Question

At 15 atm pressure, $ \text{NH}_3(g) $ is being heated in a closed container from 27°C to 347°C and as a result, it partially dissociates following the equation: $ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) $ If the volume of the container remains constant and pressure increases to 50 atm, then calculate the percentage dissociation of $ \text{NH}_3(g) $

• A. 63
• B. 38.7
• C. 61.3
• D. 37

Hint:

Use the change in pressure to relate the degree of dissociation, remembering that the dissociation of gases results in a change in total pressure.

Answer 1

The Correct Option is B

We are given: Initial pressure, \( P_1 = 15 \, \text{atm} \), Final pressure, \( P_2 = 50 \, \text{atm} \), Initial temperature = 27°C = 300 K,
Final temperature = 347°C = 620 K.
We can use the concept of partial dissociation to solve this problem. The balanced dissociation equation is: \[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \] Let the initial moles of NH\(_3\) be \( n_0 \). Let the extent of dissociation be \( x \), so: The number of moles of NH\(_3\) remaining = \( n_0 - x \), The number of moles of N\(_2\) formed = \( \frac{x}{2} \), The number of moles of H\(_2\) formed = \( \frac{3x}{2} \). The total number of moles after dissociation is: \[ n_{\text{total}} = (n_0 - x) + \frac{x}{2} + \frac{3x}{2} = n_0 + x \] Using the ideal gas law for the initial and final states: \[ \frac{P_1 V}{n_0 R T_1} = 1 \quad \text{(initial state)} \] \[ \frac{P_2 V}{n_{\text{total}} R T_2} = 1 \quad \text{(final state)} \] Since the volume and gas constant are constant, the equation simplifies to: \[ \frac{P_1}{n_0 T_1} = \frac{P_2}{n_{\text{total}} T_2} \] Substituting \( n_{\text{total}} = n_0 + x \) into the equation: \[ \frac{P_1}{n_0 T_1} = \frac{P_2}{(n_0 + x) T_2} \] Now, we substitute the known values: \[ \frac{15}{n_0 \times 300} = \frac{50}{(n_0 + x) \times 620} \] Cross-multiply: \[ 15 \times (n_0 + x) \times 620 = 50 \times n_0 \times 300 \] \[ 9300 \times (n_0 + x) = 15000 \times n_0 \] \[ 9300n_0 + 9300x = 15000n_0 \] \[ 9300x = 5700n_0 \] \[ x = \frac{5700n_0}{9300} = 0.6129n_0 \] Thus, the percentage dissociation is: \[ \text{Percentage dissociation} = \frac{x}{n_0} \times 100 = 61.29% \] Therefore, the correct percentage dissociation of NH\(_3\) is 38.7%, corresponding to (B).