Using moler concentrations, what is the unit of `K_(c)` for the reaction ?
`CH_(3)OH(g)hArrCO(g)+2H_(2)(g)`

Options

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

Answer

To determine the unit of \( K_c \) for the reaction

\[ \text{CH}_3\text{OH}(g) \rightleftharpoons \text{CO}(g) + 2\text{H}_2(g), \]

we will follow these steps:

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

The equilibrium constant \( K_c \) is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.

For the given reaction, the expression for \( K_c \) is:

\[ K_c = \frac{[\text{CO}]^{1} [\text{H}_2]^{2}}{[\text{CH}_3\text{OH}]^{1}} \]

### Step 2: Identify the units of concentration

The concentration of a substance in this context is measured in molarity (M), which is equivalent to moles per liter (mol/L).

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

Now, we can substitute the units into the \( K_c \) expression:

- The concentration of CO is in molarity: \([\text{CO}] = \text{M}\)
- The concentration of \( H_2 \) is also in molarity: \([\text{H}_2] = \text{M}\)
- The concentration of \( \text{CH}_3\text{OH} \) is in molarity: \([\text{CH}_3\text{OH}] = \text{M}\)

Substituting these into the expression for \( K_c \):

\[ K_c = \frac{[\text{CO}]^{1} [\text{H}_2]^{2}}{[\text{CH}_3\text{OH}]^{1}} = \frac{(\text{M})^{1} \cdot (\text{M})^{2}}{(\text{M})^{1}} \]

### Step 4: Simplify the units

Now, simplify the units:

\[ K_c = \frac{\text{M}^1 \cdot \text{M}^2}{\text{M}^1} = \frac{\text{M}^3}{\text{M}^1} = \text{M}^{3-1} = \text{M}^2 \]

### Conclusion

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

\[ \text{M}^2 \text{ or } \text{molar}^2 \]