The equilibrium constant for the following reaction is 10.5 at 500 K .A syatem at equilibrium has `[CO]=0.250M and [H_(2)]=0.120 M "what is the "[CH_(3)OH]`?
`CO(g)+2H_(2)(g)hArrCH_(3)OH(g)`

Options

A. `0.0378`
B. `0.435`
C. `0.546`
D. `0.0499`

Answer

To find the concentration of CH₃OH at equilibrium for the reaction:

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

given that the equilibrium constant \( K_c = 10.5 \) at 500 K, and the concentrations of CO and H₂ are 0.250 M and 0.120 M respectively, we can follow these steps:

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

The equilibrium constant expression for the reaction is given by:

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

### Step 2: Substitute the known values into the equation

We know that:

- \( K_c = 10.5 \)
- \( [\text{CO}] = 0.250 \, \text{M} \)
- \( [\text{H}_2] = 0.120 \, \text{M} \)

Substituting these values into the equation gives:

\[ 10.5 = \frac{[\text{CH}_3\text{OH}]}{(0.250)(0.120)^2} \]

### Step 3: Calculate \( [\text{H}_2]^2 \)

First, calculate \( [\text{H}_2]^2 \):

\[ [H_2]^2 = (0.120)^2 = 0.0144 \]

### Step 4: Substitute \( [\text{H}_2]^2 \) back into the equation

Now substitute this back into the equation:

\[ 10.5 = \frac{[\text{CH}_3\text{OH}]}{(0.250)(0.0144)} \]

### Step 5: Calculate the denominator

Calculate the denominator:

\[ (0.250)(0.0144) = 0.0036 \]

### Step 6: Solve for \( [\text{CH}_3\text{OH}] \)

Now, rearranging the equation to solve for \( [\text{CH}_3\text{OH}] \):

\[ [\text{CH}_3\text{OH}] = 10.5 \times 0.0036 \]

Calculating this gives:

\[ [\text{CH}_3\text{OH}] = 0.0378 \, \text{M} \]

### Conclusion

Thus, the concentration of CH₃OH at equilibrium is:

\[ [\text{CH}_3\text{OH}] = 0.0378 \, \text{M} \]