The reaction,
`2A(g) + B(g)hArr3C(g) + D(g)`
is begun with the concentration of A and B both at an intial value of `1.00` M. When equilibrium is reached, the concentration of D is measured and found to be `0.25` M. The value for the equilibrium constant for this reaction is given by the expression:

Options

A. `[(0.75)^(3)(0.25)] div [(0.50)^(2)(0.75)]`
B. `[(0.75)^(3) (0.25)] div [(0.50)^(2)(0.25)]`
C. `[(0.75)^(3)(0.25)] div [(0.75)^(2)(0.25)]`
D. `[(0.75)^(3)(0.25)]div [(1.00)^(2)(1.00)]`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ 2A(g) + B(g) \rightleftharpoons 3C(g) + D(g) \]

### Step 2: Set up the initial concentrations
At the beginning of the reaction (t = 0), the concentrations of A and B are both given as:
\[ [A] = 1.00 \, \text{M} \]
\[ [B] = 1.00 \, \text{M} \]
The initial concentrations of C and D are both 0 M since the reaction has not started yet.

### Step 3: Define the change in concentrations
Let \( x \) be the amount of B that reacts at equilibrium. According to the stoichiometry of the reaction:
- For every 2 moles of A that react, 1 mole of B reacts.
- For every \( x \) moles of B that react, \( 2x \) moles of A will react.

Thus, the changes in concentrations at equilibrium will be:
- For A: \( [A] = 1.00 - 2x \)
- For B: \( [B] = 1.00 - x \)
- For C: \( [C] = 3x \)
- For D: \( [D] = x \)

### Step 4: Use the information given about D
We know from the problem that at equilibrium, the concentration of D is measured to be:
\[ [D] = 0.25 \, \text{M} \]
This means:
\[ x = 0.25 \]

### Step 5: Calculate the equilibrium concentrations
Now we can substitute \( x = 0.25 \) into the expressions for the equilibrium concentrations:
- For A:
\[ [A] = 1.00 - 2(0.25) = 1.00 - 0.50 = 0.50 \, \text{M} \]
- For B:
\[ [B] = 1.00 - 0.25 = 0.75 \, \text{M} \]
- For C:
\[ [C] = 3(0.25) = 0.75 \, \text{M} \]
- For D:
\[ [D] = 0.25 \, \text{M} \] (as given)

### Step 6: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[C]^3[D]}{[A]^2[B]} \]

### Step 7: Substitute the equilibrium concentrations into the expression
Now we can substitute the equilibrium concentrations into the expression:
\[ K_c = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)} \]

### Step 8: Calculate the value of \( K_c \)
Calculating the numerator:
\[ (0.75)^3 = 0.421875 \]
So,
\[ 0.421875 \times 0.25 = 0.10546875 \]

Calculating the denominator:
\[ (0.50)^2 = 0.25 \]
So,
\[ 0.25 \times 0.75 = 0.1875 \]

Now, substituting back into the expression for \( K_c \):
\[ K_c = \frac{0.10546875}{0.1875} \approx 0.5625 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is approximately:
\[ K_c \approx 0.5625 \]

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