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

Step 1: Write the expression for K$_p$.
For the given reaction: \[ \text{NH}_3(g) \longrightarrow \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \] At equilibrium, let the dissociation be $\alpha$. At time $t = 0$, the initial amount of NH$_3$ is 1 mole, and at equilibrium, the amounts of the products will be as follows: - NH$_3$ = (1 - $\alpha$) moles - N$_2$ = $\frac{\alpha}{2}$ moles - H$_2$ = $\frac{3\alpha}{2}$ moles The total pressure at equilibrium is given as $\sqrt{5}$ atm. Using the expression for the total pressure $P_T$, we can write: \[ P_T = \left( \frac{\alpha}{2} + \frac{3\alpha}{2} \right) \cdot \frac{RT}{V} = \sqrt{5} \text{ atm} \]
Step 2: Express K$_p$.
We know that: \[ K_p = \frac{(P_{\text{N}_2})^{1/2} (P_{\text{H}_2})^{3/2}}{(P_{\text{NH}_3})} \] Substitute the equilibrium concentrations in terms of $\alpha$: \[ K_p = \frac{(\frac{\alpha}{2})^{1/2} (\frac{3\alpha}{2})^{3/2}}{(1-\alpha)} \] Simplify this expression using the given value for $K_p = 9$ atm: \[ 9 = \left( \frac{\alpha}{2} \right)^{1/2} \left( \frac{3\alpha}{2} \right)^{3/2} \cdot \frac{P_T}{1+\alpha} \] Substitute $P_T = \sqrt{5}$: \[ 9 = \left( \frac{\alpha}{2} \right)^{1/2} \left( \frac{3\alpha}{2} \right)^{3/2} \times \frac{\sqrt{5}}{1 + \alpha} \]
Step 3: Solve for $\alpha$.
After simplifying, we obtain: \[ 9 = \frac{(\alpha^{1/2}) (3\alpha^{3/2})}{(1-\alpha)} \] Simplifying further: \[ 9 = \frac{\alpha^2}{4(1-\alpha)} \] Solving for $\alpha$, we get: \[ \alpha^2 = 0.8 \] So, $\alpha = \sqrt{0.8}$.
Step 4: Conclusion.
Finally, to find the value of 7$\alpha^2$: \[ 7\alpha^2 = 7 \times 0.8 = 5.6 \]