The equlibrium constant for the reaction between `CH_(4)(g)and H_(2)S(g)` to from `CS_(2)(g)and H_(2)(g),` at 1173 K is 3.6. for the following composition of th reaciton mixture, decide which of the following option is correct
`[CH_(4)]=1.07M,[H_(2)S]=1.20M,`
`[CS_(2)]=0.09 M, [H_(2)]=1.78M`

Options

A. Reaction is in equlibrium
B. Reaction will shift to from more of `CS_(2)`
C. Reaction will shift to from more of `H_(2)S`
D. No reaction takes place

Explanation

To solve the problem, we need to determine the reaction quotient (Q) for the given concentrations and compare it to the equilibrium constant (Kc) to predict the direction of the reaction.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
The reaction is:
\[ CH_4(g) + 2H_2S(g) \rightleftharpoons CS_2(g) + 2H_2(g) \]

2. **Identify the equilibrium constant expression (Kc):**
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[CS_2][H_2]^2}{[CH_4][H_2S]^2} \]

3. **Substitute the given concentrations into the Kc expression:**
Given:
- \([CH_4] = 1.07 \, M\)
- \([H_2S] = 1.20 \, M\)
- \([CS_2] = 0.09 \, M\)
- \([H_2] = 1.78 \, M\)

Substitute these values into the Kc expression:
\[ Q = \frac{[0.09][1.78]^2}{[1.07][1.20]^2} \]

4. **Calculate Q:**
First, calculate \([1.78]^2\) and \([1.20]^2\):
\[ [1.78]^2 = 3.1684 \]
\[ [1.20]^2 = 1.44 \]

Now substitute these values into the equation for Q:
\[ Q = \frac{0.09 \times 3.1684}{1.07 \times 1.44} \]
Calculate the numerator:
\[ 0.09 \times 3.1684 = 0.286156 \]
Calculate the denominator:
\[ 1.07 \times 1.44 = 1.5408 \]

Now calculate Q:
\[ Q = \frac{0.286156}{1.5408} \approx 0.186 \]

5. **Compare Q with Kc:**
Given \(K_c = 3.6\) and we calculated \(Q \approx 0.186\).

Since \(Q < K_c\), the reaction will shift to the right (towards the products) to reach equilibrium.

6. **Conclusion:**
More \(CS_2\) and \(H_2\) will be formed as the reaction proceeds forward.

### Final Answer:
The correct option is that the reaction will shift to the right, producing more \(CS_2\) and \(H_2\).

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