Four substances, HCl (g), I2 (s), HI (g), and Cl2 (g), are mixed in a reaction vessel and allowed to reach equilibrium in the reaction: 2HCl (g) + I2 (s) ⇄ 2HI (g) + Cl2 (g) Which of the following changes will result in an increased amount of HI (g) when equilibrium is reached again? X. Add HCl (g) to the reaction vessel. Y. Add I2 (s) to the reaction vessel. Z. Increase the volume of the reaction vessel. A. X only B. X and Y only C. X and Z only D. Z only E. X, Y, and Z

Question

Emely West

Chemistry

3 December, 00:55

Four substances, HCl (g), I2 (s), HI (g), and Cl2 (g), are mixed in a reaction vessel and allowed to reach equilibrium in the reaction: 2HCl (g) + I2 (s) ⇄ 2HI (g) + Cl2 (g) Which of the following changes will result in an increased amount of HI (g) when equilibrium is reached again? X. Add HCl (g) to the reaction vessel. Y. Add I2 (s) to the reaction vessel. Z. Increase the volume of the reaction vessel. A. X only B. X and Y only C. X and Z only D. Z only E. X, Y, and Z

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Answers (1)

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Amari Mendoza · Answer 1

B. X and Y only

Explanation:

Hello,

In this case, by considering the Le Chatelier's principle, if we desire an increase in the hydrogen iodide amount, based on the given reaction, two procedures are needed:

- Add HCl (g) to the reaction vessel: because the addition of a reagent shifts the reaction rightwards.

- Add I2 (s) to the reaction vessel: by cause of the same aforesaid reason.

Therefore, the answer is: B. X and Y only.

Z statement is not true since the change in the mole number for the reaction is zero (2+1-2-1=0) considering each species' stoichiometric coefficients.

Best regards.