The standard Gibbs energy change at `300K` for the reaction `2AhArrB+C` is 2494. `2J`. At a given time, the composition of the reaction mixture is `[A]=1/2, [B]=2` and `[C]=1/2`. The reaction proceeds in the
`(R=8.314JK//"mol"e=2.718)`

Options

A. forward direction because `QgtK_(e)`
B. reverse direction because `QgtK_(e)`
C. forward direction because `QltK_(e)`
D. reverse direction because `QltK_(e)`

Answer

To determine the direction in which the reaction proceeds, we will follow these steps:

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

### Step 2: Identify the standard Gibbs energy change
The standard Gibbs energy change (\( \Delta G^\circ \)) at 300 K is given as:
\[ \Delta G^\circ = 2494.2 \, \text{J} \]

### Step 3: Use the relationship between Gibbs energy and equilibrium constant
The relationship between the standard Gibbs energy change and the equilibrium constant (\( K_{eq} \)) is given by:
\[ \Delta G^\circ = -RT \ln K_{eq} \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 300 \, \text{K} \)

### Step 4: Rearrange the equation to find \( K_{eq} \)
Rearranging the equation gives:
\[ K_{eq} = e^{-\Delta G^\circ / (RT)} \]

### Step 5: Substitute the values into the equation
Substituting the known values:
\[ K_{eq} = e^{-2494.2 / (8.314 \times 300)} \]

### Step 6: Calculate \( K_{eq} \)
Calculating the exponent:
\[ K_{eq} = e^{-2494.2 / 2494.2} = e^{-1} \]
\[ K_{eq} \approx 0.3678 \]

### Step 7: Calculate the reaction quotient \( Q \)
The reaction quotient \( Q \) is calculated using the concentrations at the given time:
\[ Q = \frac{[B]^1 \cdot [C]^1}{[A]^2} \]
Given concentrations:
- \([A] = \frac{1}{2}\)
- \([B] = 2\)
- \([C] = \frac{1}{2}\)

Substituting these values into the expression for \( Q \):
\[ Q = \frac{(2)^1 \cdot \left(\frac{1}{2}\right)^1}{\left(\frac{1}{2}\right)^2} \]
\[ Q = \frac{2 \cdot \frac{1}{2}}{\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4 \]

### Step 8: Compare \( Q \) and \( K_{eq} \)
Now, we compare \( Q \) and \( K_{eq} \):
- \( Q = 4 \)
- \( K_{eq} \approx 0.3678 \)

Since \( Q > K_{eq} \), the reaction will proceed in the reverse direction to reach equilibrium.

### Conclusion
The reaction proceeds in the backward direction.

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