The dissociation of molecular iodine into iodine atoms is represented as $1_{2} (g) ? 21 (g)$ At $1000.0 K$ , the equilibrium constant $K_{e}$ for the reaction is $3.80 \times 1 0^{? 5}$ . Suppose you start with $0.0456 mol$ of $I_{2}$ in a $2.34 L$ flark at $1000.0 K$ . What are the concentrations of the gases at equilibrlum? Part 1 of 2 . What is the equilibrlum concentration of $I_{2}$ ? Round your answer to 3 significant digits. $M$ Part 2 of 2 What is the equibrium concentration of 1? Round your answer to 3 significant digits:

Question

The dissociation of molecular iodine into iodine atoms is represented as

1_{2} (g) ? 21 (g)

At

1000.0 K

, the equilibrium constant

K_{e}

for the reaction is

3.80 \times 1 0^{? 5}

. Suppose you start with

0.0456 mol

of

I_{2}

in a

2.34 L

flark at

1000.0 K

. What are the concentrations of the gases at equilibrlum? Part 1 of 2 . What is the equilibrlum concentration of

I_{2}

? Round your answer to 3 significant digits.

M

Part 2 of 2 What is the equibrium concentration of 1? Round your answer to 3 significant digits:

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