Suppose the reaction system below has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected.

UO2(s) + 4 HF(g) ↔ UF4(g) + 2 H2O(g)

1. Water vapor is removed

2. UF4(g) is added

3. The reaction is done in a glass reaction vessel. HF(g) attacks and reacts with the glass

4. If the reaction is endothermic, and you want to make the equilibrium shift to the right to maximize the products, would you heat or cool the reaction vessel?

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

1. Water vapor is removed

On removal of water vapor from equilibrium will decrease the product which will lead to moving forward of an equilibrium that is equilibrium will shift to right direction.

2. is added

On addition of from equilibrium will increase the reactant which will lead to moving forward of an equilibrium that is equilibrium will shift to right direction.

3.The reaction is done in a glass reaction vessel. HF(g) attacks and reacts with the glass

Due to reaction of HF with glass will result in decrease in amount of HF present at equilibrium. So, order to counter this equilibrium will shift in backward direction or in left direction.

4. If the reaction is endothermic, and you want to make the equilibrium shift to the right to maximize the products, would you heat or cool the reaction vessel.

I endothermic reaction , heat is added to the reaction for which we can treat heat as a reactant.Now, we want to shift the equilibrium to right which means in the forward direction.

And this can be done by increasing the reactant amount, here by adding more heat to the equilibrium will shift the reaction to the right side.

So, for increasing heat we will heat the reaction vessel.

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'