CBSE Class 11 Chemistry Chapter 7: Equilibrium NCERT Solutions

1. Initially, when the volume of the container is suddenly increased, the vapour pressure will decrease. This is because the existing amount of vapour is now distributed over a larger volume, leading to a lower concentration of vapour molecules per unit volume.

1. Since the temperature remains constant, the rate of evaporation, which depends on the surface area of the liquid and temperature, remains unchanged initially. However, when the volume increases, the density of the vapour decreases. This leads to fewer collisions between vapour molecules and the liquid surface, causing the rate of condensation to decrease initially.

1. When equilibrium is restored, the system will adjust such that the rate of evaporation becomes equal to the rate of condensation again. Since the temperature is constant and the vapour pressure is primarily dependent on temperature for a given liquid, the final vapour pressure will return to its original value. The volume change does not affect the equilibrium vapour pressure at a constant temperature.

The equilibrium constant $(K\_c)$ for the given reaction is expressed based on the law of mass action. 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:

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

Calculating the values:

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

$\backslash approx\; 12.239\; \backslash text\{\; M\}^\{-1\}$

Therefore, the value of $K\_c$ for the equilibrium is approximately $12.239\; \backslash text\{\; M\}^\{-1\}$.

Given the total pressure $p\_\{\backslash text\{total\}\}\; =\; 10^5\; \backslash text\{\; Pa\}$. The iodine vapour contains 40% by volume of I atoms, which means the mole fraction of I atoms is 0.40. Consequently, the mole fraction of $I\_2$ molecules is $1\; -\; 0.40\; =\; 0.60$.

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

$p\_I\; =\; (\backslash text\{mole\; fraction\; of\; I\})\; \backslash times\; p\_\{\backslash text\{total\}\}\; =\; \backslash frac\{40\}\{100\}\; \backslash times\; 10^5\; \backslash text\{\; Pa\}\; =\; 0.40\; \backslash times\; 10^5\; \backslash text\{\; Pa\}\; =\; 4\; \backslash times\; 10^4\; \backslash text\{\; Pa\}$

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

$p\_\{I\_2\}\; =\; (\backslash text\{mole\; fraction\; of\; \}\; I\_2)\; \backslash times\; p\_\{\backslash text\{total\}\}\; =\; \backslash frac\{60\}\{100\}\; \backslash times\; 10^5\; \backslash text\{\; Pa\}\; =\; 0.60\; \backslash times\; 10^5\; \backslash text\{\; Pa\}\; =\; 6\; \backslash times\; 10^4\; \backslash text\{\; Pa\}$

For the given equilibrium reaction $I\_2(g)\; \backslash longleftrightarrow\; 2I(g)$, the expression for $K\_p$ is:

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

Substituting the calculated partial pressures:

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

$K\_p\; =\; \backslash frac\{16\}\{6\}\; \backslash times\; 10^\{8-4\}\; \backslash text\{\; Pa\}\; =\; 2.667\; \backslash times\; 10^4\; \backslash text\{\; Pa\}$

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