The equilibrium constant for the reaction `CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g) is 5` how many moles of `CO_(2)` must be added to 1 litre container alrady containing 3 moles each of CO and `H_(2)O` to make 2 M equilibrium conentration of CO?

Options

A. 15
B. 19
C. 5
D. 20

Answer

To solve the problem step by step, we will analyze the given reaction and the equilibrium conditions.

**Step 1: Write the balanced chemical equation.**
The reaction is:
\[ \text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)} \]

**Step 2: Identify the initial conditions.**
We are given:
- Initial moles of CO = 3 moles
- Initial moles of H2O = 3 moles
- Initial moles of CO2 = 0 moles (since we are adding it)
- Initial moles of H2 = 0 moles (since no information is given)

**Step 3: Define the change in moles at equilibrium.**
Let \( x \) be the number of moles of CO2 added. At equilibrium, we want the concentration of CO to be 2 M in a 1 L container. Thus:
- Moles of CO at equilibrium = 2 moles
- Moles of H2O at equilibrium = 3 - 1 = 2 moles (since 1 mole of H2O reacts)
- Moles of CO2 at equilibrium = \( x + 1 \) (1 mole formed from the reaction)
- Moles of H2 at equilibrium = 1 mole (1 mole formed from the reaction)

**Step 4: Set up the equilibrium expression.**
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(x + 1)(1)}{(2)(2)} = \frac{x + 1}{4} \]
Given \( K_c = 5 \):
\[ \frac{x + 1}{4} = 5 \]

**Step 5: Solve for \( x \).**
Multiply both sides by 4:
\[ x + 1 = 20 \]
Subtract 1 from both sides:
\[ x = 19 \]

**Conclusion:**
To achieve an equilibrium concentration of 2 M for CO, we need to add **19 moles of CO2** to the container.