of products and reactants. For the reaction $2SO\_2(g)+O\_2(g)\; \backslash longleftrightarrow\; 2SO\_3(g)$, the expression for $K\_c$ is:

$$K\_\{c\}\; =\; \backslash frac\{\backslash left[SO\_\{3\}\backslash right]^\{2\}\}\{\backslash left[SO\_\{2\}\backslash right]^\{2\}\backslash left[O\_\{2\}\backslash right]\}$$

Now, we substitute the given equilibrium concentrations into the expression:

$$K\_\{c\}\; =\; \backslash frac\{\backslash left(1.90\; \backslash text\{\; M\}\backslash right)^2\}\{\backslash left(0.60\; \backslash text\{\; M\}\backslash right)^2\; \backslash left(0.82\; \backslash text\{\; M\}\backslash right)\}$$

Calculating the values:

$$K\_\{c\}\; =\; \backslash frac\{3.61\; \backslash text\{\; M\}^2\}\{0.36\; \backslash text\{\; M\}^2\; \backslash times\; 0.82\; \backslash text\{\; M\}\}\; =\; \backslash frac\{3.61\; \backslash text\{\; M\}^2\}\{0.2952\; \backslash text\{\; M\}^3\}$$

$K\_\{c\}\; \backslash approx\; 12.23\; \backslash text\{\; M\}^\{-1\}$

Therefore, the equilibrium constant $K\_c$ for the given equilibrium is approximately $12.23\; \backslash text\{\; M\}^\{-1\}$.

The reaction is $I\_2(g)\; \backslash longleftrightarrow\; 2I(g)$. The total pressure is given as $P\_\{total\}\; =\; 10^5$ Pa. The iodine vapour contains 40% by volume of I atoms, which means the remaining 60% is $I\_2$ molecules by volume.

The partial pressure of I atoms ($p\_I$) is:

$p\_I\; =\; (\backslash text\{Volume\; \%\; of\; I\; atoms\})\; \backslash times\; P\_\{total\}$

$p\_I\; =\; \backslash frac\{40\}\{100\}\; \backslash times\; 10^5\; \backslash text\{\; Pa\}\; =\; 0.4\; \backslash times\; 10^5\; \backslash text\{\; Pa\}\; =\; 4\; \backslash times\; 10^4\; \backslash text\{\; Pa\}$

The partial pressure of $I\_2$ molecules ($p\_\{I\_2\}$) is:

$p\_\{I\_2\}\; =\; (\backslash text\{Volume\; \%\; of\; \}\; I\_2\; \backslash text\{\; molecules\})\; \backslash times\; P\_\{total\}$

$p\_\{I\_2\}\; =\; \backslash frac\{60\}\{100\}\; \backslash times\; 10^5\; \backslash text\{\; Pa\}\; =\; 0.6\; \backslash times\; 10^5\; \backslash text\{\; Pa\}\; =\; 6\; \backslash times\; 10^4\; \backslash text\{\; Pa\}$

The expression for the equilibrium constant $K\_p$ for the given reaction is:

$$K\_p\; =\; \backslash frac\{(p\_I)^2\}\{p\_\{I\_2\}\}$$

Now, substitute the partial pressures into the $K\_p$ expression:

$$K\_p\; =\; \backslash frac\{(4\; \backslash times\; 10^4\; \backslash text\{\; Pa\})^2\}\{6\; \backslash times\; 10^4\; \backslash text\{\; Pa\}\}$$

$$K\_p\; =\; \backslash frac\{16\; \backslash times\; 10^8\; \backslash text\{\; Pa\}^2\}\{6\; \backslash times\; 10^4\; \backslash text\{\; Pa\}\}$$

$$K\_p\; =\; \backslash frac\{16\}\{6\}\; \backslash times\; 10^\{(8-4)\}\; \backslash text\{\; Pa\}\; =\; 2.666...\; \backslash times\; 10^4\; \backslash text\{\; Pa\}$$

Therefore, $K\_p$ for the equilibrium is approximately $2.67\; \backslash times\; 10^4\; \backslash text\{\; Pa\}$.