Question

the formation of ammonia is a reactant-favored process, as shown below.
\\(\text{n}_2(g) + 3\text{h}_2(g) \
ightleftharpoons 2\text{nh}_3(g)\\)
\\(1\text{ atm}\\) \\(7\text{ atm}\\)
\\(k = 2.8 \times 10^{-5}\\) @ \\(745\text{ k}\\)
the initial conditions are shown above. calculate the equilibrium pressures of the components. what is the equilibrium pressure of \\(\text{nh}_3\\)?
\\(p_{\text{nh}_3} = ?\text{ atm}\\)

Explanation:

Set up the RICE table

Using the RICE Table Calculations and Chemical Equilibrium knowledge points

$$
LATEXBLOCK0
$$

Express the equilibrium constant

Using the Equilibrium Constant and Partial Pressure Equilibrium knowledge points

$$ K_p = \frac{P_{\text{NH}_3}^2}{P_{\text{N}_2} \cdot P_{\text{H}_2}^3} = 2.8 \times 10^{-5} $$

$$ \frac{(2x)^2}{(1-x)(7-3x)^3} = 2.8 \times 10^{-5} $$

Solve for x using approximation

Using the Chemical Equilibrium knowledge point

$$
LATEXBLOCK1
$$

Calculate the equilibrium pressure of NH3

Using the Partial Pressure Equilibrium knowledge point

$$ P_{\text{NH}_3} = 2x = 2(0.049) = 0.098\text{ atm} $$

Answer:

The formation of ammonia is a reactant-favored process, as shown below.

$$\text{N}_2(g) + 3\text{H}_2(g) ightleftharpoons 2\text{NH}_3(g)$$

\(1\text{ atm}\quad 7\text{ atm}\)

\(K = 2.8 \times 10^{-5}\) @ \(745\text{ K}\)

The initial conditions are shown above. Calculate the equilibrium pressures of the components. What is the equilibrium pressure of \(\text{NH}_3\)?
\(P_{\text{NH}_3} =\) <blank>\(0.098\)</blank> atm