To solve this problem, we can use the equilibrium constant expression:

The reaction is: $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$

The equilibrium constant expression is: [ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} ]

We are given:

• $K_c = 44.0$
• $[\text{HI}] = 0.50 \, M$ at equilibrium
• The initial concentrations of $[\text{H}_2]$ and $[\text{I}_2]$ are the same, say $x$.

Steps:

1. Define the changes in concentration: Let the decrease in $[\text{H}_2]$ and $[\text{I}_2]$ be $y$. Since the reaction produces $2\text{HI}$, the increase in $[\text{HI}]$ is $2y$.

   At equilibrium: [ [\text{H}_2] = x - y, \quad [\text{I}_2] = x - y, \quad [\text{HI}] = 2y ]

2. Relate equilibrium concentrations to $K_c$: At equilibrium: [ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} ]

   Substitute the expressions: $44.0 = \frac{(0.50)^2}{(x - y)(x - y)}$

3. Determine $y$: Since $[\text{HI}] = 0.50 \, M$, $2y = 0.50$, so $y = 0.25$.

4. Substitute $y$ into the equation: $44.0 = \frac{(0.50)^2}{(x - 0.25)(x - 0.25)}$ Simplify: $44.0 = \frac{0.25}{(x - 0.25)^2}$

   Multiply through by $(x - 0.25)^2$: $44.0(x - 0.25)^2 = 0.25$

5. Solve for $x$: Divide both sides by 44.0: $(x - 0.25)^2 = \frac{0.25}{44.0}$ Simplify: $(x - 0.25)^2 = 0.00568$

   Take the square root: $x - 0.25 = \sqrt{0.00568}$

   $x - 0.25 \approx 0.0753$

   $x \approx 0.25 + 0.0753 = 0.3253$

6. Equilibrium concentration of $\text{H}_2$: At equilibrium: $[\text{H}_2] = x - y = 0.3253 - 0.25 = 0.0753 \, M$

Final Answer:

The equilibrium concentration of $\text{H}_2$ is approximately 0.075 M.

Let me know if you'd like me to elaborate or if you have additional questions!

Related Questions:

1. How does the equilibrium constant $K_c$ depend on temperature?
2. What would happen to the equilibrium concentrations if $K_c$ were smaller?
3. How does the initial concentration of reactants affect the final equilibrium concentrations?
4. Can $K_c$ predict whether reactants or products dominate at equilibrium?
5. How would the introduction of a catalyst affect the equilibrium concentrations?

Tip:

Always double-check the units and ensure that all equilibrium concentrations are in the same units before substituting into the $K_c$ expression.