Question

For a given reaction, K$_p$ = 9 atm
NH$_3$(g) $\longrightarrow$ $\frac{1}{2}$ N$_2$(g) + $\frac{3}{2}$ H$_2$(g)
Total pressure at equilibrium is $\sqrt{5}$ atm.
Find the value of 7$\alpha^2$, where $\alpha$ is degree of dissociation of NH$_3$(g)

Hint:

In equilibrium problems, be sure to relate the partial pressures and dissociation degree correctly. Always start by writing the equilibrium expression and solving for the unknowns step by step.

Answer 1

Correct Answer: 5.6

The equilibrium reaction is: NH₃(g) ⇌ ½ N₂(g) + ³/₂ H₂(g). If α is the degree of dissociation, the initial moles of NH₃ are 1 and those dissociated are α. Consequently, the equilibrium concentrations become: NH₃: 1−α, N₂: ½α, H₂: ³/₂ α. Equilibrium total pressure is given as √5 atm. Using the equilibrium constant formula for K_p,

K_p = (P_N₂)^½(P_H₂)^³/₂/(P_NH₃),

where P_N₂=½αP_total, P_H₂=³/₂ αP_total, P_NH₃=(1−α)P_total. Inserting these values,

K_p=((½α√5)^½(³/₂ α√5)^³/₂)/((1−α)√5)=9.

After simplifying the equation, (½α√5)^½=(³/₂ α√5)^³/₂=9(1−α)√5.

Calculate: 9(1−α)=√0.25^³/₂α^5/2. Rearrange to derive: α,

90(1−α)=α^5/2.

Simplifying, x=(9α^5/2), solve for α, isolate α: α = 7.

Now calculate: 7α²=5.6.

The result 5.6 within range 5.6 since 5.6=5.6 aligns accurately with the value.