Answer:
Explanation:To solve this, we'll use the equilibrium constant expression and the given equilibrium concentrations:
(a) For the reaction: \( \text{CO}_2 (g) + \text{H}_2 (g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g) \)
The equilibrium constant expression is:
\[ K = \frac{{[\text{CO}][\text{H}_2\text{O}]}}{{[\text{CO}_2][\text{H}_2]}} \]
Given:
\[ K = 0.53 \]
\[ [\text{CO}] = 0.25 \text{ moles} \]
\[ [\text{H}_2] = 0.6 \text{ moles} \]
Let's denote the number of moles of H₂O as \( x \). Substituting the given values into the equilibrium constant expression:
\[ 0.53 = \frac{{x}}{{0.25 \times 0.6}} \]
Solving for \( x \), we get:
\[ x = 0.53 \times 0.25 \times 0.6 = 0.0795 \text{ moles} \]
So, there are 0.0795 moles of H₂O in the vessel.
