Which of the following are true statements about equilibrium systems? for the following reaction at equilibrium: 2 h2(g) + o2(g) ? 2 h2o(g)the equilibrium will shift to the left if the volume is doubled. for the following reaction at equilibrium: h2(g) + f2(g) ? 2hf(g)removing h2 will decrease the amount of f2 present once equilibrium is reestablished. increasing the temperature of an exothermic reaction shifts the equilibrium position to the right. for the following reaction at equilibrium: caco3(s) ? cao(s) + co2(g)adding more caco3 will shift the equilibrium to the right. for the following reaction at equilibrium: caco3(s) ? cao(s) + co2(g)increasing the total pressure by adding ar(g) will have no effect on the equilibrium position.

These are five questions about equilibrium systems each with its complete answer.

Question 1. For the following reaction at equilibrium:

2 H₂(g) + O₂(g) ⇄ 2 H2O(g),  the equilibrium will shift to the left if the volume is doubled?

TRUE

Explanation:

When a force disturbs a chemical equillibrium, the system will shift toward the direction that reduces the effect. This is Le Chatelier's principle.

As per Bolye's law, at constant temperature, the volume and the pressure of a fixed amount of gas are inversely related.

Also, the pressure of the system is directly related to the number of particles (atoms or molecules). Hence, more molecules, more pressure; less molecules, less pressure.

Now, you can reason in this way: if the volume of the given system is doubled, then the pressure is lowered, and the system will try to alleviate this disturbance by shifting the reaction to the side that produces more molecules, to restore the pressure.  Because on the left side three molecules can be produced from the reaction of two molecules of H₂O on the rihgt, the system will shift to the left. And this proves the truth of the statement.

Question 2. For the following reaction at equilibrium:

H₂(g) + F₂(g) ⇄  2HF(g), removing H₂ will decrease the amount of F₂ present once equilibrium is reestablished.

FALSE.

Explanation:

Note that, since the temperature and other conditions have not changed, the equilibrium constant, Ke, has not changed. And, for the given equilibrium, Ke is given by the following equation.

Hence, to keep Ke unchanged, when removing H₂, the amount of F₂ present once equilibrium is reestablished will have to increase.

This is the opposite of the stated on the question, so the statement is false.

Question 3. Increasing the temperature of an exothermic reaction shifts the equilibrium position to the right.

FALSE.

Explanation:

You can write an exothermic equlibrium placing heat as a product on the right side of the equation; in this way:

There, treating the heat as another product, you can reason that increasing the temperature, which is equivalent to supplying heat, will shift the equilibrium to the left side to consume heat, instead to the proposed by the statement. So, this is a false statement.

Question 4. For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), adding more CaCO₃ will shift the equilibrium to the right.

TRUE.

Explanation:

CaCO₃(g) is the only reactant of the forward reaction.

Adding more CaCO₃ may be seen as a disturbance against which the system will act by consuming it and producing more CaO and CO₂.

So, the forward reation will be favored and you conclude that adding more CaCO₃ will shift the equilibrium to the right.

Question 5. For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), increasing the total pressure by adding Ar(g) will have no effect on the equilibrium position.

TRUE.

Explanation:

In accordance to Le Chatelier's principle, increasing the pressure should be addresed by the equilibrium by shifting to the side where such pressure increase could be released.

That is possible when the number of molecules of gases on both sides are different: the equilibrium will shift to the side where more molecules less molecules are produced.

But, when the stoichiometry of the reaction shows the same number of molecules on both sides, which is the case in the given equilibrium, increasiing (or decreasing) the pressure will have no effect on the equilibrium position. Then, the answer is true.

5.6⋅10−^19  j

do you need the explaination?

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.