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Consider other possibilities to achieve the result, do not be limited by the prompt.

Solve this step by step: A technician is producing hydrogen gas. He adds methane gas, steam, and a nickel catalyst to an empty 3.00L reaction container and allows the system to reach equilibrium. The reaction is represented by the following equation.

CH4(g)+H2O(g) 3H2(g)+CO(g)

At equilibrium, the amount of hydrogen gas is 1.8 mol, carbon monoxide is 0.60 mol, methane gas is 0.40 mol, and water vapour is 0.20 mol.

Numeric Response

4. The equilibrium constant for the reaction represented in the reaction above is

To find the equilibrium constant (\(K_c\)) for the reaction, we need to use the equilibrium concentrations of the reactants and products. The reaction is given by:

\[ \text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons 3\text{H}_2(g) + \text{CO}(g) \]

The equilibrium constant expression for this reaction is:

\[ K_c = \frac{[\text{H}_2]^3 [\text{CO}]}{[\text{CH}_4] [\text{H}_2\text{O}]} \]

Given:
- Volume of the reaction container = 3.00 L
- Amount of \(\text{H}_2\) at equilibrium = 1.8 mol
- Amount of \(\text{CO}\) at equilibrium = 0.60 mol
- Amount of \(\text{CH}_4\) at equilibrium = 0.40 mol
- Amount of \(\text{H}_2\text{O}\) at equilibrium = 0.20 mol

First, we need to convert these amounts to concentrations (molarity, \(M\)), which is moles per liter (mol/L):

\[ [\text{H}_2] = \frac{1.8 \text{ mol}}{3.00 \text{ L}} = 0.60 \text{ M} \]
\[ [\text{CO}] = \frac{0.60 \text{ mol}}{3.00 \text{ L}} = 0.20 \text{ M} \]
\[ [\text{CH}_4] = \frac{0.40 \text{ mol}}{3.00 \text{ L}} = 0.1333 \text{ M} \]
\[ [\text{H}_2\text{O}] = \frac{0.20 \text{ mol}}{3.00 \text{ L}} = 0.0667 \text{ M} \]

Now, substitute these concentrations into the equilibrium constant expression:

\[ K_c = \frac{(0.60)^3 (0.20)}{(0.1333)(0.0667)} \]

Calculate the numerator:

\[ (0.60)^3 = 0.216 \]
\[ 0.216 \times 0.20 = 0.0432 \]

Calculate the denominator:

\[ 0.1333 \times 0.0667 = 0.008888 \]

Finally, compute \(K_c\):

\[ K_c = \frac{0.0432}{0.008888} \approx 4.86 \]

Therefore, the equilibrium constant \(K_c\) for the reaction is approximately 4.86.

Do you want details or have any questions?

Here are five questions you can ask:
1. How do you determine equilibrium concentrations in a chemical reaction?
2. What are the units of the equilibrium constant \(K_c\)?
3. How does the equilibrium constant change with temperature?
4. Can you explain the significance of the equilibrium constant value?
5. What happens to the equilibrium position if the concentration of a reactant is increased?

**Tip:** Always ensure to convert moles to molarity (concentration) when calculating the equilibrium constant for reactions occurring in a solution.

Learn how to calculate the equilibrium constant (Kc) for the chemical reaction CH4 + H2O ⇌ 3H2 + CO with detailed step-by-step instructions. Understand the significance of equilibrium concentrations and the application of the equilibrium constant formula. Suitable for advanced high school chemistry students.

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