In this image, we are asked to predict the equilibrium pressure of ammonia ($\text{NH}_3$) using the provided equilibrium data and tools.

We have the following chemical equilibrium:

$\text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g)$ with the equilibrium constant $K_p = 3.1 \times 10^{-7}$.

Initial conditions are:

• $\text{N}_2$ partial pressure = 1.1 atm
• $\text{H}_2$ partial pressure = 4.1 atm.

To solve:

1. Set up the ICE table (Initial, Change, Equilibrium) for the reaction:

   \text{Species} & \text{N}_2 & \text{H}_2 & \text{NH}_3 \\ \hline \text{Initial (atm)} & 1.1 & 4.1 & 0 \\ \text{Change (atm)} & -x & -3x & +2x \\ \text{Equilibrium (atm)} & 1.1 - x & 4.1 - 3x & 2x \\ \end{array}$$

2. Expression for $K_p$: $K_p = \frac{(P_{\text{NH}_3})^2}{(P_{\text{N}_2})(P_{\text{H}_2})^3} = 3.1 \times 10^{-7}$ Substituting equilibrium pressures: $K_p = \frac{(2x)^2}{(1.1 - x)(4.1 - 3x)^3} = 3.1 \times 10^{-7}$

3. Small $x$ approximation: Since $K_p$ is very small, we assume that $x$ is small, so $1.1 - x \approx 1.1$ and $4.1 - 3x \approx 4.1$. Simplifying: $K_p = \frac{(2x)^2}{(1.1)(4.1)^3}$ Solving for $x$: $3.1 \times 10^{-7} = \frac{4x^2}{1.1 \times 69.921}$ $3.1 \times 10^{-7} = \frac{4x^2}{76.9131}$ $x^2 = \frac{3.1 \times 10^{-7} \times 76.9131}{4}$ $x^2 \approx 5.96 \times 10^{-6}$ $x \approx 0.00244$

4. Equilibrium pressure of $\text{NH}_3$: $P_{\text{NH}_3} = 2x = 2 \times 0.00244 = 0.00488 \, \text{atm}$ Rounding to 1 significant digit: $P_{\text{NH}_3} \approx 0.005 \, \text{atm}$.

Would you like more details or have any questions?

Further questions:

1. How does the ICE table help in equilibrium calculations?
2. What is the purpose of the small $x$ approximation?
3. When should the small $x$ approximation be used?
4. Can the approximation be used if $K_p$ is larger?
5. How do you determine when the small $x$ approximation is valid?

Tip:

Always verify if your small $x$ approximation is valid by checking that the calculated value of $x$ is much smaller than the initial concentrations or pressures.