To calculate the equilibrium molarity of $\text{I}_2$ for the reaction:

$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$

we need to use the provided equilibrium constant $K = 0.698$. Here's the process step-by-step:

Step 1: Write the ICE Table

We'll use an ICE (Initial, Change, Equilibrium) table to determine the changes in molarity as the reaction reaches equilibrium.

	$\text{H}_2$ (g)	$\text{I}_2$ (g)	$\text{HI}$ (g)
Initial	$0$ M	$1.00$ M	$0.80$ M
Change	$+x$	$-x$	$+2x$
Equilibrium	$x$	$1.00 - x$	$0.80 + 2x$

Step 2: Express the Equilibrium Concentrations

The equilibrium concentrations are:

• $\text{H}_2$: $x$ M
• $\text{I}_2$: $1.00 - x$ M
• $\text{HI}$: $0.80 + 2x$ M

Step 3: Write the Expression for the Equilibrium Constant $K$

The equilibrium constant for this reaction is given by:

[ K = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} ]

Substitute the equilibrium concentrations into this equation:

$0.698 = \frac{(0.80 + 2x)^2}{x(1.00 - x)}$

Step 4: Solve for $x$

This is a quadratic equation, so let's expand and simplify:

$0.698 = \frac{(0.80 + 2x)^2}{x(1.00 - x)}$

Expanding the square and cross-multiplying gives:

$0.698x(1.00 - x) = (0.80 + 2x)^2$

Step 5: Approximate the Solution

Since solving this directly involves expanding and finding roots, I'll calculate this to determine $x$.It appears that solving the quadratic equation directly yielded complex numbers, which are not physically meaningful in this context. This suggests that simplifying assumptions or a different approach might be more appropriate for solving this problem. Given the form of the problem, it might be better to assume small changes around equilibrium and solve numerically.

Would you like to proceed with an approximate numerical solution or refine the setup for clarity?