For the equilibrium `H_(2)(g)+CO_(2)(g)hArr hArr H_(2)O(g)+CO(g), K_(c)=16` at 1000 K. If 1.0 mole of `CO_(2)` and 1.0 mole of `H_(2)` are taken in a l L flask, the final equilibrium concentration of CO at 1000 K will be

Options

A. 0.8 M
B. 0.08 M
C. 1.6 M
D. 1.0 atm

Answer

To solve the problem, we need to determine the final equilibrium concentration of CO in the reaction:

\[ H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g) \]

Given that the equilibrium constant \( K_c = 16 \) at 1000 K, and we start with 1.0 mole of \( CO_2 \) and 1.0 mole of \( H_2 \) in a 1 L flask, we can follow these steps:

### Step-by-Step Solution:

1. **Initial Concentrations**:
Since the volume of the flask is 1 L, the initial concentrations (in molarity) of the reactants are:
- \([H_2] = 1.0 \, M\)
- \([CO_2] = 1.0 \, M\)
- \([H_2O] = 0 \, M\)
- \([CO] = 0 \, M\)

2. **Change in Concentrations**:
Let \( x \) be the amount of \( H_2 \) and \( CO_2 \) that reacts at equilibrium. Therefore, at equilibrium, the concentrations will be:
- \([H_2] = 1.0 - x \, M\)
- \([CO_2] = 1.0 - x \, M\)
- \([H_2O] = x \, M\)
- \([CO] = x \, M\)

3. **Equilibrium Expression**:
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[H_2O][CO]}{[H_2][CO_2]} \]
Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{x \cdot x}{(1.0 - x)(1.0 - x)} = \frac{x^2}{(1.0 - x)^2} \]

4. **Setting Up the Equation**:
We know that \( K_c = 16 \), so we can set up the equation:
\[ \frac{x^2}{(1.0 - x)^2} = 16 \]

5. **Solving for \( x \)**:
Taking the square root of both sides:
\[ \frac{x}{1.0 - x} = 4 \]
Cross-multiplying gives:
\[ x = 4(1.0 - x) \]
Expanding the equation:
\[ x = 4 - 4x \]
Combining like terms:
\[ 5x = 4 \quad \Rightarrow \quad x = \frac{4}{5} = 0.8 \]

6. **Final Equilibrium Concentration of CO**:
At equilibrium, the concentration of CO is equal to \( x \):
\[ [CO] = x = 0.8 \, M \]

### Final Answer:
The final equilibrium concentration of CO at 1000 K is \( 0.8 \, M \).