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This problem involves equilibrium and asks for the equilibrium pressure of \(C_2H_2\) based on the given reaction:

\[
2CH_4(g) \rightleftharpoons C_2H_2(g) + 3H_2(g)
\]

with an equilibrium constant \(K_p = 3.0 \times 10^{-7}\) at a certain temperature. We are also provided that the initial pressure of methane (\(CH_4\)) is 12 atm.

### Step-by-step process:

1. **Define the change in pressures**:
   - Let the change in pressure of \(CH_4\) be \(2x\) (since the stoichiometric coefficient is 2). 
   - Therefore, the pressure of \(C_2H_2\) will be \(x\) and the pressure of \(H_2\) will be \(3x\) at equilibrium.

2. **Initial and equilibrium pressures**:
   - Initial pressure of \(CH_4 = 12\) atm.
   - At equilibrium, the pressure of \(CH_4\) will be \(12 - 2x\) atm.

3. **Write the expression for \(K_p\)**:
   \[
   K_p = \frac{P_{C_2H_2} \cdot P_{H_2}^3}{P_{CH_4}^2}
   \]
   Substituting in terms of \(x\):
   \[
   3.0 \times 10^{-7} = \frac{x \cdot (3x)^3}{(12 - 2x)^2}
   \]

4. **Approximate using the small \(x\) assumption**:
   Since \(K_p\) is very small, we can assume that \(x\) is small, meaning \(12 - 2x \approx 12\). The equation simplifies to:
   \[
   3.0 \times 10^{-7} = \frac{x \cdot (27x^3)}{144}
   \]
   Simplifying further:
   \[
   3.0 \times 10^{-7} = \frac{27x^4}{144}
   \]
   \[
   3.0 \times 10^{-7} = \frac{x^4}{16}
   \]
   Solving for \(x^4\):
   \[
   x^4 = 16 \times 3.0 \times 10^{-7} = 4.8 \times 10^{-6}
   \]
   \[
   x = \sqrt[4]{4.8 \times 10^{-6}} = 0.027
   \]

Thus, the equilibrium pressure of \(C_2H_2\) is approximately **0.03 atm** (rounded to 1 significant digit).

Would you like more details on any step?

Here are some related questions for further understanding:
1. How do changes in temperature affect the equilibrium constant \(K_p\)?
2. Why is the small \(x\) approximation used in this problem?
3. How would the equilibrium shift if more \(CH_4\) were added to the system?
4. What would happen to the equilibrium pressure of \(C_2H_2\) if \(K_p\) were larger?
5. Can you explain the difference between \(K_p\) and \(K_c\)?

**Tip:** When using the small \(x\) approximation, always check whether the approximation is valid by comparing \(x\) to the initial concentrations or pressures. If \(x\) is very small compared to the initial values, the approximation holds.

A chemist is studying the following equilibrium: 2CH4(g) ↔ C2H2(g) + 3H2(g), with an equilibrium constant Kp = 3.0 x 10^-7. The reaction vessel is filled with 12 atm of methane gas. Can you predict the equilibrium pressure of C2H2 using only the tools available in ALEKS?

Learn how to predict the equilibrium pressure of C2H2 in a reaction vessel with 12 atm of methane gas, using the equilibrium constant Kp and the small x approximation method. This step-by-step solution is ideal for high school chemistry students.

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