The dissociation of molecular iodine into iodine atoms is represented as: $I_{2} (g) ? 21 (g)$ At $1000.0 K$ , the equilibrium constant $K_{f}$ for the reaction is $3, 80 \times 1 0^{? 5}$ . Suppose you start with 0.0454 mol of $I_{2}$ in a $2.28 L$ . flask at $1000, 0 K$ . What are the concentrations of the gases at equibrium? Patt 1 of 2 What is the equilibrium concontration of 1,2 Round your answer to 3 significant digits M Pait 2 of 2 Whar is the enulibrium concentration of IP Round your answer to 3 segnificant dights.

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The dissociation of molecular iodine into iodine atoms is represented as:

I_{2} (g) ? 21 (g)

At

1000.0 K

, the equilibrium constant

K_{f}

for the reaction is

3, 80 \times 1 0^{? 5}

. Suppose you start with 0.0454 mol of

I_{2}

in a

2.28 L

. flask at

1000, 0 K

. What are the concentrations of the gases at equibrium? Patt 1 of 2 What is the equilibrium concontration of 1,2 Round your answer to 3 significant digits M Pait 2 of 2 Whar is the enulibrium concentration of IP Round your answer to 3 segnificant dights.

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