A) how will adding more s(s) shift the equilibrium?

Consider the following system at equilibrium. S(s)+O2(g)<--->SO2(g) 1. How will adding more S(s) shift the equilibrium? A) to the right B) to the left C) no effect 2. How will removing some SO2(g) shift the equilibrium? A) to the right B) to the left C) no effect 3. How will decreasing the volume of the container shift the equilibrium? A) to the right B) to the left C) no effect

Answer

General guidance

Concepts and reason

The concept used to solve this question is to determine the equilibrium shift due to changes in the conditions.

Fundamentals

According to Le Chatelier’s principle, if equilibrium of a system is disturbed, the position of the equilibrium shifts so as to counteract that change.

For an example, if the temperature of an exothermic reaction is increased, the equilibrium shifts so that the temperature of the reaction decreases. So, the reaction moves in backward direction.

If the pressure of the system is increased, the equilibrium shifts so as to decrease the pressure. The reaction proceeds to the side with fewer moles.

Step-by-step

Step 1 of 3

The given reaction is as follows:

S(s)+02(8)
SO2(g)

The expression for the equilibrium constant for the reaction is as follows:

K
(S0,7
[02]

Although is a reactant, it is not considered because it is a solid reactant.

When more is added, there is effect in the position of the equilibrium.

Therefore, there is no effect when more is added to the reaction.


Since, is a solid reactant, it is not considered for the equilibrium constant expression and there is no effect on the equilibrium position where more is added.

Step 2 of 3

The given reaction is as follows:

S(s)+02(8)
SO2(g)

The expression for the equilibrium constant for the reaction is as follows:

K
(S0,7
[02]

Although is a reactant, it is not considered because it is a solid reactant.

When SO2(g)is removed, the position of the equilibrium shifts to the right.

Therefore, when SO2(g)is removed from the reaction, the position of the equilibrium shifts to the right.


When SO2(g) is removed, the concentration of SO2(g) decreases. So, the reaction shifts so that more SO2(g) is formed. Hence, the equilibrium shifts to the right.

Step 3 of 3

The given reaction is as follows:

S(s)+02(8)
SO2(g)

The expression for the equilibrium constant for the reaction is as follows:

K
(S0,7
[02]

Although is a reactant, it is not considered because it is a solid reactant.

When the volume of the container is decreased, there is no effect in the position of the equilibrium.

Therefore, there is no effect when the volume of the container is decreased.


When the volume of the container is decreased, the pressure on the gas molecules increases. So, the equilibrium shifts to the side with fewer moles. Since both sides of the reaction have same number of moles, there is no effect on the equilibrium position.

Answer

Therefore, there is no effect when more is added to the reaction.

Therefore, when SO2(g)is removed from the reaction, the position of the equilibrium shifts to the right.

Therefore, there is no effect when the volume of the container is decreased.

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