For the reaction A + 2B hArr 2C at equilibrium `[C] = 1.4 M, [A]_(o) = 1 M, [B]_(o) = 2 M, [C]_(o) = 3 M` . The value of `K_(c)` is

Options

A. `0.084`
B. `8.4`
C. 84
D. 840

Explanation

To find the equilibrium constant \( K_c \) for the reaction \( A + 2B \rightleftharpoons 2C \) given the equilibrium concentration of \( C \) and the initial concentrations of \( A \), \( B \), and \( C \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ A + 2B \rightleftharpoons 2C \]

### Step 2: Identify initial concentrations
From the problem, we have:
- Initial concentration of \( A \) (\([A]_o\)) = 1 M
- Initial concentration of \( B \) (\([B]_o\)) = 2 M
- Initial concentration of \( C \) (\([C]_o\)) = 3 M

### Step 3: Set up the change in concentrations
Let \( x \) be the amount of \( A \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- The change in concentration of \( A \) will be \( -x \).
- The change in concentration of \( B \) will be \( -2x \) (since 2 moles of \( B \) react for every mole of \( A \)).
- The change in concentration of \( C \) will be \( +2x \) (since 2 moles of \( C \) are produced for every mole of \( A \)).

### Step 4: Write the equilibrium concentrations
At equilibrium, the concentrations will be:
- \([A] = [A]_o - x = 1 - x\)
- \([B] = [B]_o - 2x = 2 - 2x\)
- \([C] = [C]_o + 2x = 3 + 2x\)

### Step 5: Use the equilibrium concentration of \( C \)
We are given that at equilibrium, \([C] = 1.4 M\). Therefore, we can set up the equation:
\[ 3 + 2x = 1.4 \]

### Step 6: Solve for \( x \)
Rearranging the equation gives:
\[ 2x = 1.4 - 3 \]
\[ 2x = -1.6 \]
\[ x = -0.8 \]

### Step 7: Calculate equilibrium concentrations for \( A \) and \( B \)
Now we can find the equilibrium concentrations:
- \([A] = 1 - (-0.8) = 1 + 0.8 = 1.8 M\)
- \([B] = 2 - 2(-0.8) = 2 + 1.6 = 3.6 M\)

### Step 8: Write the expression for \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2}{[A][B]^2} \]

### Step 9: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we found:
\[ K_c = \frac{(1.4)^2}{(1.8)(3.6)^2} \]

### Step 10: Calculate \( K_c \)
Calculating the values:
- \( (1.4)^2 = 1.96 \)
- \( (3.6)^2 = 12.96 \)
- Therefore,
\[ K_c = \frac{1.96}{(1.8)(12.96)} \]
\[ K_c = \frac{1.96}{23.328} \approx 0.084 \]

Thus, the value of \( K_c \) is approximately \( 0.084 \).

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