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 find the final equilibrium concentration of CO at 1000 K for the reaction:

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

Given that \( K_c = 16 \) 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 1: Write the expression for the equilibrium constant \( K_c \)

The equilibrium constant expression for the given reaction is:

\[ K_c = \frac{[H_2O][CO]}{[H_2][CO_2]} \]

### Step 2: Set up the initial concentrations

Since we have 1 mole of \( CO_2 \) and 1 mole of \( H_2 \) in a 1 L flask, the initial concentrations are:

\[ [H_2] = 1 \, \text{M}, \quad [CO_2] = 1 \, \text{M}, \quad [H_2O] = 0 \, \text{M}, \quad [CO] = 0 \, \text{M} \]

### Step 3: Define the change in concentrations at equilibrium

Let \( x \) be the amount of \( H_2 \) and \( CO_2 \) that react to form products. Therefore, at equilibrium, the concentrations will be:

\[ [H_2] = 1 - x, \quad [CO_2] = 1 - x, \quad [H_2O] = x, \quad [CO] = x \]

### Step 4: Substitute into the equilibrium expression

Substituting the equilibrium concentrations into the \( K_c \) expression:

\[ K_c = \frac{x \cdot x}{(1 - x)(1 - x)} = \frac{x^2}{(1 - x)^2} \]

Given that \( K_c = 16 \):

\[ 16 = \frac{x^2}{(1 - x)^2} \]

### Step 5: Solve for \( x \)

Taking the square root of both sides gives:

\[ 4 = \frac{x}{1 - x} \]

Cross-multiplying yields:

\[ 4(1 - x) = x \]

Expanding and rearranging gives:

\[ 4 - 4x = x \implies 4 = 5x \implies x = \frac{4}{5} = 0.8 \]

### Step 6: Find the final equilibrium concentration of CO

Since \( x \) represents the concentration of \( CO \) at equilibrium:

\[ [CO] = x = 0.8 \, \text{M} \]

### Final Answer

The final equilibrium concentration of CO at 1000 K is:

\[ \boxed{0.8 \, \text{M}} \]