At 1400 K, `K_c=2.5xx10^(-3)` for the reaction `CH_4(g)+2H_2S(g)hArrCS_2(g)+4H_2(g)`.A 10.0 L reaction vessel at 1400 K contains 2.00 mole of `CH_4`.3.0 mole of `CS_2`,3.0 mole of `H_2` and 4.0 mole of `H_2S`.Then

Options

A. This reaction is at equilibrium with above concentrations
B. The reaction will proceed in forward direction to reach equilibrium
C. The reaction will proceed in backward direction to reach equilibrium
D. The information is insufficient to decide the direction of progress of reaction

Explanation

To solve the problem, we need to determine the reaction quotient \( Q_c \) for the given reaction and compare it with the equilibrium constant \( K_c \) to find the direction in which the reaction will proceed.

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ \text{CH}_4(g) + 2 \text{H}_2\text{S}(g) \rightleftharpoons \text{CS}_2(g) + 4 \text{H}_2(g) \]

### Step 2: Write the expression for \( Q_c \)
The expression for the reaction quotient \( Q_c \) is given by:
\[ Q_c = \frac{[\text{CS}_2][\text{H}_2]^4}{[\text{CH}_4][\text{H}_2\text{S}]^2} \]

### Step 3: Calculate the concentrations of each species
Given the number of moles and the volume of the reaction vessel (10.0 L), we can calculate the concentrations:
- Concentration of \( \text{CH}_4 \):
\[ [\text{CH}_4] = \frac{2.00 \, \text{moles}}{10.0 \, \text{L}} = 0.20 \, \text{M} \]
- Concentration of \( \text{CS}_2 \):
\[ [\text{CS}_2] = \frac{3.00 \, \text{moles}}{10.0 \, \text{L}} = 0.30 \, \text{M} \]
- Concentration of \( \text{H}_2} \):
\[ [\text{H}_2] = \frac{4.00 \, \text{moles}}{10.0 \, \text{L}} = 0.40 \, \text{M} \]
- Concentration of \( \text{H}_2\text{S} \):
\[ [\text{H}_2\text{S}] = \frac{3.00 \, \text{moles}}{10.0 \, \text{L}} = 0.30 \, \text{M} \]

### Step 4: Substitute the concentrations into the \( Q_c \) expression
Now we can substitute the concentrations into the \( Q_c \) expression:
\[ Q_c = \frac{(0.30)(0.40)^4}{(0.20)(0.30)^2} \]

### Step 5: Calculate \( Q_c \)
Calculating \( Q_c \):
\[ Q_c = \frac{(0.30)(0.40^4)}{(0.20)(0.30^2)} = \frac{(0.30)(0.0256)}{(0.20)(0.09)} = \frac{0.00768}{0.018} \approx 0.4267 \]

### Step 6: Compare \( Q_c \) with \( K_c \)
Given \( K_c = 2.5 \times 10^{-3} \):
- Since \( Q_c (0.4267) > K_c (2.5 \times 10^{-3}) \), the reaction will proceed in the backward direction to reach equilibrium.

### Conclusion
The correct option is that the reaction will proceed in the backward direction to reach equilibrium.

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