To find the numerical value of the equilibrium constant [tex]\( K \)[/tex] for the given reaction, we need to use the concentrations of the reactants and products at equilibrium:
The given reaction is:
[tex]\[
\text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g)
\][/tex]
The equilibrium constant expression ([tex]\( K \)[/tex]) for this reaction is given by:
[tex]\[
K = \frac{[\text{CO}] \cdot [\text{H}_2\text{O}]}{[\text{CO}_2] \cdot [\text{H}_2]}
\][/tex]
Where:
- [tex]\([\text{CO}_2]\)[/tex] is the concentration of [tex]\(\text{CO}_2\)[/tex]
- [tex]\([\text{H}_2]\)[/tex] is the concentration of [tex]\(\text{H}_2\)[/tex]
- [tex]\([\text{CO}]\)[/tex] is the concentration of [tex]\(\text{CO}\)[/tex]
- [tex]\([\text{H}_2\text{O}]\)[/tex] is the concentration of [tex]\(\text{H}_2\text{O}\)[/tex]
Given concentrations at equilibrium are:
- [tex]\([\text{CO}_2] = 0.32 \, \text{M}\)[/tex]
- [tex]\([\text{H}_2] = 0.035 \, \text{M}\)[/tex]
- [tex]\([\text{CO}] = 0.17 \, \text{M}\)[/tex]
- [tex]\([\text{H}_2\text{O}] = 0.29 \, \text{M}\)[/tex]
Substituting these values into the equilibrium constant expression:
[tex]\[
K = \frac{(0.17) \cdot (0.29)}{(0.32) \cdot (0.035)}
\][/tex]
Perform the multiplication in the numerator and the denominator first:
[tex]\[
\text{Numerator} = 0.17 \times 0.29 = 0.0493
\][/tex]
[tex]\[
\text{Denominator} = 0.32 \times 0.035 = 0.0112
\][/tex]
Now, divide the numerator by the denominator:
[tex]\[
K = \frac{0.0493}{0.0112} \approx 4.4017857
\][/tex]
Finally, round the result to two significant figures:
[tex]\[
K \approx 4.4
\][/tex]
Therefore, the numerical value of [tex]\( K \)[/tex] for the reaction is [tex]\( 4.4 \)[/tex].
