The value of K for `H_(2)(g)+CO_(2)(g)hArr H_(2)O(g)+CO(g)` is 1.80 at `1000^(@)C`. If 1.0 mole of each `H_(2) and CO_(2)` placed in 1 litre flask, the final equilibrium concentration of CO at `1000^(@)C` will be

Options

A. 0.295 M
B. 0.385 M
C. 0.572 M
D. 0.473 M

Answer

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

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

Given that the equilibrium constant \( K_c \) is 1.80 at \( 1000^\circ C \) and that we start with 1 mole of \( H_2 \) and 1 mole of \( CO_2 \) in a 1-liter flask, we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)

The expression for the equilibrium constant \( K_c \) for the reaction is:

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

### Step 2: Set up the initial concentrations

Initially, we have:

- \([H_2] = 1 \, \text{mol/L}\)
- \([CO_2] = 1 \, \text{mol/L}\)
- \([H_2O] = 0 \, \text{mol/L}\)
- \([CO] = 0 \, \text{mol/L}\)

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

Let \( x \) be the change in concentration of \( H_2 \) and \( CO_2 \) that reacts to reach equilibrium. Therefore, at equilibrium, the concentrations will be:

- \([H_2] = 1 - x\)
- \([CO_2] = 1 - x\)
- \([H_2O] = x\)
- \([CO] = x\)

### Step 4: Substitute the equilibrium concentrations into the \( K_c \) expression

Substituting these values into the expression for \( K_c \):

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

Given that \( K_c = 1.80 \), we set up the equation:

\[ 1.80 = \frac{x^2}{(1 - x)^2} \]

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

Cross-multiplying gives:

\[ 1.80(1 - x)^2 = x^2 \]

Expanding the left side:

\[ 1.80(1 - 2x + x^2) = x^2 \]

This simplifies to:

\[ 1.80 - 3.60x + 1.80x^2 = x^2 \]

Rearranging gives:

\[ 0.80x^2 - 3.60x + 1.80 = 0 \]

### Step 6: Use the quadratic formula to find \( x \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 0.80 \), \( b = -3.60 \), and \( c = 1.80 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-3.60)^2 - 4 \cdot 0.80 \cdot 1.80 = 12.96 - 5.76 = 7.20 \]

Now substituting into the quadratic formula:

\[ x = \frac{3.60 \pm \sqrt{7.20}}{2 \cdot 0.80} \]

Calculating \( \sqrt{7.20} \approx 2.68 \):

\[ x = \frac{3.60 \pm 2.68}{1.60} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{3.60 + 2.68}{1.60} \approx 3.36 \) (not valid since it exceeds initial moles)
2. \( x = \frac{3.60 - 2.68}{1.60} \approx 0.57 \)

### Step 7: Find the equilibrium concentration of CO

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

\[ [CO] = x \approx 0.57 \, \text{mol/L} \]

### Final Answer

The final equilibrium concentration of \( CO \) at \( 1000^\circ C \) is approximately \( 0.57 \, \text{mol/L} \).

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