What is the unit of `K_(p)` for the reaction ?
`CS_(2)(g)+4H_(4)(g)hArrCH_(4)(g)+2H_(2)S(g)`

Options

A. atm
B. `atm^(-2)`
C. `atm^(2)`
D. `atm^(-1)`

Answer

To determine the unit of \( K_p \) 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 \( K_p \)

The equilibrium constant \( K_p \) is defined in terms of the partial pressures of the products and reactants. For the given reaction, the expression for \( K_p \) is:

\[ K_p = \frac{(P_{CH_4})(P_{H_2S})^2}{(P_{CS_2})(P_{H_2})^4} \]

where \( P_{CH_4} \), \( P_{H_2S} \), \( P_{CS_2} \), and \( P_{H_2} \) are the partial pressures of methane, hydrogen sulfide, carbon disulfide, and hydrogen, respectively.

### Step 2: Identify the powers in the expression

From the expression for \( K_p \):
- The partial pressure of \( CH_4 \) is raised to the power of 1.
- The partial pressure of \( H_2S \) is raised to the power of 2.
- The partial pressure of \( CS_2 \) is raised to the power of 1.
- The partial pressure of \( H_2 \) is raised to the power of 4.

### Step 3: Substitute the units of pressure

The unit of pressure is typically expressed in atmospheres (atm). Therefore, we substitute the units into the \( K_p \) expression:

\[ K_p = \frac{(atm)^1 \cdot (atm)^2}{(atm)^1 \cdot (atm)^4} \]

### Step 4: Simplify the expression

Now, we simplify the expression:

\[ K_p = \frac{(atm)^{1+2}}{(atm)^{1+4}} = \frac{(atm)^3}{(atm)^5} \]

This simplifies to:

\[ K_p = (atm)^{3-5} = (atm)^{-2} \]

### Conclusion

Thus, the unit of \( K_p \) for the given reaction is:

\[ \text{Unit of } K_p = atm^{-2} \]

### Final Answer

The correct option is (b) \( atm^{-2} \).

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