Question

H2(g), I2(g), and HI(g) are at equilibrium at a different temperature in a different vessel. (c) When the temperature in the vessel is decreased, does the equilibrium shift to the right, favoring the product, or to the left, favoring the reactants? Justify your answer. Rigid vessel

Answer 1

Answer:

The reaction will shift rightwards favoring the product.

Explanation:

Hello,

In this case, considering the chemical reaction to be:

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g) ;\Delta H^o rxn=-9.4kJ/mol;Kc=49$

We notice that the sign of the change in the enthalpy of the reaction is negative, therefore this an exothermic chemical reaction in which the heat is a product as the reaction vessel heats up as the reaction undergoes. In such a way, by decreasing the temperature we are removing heat and by means of the Le Chatelier's principle, when removing a product, the reaction will shift rightwards favoring the product, HI.

Best regards.

Answer 2

Answer:

the balance will shift to the right, favoring the products. More HI will be formed.

Explanation:

Step 1: Data given

H2(g), I2(g), and HI(g) are at equilibrium at a different temperature in a different vessel.

Step 2: The balanced equation

H2(g) + I2(g) ⇆ 2HI(g)

Step 3:

If the temperature is increased, the system will release less heat as reaction.

Since formation of HI is exothermic

When the temperature drops, the system will produce more heat. With the result, the balance will shift to the right, favoring the products. More HI will be formed.