Given `[CS_(2)]=0.120M,[H_(2)]=0.10,[H_(2)S]=0.20 and [CH_(4)]=8.40xx10^(-5)M ` for the following reaction at `900^(@)`C at eq.
Calculate the equilibrium constant `(K_(c))`.
`CS_(2)(g) +4H_(2)(g) harr CH_(4) (g) + 2H_(2)S(g)`

Options

A. (a) `0.0120`
B. (b) `0.0980`
C. (c) `0.280`
D. (d) `0.120`

Answer

To calculate the equilibrium constant \( K_c \) for the reaction:

\[ CS_2(g) + 4H_2(g) \rightleftharpoons CH_4(g) + 2H_2S(g) \]

we will follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)

The expression for the equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[CH_4]^{\text{coeff.}} \cdot [H_2S]^{\text{coeff.}}}{[CS_2]^{\text{coeff.}} \cdot [H_2]^{\text{coeff.}}} \]

Where:
- The coefficients are the stoichiometric coefficients from the balanced equation.
- For this reaction, the coefficients are:
- \( [CH_4] \) has a coefficient of 1
- \( [H_2S] \) has a coefficient of 2
- \( [CS_2] \) has a coefficient of 1
- \( [H_2] \) has a coefficient of 4

Thus, the expression becomes:

\[ K_c = \frac{[CH_4]^1 \cdot [H_2S]^2}{[CS_2]^1 \cdot [H_2]^4} \]

### Step 2: Substitute the given concentrations into the expression

We are given the following concentrations:
- \( [CS_2] = 0.120 \, M \)
- \( [H_2] = 0.10 \, M \)
- \( [H_2S] = 0.20 \, M \)
- \( [CH_4] = 8.40 \times 10^{-5} \, M \)

Now substituting these values into the expression for \( K_c \):

\[ K_c = \frac{(8.40 \times 10^{-5})^1 \cdot (0.20)^2}{(0.120)^1 \cdot (0.10)^4} \]

### Step 3: Calculate the numerator and denominator

**Numerator:**
\[ (8.40 \times 10^{-5}) \cdot (0.20)^2 = (8.40 \times 10^{-5}) \cdot (0.04) = 3.36 \times 10^{-6} \]

**Denominator:**
\[ (0.120) \cdot (0.10)^4 = (0.120) \cdot (0.0001) = 1.2 \times 10^{-5} \]

### Step 4: Divide the numerator by the denominator

Now we can calculate \( K_c \):

\[ K_c = \frac{3.36 \times 10^{-6}}{1.2 \times 10^{-5}} = 0.28 \]

### Final Answer

Thus, the equilibrium constant \( K_c \) is:

\[ K_c = 0.28 \]

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