Certainly! Let's solve the problem step-by-step.
The equilibrium reaction is given by:
[tex]\[ \text{CO} (g) + 2 \text{H}_2 (g) \leftrightarrow \text{CH}_3\text{OH} (g) \][/tex]
The equilibrium constant ([tex]\( K_{eq} \)[/tex]) for this reaction is 26.
Given data:
- The equilibrium concentration of CO ([tex]\([ \text{CO} ]\)[/tex]) is [tex]\( 3.4 \times 10^{-1} M \)[/tex].
- The equilibrium concentration of [tex]\(\text{H}_2\)[/tex] ([tex]\([ \text{H}_2 ]\)[/tex]) is [tex]\( 6.8 \times 10^{-1} M \)[/tex].
We need to determine the equilibrium concentration of [tex]\(\text{CH}_3\text{OH}\)[/tex] ([tex]\([ \text{CH}_3\text{OH} ]\)[/tex]).
The expression for the equilibrium constant ([tex]\( K_{eq} \)[/tex]) for the given reaction is:
[tex]\[ K_{eq} = \frac{[ \text{CH}_3\text{OH} ]}{[ \text{CO} ] \times [ \text{H}_2 ]^2} \][/tex]
We are given [tex]\( K_{eq} = 26 \)[/tex], [tex]\([ \text{CO} ] = 0.34 M \)[/tex], and [tex]\([ \text{H}_2 ] = 0.68 M\)[/tex].
Let's substitute these values into the equilibrium expression to find [tex]\([ \text{CH}_3\text{OH} ]\)[/tex]:
[tex]\[ 26 = \frac{[ \text{CH}_3\text{OH} ]}{0.34 \times (0.68)^2} \][/tex]
First, calculate [tex]\( (0.68)^2 \)[/tex]:
[tex]\[ (0.68)^2 = 0.4624 \][/tex]
Now, substitute [tex]\( 0.4624 \)[/tex] back into the equation:
[tex]\[ 26 = \frac{[ \text{CH}_3\text{OH} ]}{0.34 \times 0.4624} \][/tex]
Next, calculate [tex]\( 0.34 \times 0.4624 \)[/tex]:
[tex]\[ 0.34 \times 0.4624 = 0.157216 \][/tex]
Now, the equation becomes:
[tex]\[ 26 = \frac{[ \text{CH}_3\text{OH} ]}{0.157216} \][/tex]
To find [tex]\([ \text{CH}_3\text{OH} ]\)[/tex], multiply both sides by [tex]\( 0.157216 \)[/tex]:
[tex]\[ [ \text{CH}_3\text{OH} ] = 26 \times 0.157216 \][/tex]
Finally, calculate the right-hand side:
[tex]\[ [ \text{CH}_3\text{OH} ] = 4.087616 \][/tex]
Therefore, the equilibrium concentration of [tex]\(\text{CH}_3\text{OH}\)[/tex] is [tex]\( 4.0876 \, M \)[/tex] (rounded to four decimal places).
