Question

Incorrect Your answer is incorrect. Suppose a 500. mL flask is filled with 0.30 mol of $I_2$ and 2.0 mol of HI. The following reaction becomes possible: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ The equilibrium constant $K$ for this reaction is 0.342 at the temperature of the flask. Calculate the equilibrium molarity of HI. Round your answer to two decimal places. 4.00 M

Name: incorrect your answer is incorrect suppose a 500 ml flask is filled with 030 mol of i2 and 20 mol of hi the following reaction becomes possible h2g i2g rightleftharpoons 2hig the equilibr 69664
Uploaded: 2023-07-24T04:27:15-08:00
Duration: 4 min 8 s
Channel: Madhur L
Description: incorrect your answer is incorrect suppose a 500 ml flask is filled with 030 mol of i2 and 20 mol of hi the following reaction becomes possible h2g i2g rightleftharpoons 2hig the equilibr 69664

          Incorrect
Your answer is incorrect.
Suppose a 500. mL flask is filled with 0.30 mol of $I_2$ and 2.0 mol of HI. The following reaction becomes possible:
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$
The equilibrium constant $K$ for this reaction is 0.342 at the temperature of the flask.
Calculate the equilibrium molarity of HI. Round your answer to two decimal places.
4.00 M

incorrect your answer is incorrect suppose a 500 ml flask is filled with 030 mol of i2 and 20 mol of hi the following reaction becomes possible h2g i2g rightleftharpoons 2hig the equilibr 69664

Added by Richard B.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

07/24/2023

Step 1

The volume of the flask is 500 mL = 0.500 L. Initial molarity of $I_2$ = $\frac{0.30 \text{ mol}}{0.500 \text{ L}} = 0.60 \text{ M}$ Initial molarity of HI = $\frac{2.0 \text{ mol}}{0.500 \text{ L}} = 4.0 \text{ M}$ Initial molarity of $H_2$ = 0 M Show more…

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Incorrect Your answer is incorrect. Suppose a 500. mL flask is filled with 0.30 mol of $I_2$ and 2.0 mol of HI. The following reaction becomes possible: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ The equilibrium constant $K$ for this reaction is 0.342 at the temperature of the flask. Calculate the equilibrium molarity of HI. Round your answer to two decimal places. 4.00 M