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

Options

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

Answer

To find 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 expression for \( K_p \) is given by the ratio of the partial pressures of the products to the reactants, each raised to the power of their respective coefficients in the balanced equation. For the given reaction, the expression for \( K_p \) is:

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

### Step 2: Identify the units of partial pressure

The unit of partial pressure is typically expressed in atmospheres (atm).

### Step 3: Substitute the units into the \( K_p \) expression

Now, substituting the units into the \( K_p \) expression, we have:

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

### Step 4: Simplify the expression

Now, simplify the expression:

- The numerator becomes \( atm^3 \) (since \( atm \cdot atm^2 = atm^3 \)).
- The denominator becomes \( atm^5 \) (since \( atm \cdot atm^4 = atm^5 \)).

So, we have:

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

### Conclusion

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

\[ \text{atm}^{-2}. \]