QuestionPart B Review I Constants I Periodic Table
When a chemical reaction is at equilibrium, $Q$ (the reaction quotient) is equal to $K$ (the equilibrium constant). If a stress is applied to the mixture that changes the value of $Q$ , then the system is no longer at equilibrium. To regain equilibrium, the reaction will either proceed forward or in reverse until $Q$ is equal to $K$ once again. Alternatively, equilibrium can be disrupted by a change in temperature, which changes the value of $K$ . The result however is the same, and the reaction will proceed forward or in reverse until $Q$ is equal to the new $K$ . Le Châtelier's principle summarizes this idea:
If a stress is applied to a reaction mixture at equilibrium, a net reaction occurs in the direction that relieves the stress. - The following system is at equilibrium: $\mathrm{A}(\mathrm{s})+4 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons \mathrm{C}(\mathrm{g})$ Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction no shift in the direction of the net reaction.
Drag the appropriate items to their respective bins. View Available Hint(s) Reset Help
Halve the volume Remove some A Double the volume Add more A
Leftward shift Rightward shift No shift

Studdy Solution

STEP 1

What is this asking? We need to figure out how changing things like volume and amount of a solid reactant will shift the equilibrium of a given chemical reaction. Watch out! Don't forget that solids don't affect the equilibrium position!
Also, remember that changing the volume affects the concentrations of gases, so be mindful of the number of gas molecules on each side of the equation.

STEP 2

1. Analyze the reaction
2. Halve the volume
3. Remove some A
4. Double the volume
5. Add more A

STEP 3

We're given the reaction: $A(s) + 4B(g) \rightleftharpoons C(g)$ .
This tells us that one mole of solid A reacts with four moles of gaseous B to produce one mole of gaseous C.

STEP 4

When we halve the volume, we're effectively increasing the pressure.
Le Châtelier's principle tells us the system will try to reduce this stress by shifting towards the side with fewer gas molecules.

STEP 5

There are four moles of gas on the left side and one mole of gas on the right side.
So, halving the volume will cause a rightward shift towards the side with fewer gas molecules.

STEP 6

Since A is a solid, changing its amount doesn't affect the equilibrium position.
Think of it like this: the concentration of a solid is constant!

STEP 7

Therefore, removing some A will cause no shift.

STEP 8

Doubling the volume effectively decreases the pressure.
The system will try to increase the pressure by shifting to the side with more gas molecules.

STEP 9

Since there are four moles of gas on the left and one mole on the right, doubling the volume will cause a leftward shift.

STEP 10

Just like removing A, adding more of a solid reactant won't change the equilibrium position.
The concentration of a solid is constant.

STEP 11

So, adding more A will cause no shift.

STEP 12

* Leftward shift: Double the volume * Rightward shift: Halve the volume * No shift: Remove some A, Add more A

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Studdy Solution

STEP 1

STEP 2

1. Analyze the reaction
2. Halve the volume
3. Remove some A
4. Double the volume
5. Add more A

STEP 3

We're given the reaction: $A(s) + 4B(g) \rightleftharpoons C(g)$ .
This tells us that one mole of solid A reacts with four moles of gaseous B to produce one mole of gaseous C.

STEP 4

When we halve the volume, we're effectively increasing the pressure.
Le Châtelier's principle tells us the system will try to reduce this stress by shifting towards the side with fewer gas molecules.

STEP 5

There are four moles of gas on the left side and one mole of gas on the right side.
So, halving the volume will cause a rightward shift towards the side with fewer gas molecules.

STEP 6

Since A is a solid, changing its amount doesn't affect the equilibrium position.
Think of it like this: the concentration of a solid is constant!

STEP 7

Therefore, removing some A will cause no shift.

STEP 8

Doubling the volume effectively decreases the pressure.
The system will try to increase the pressure by shifting to the side with more gas molecules.

STEP 9

Since there are four moles of gas on the left and one mole on the right, doubling the volume will cause a leftward shift.

STEP 10

Just like removing A, adding more of a solid reactant won't change the equilibrium position.
The concentration of a solid is constant.

STEP 11

So, adding more A will cause no shift.

STEP 12

* Leftward shift: Double the volume * Rightward shift: Halve the volume * No shift: Remove some A, Add more A

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Studdy Solution

STEP 1

STEP 2

1. Analyze the reaction2. Halve the volume3. Remove some A4. Double the volume5. Add more A

STEP 3

We're given the reaction: A(s)+4B(g)⇌C(g)A(s) + 4B(g) \rightleftharpoons C(g)A(s)+4B(g)⇌C(g).This tells us that **one mole** of solid A reacts with **four moles** of gaseous B to produce **one mole** of gaseous C.

STEP 4

When we **halve the volume**, we're effectively **increasing the pressure**.Le Châtelier's principle tells us the system will try to **reduce this stress** by shifting towards the side with **fewer gas molecules**.

STEP 5

There are **four moles** of gas on the left side and **one mole** of gas on the right side.So, halving the volume will cause a **rightward shift** towards the side with fewer gas molecules.

STEP 6

Since A is a **solid**, changing its amount *doesn't affect the equilibrium position*.Think of it like this: the concentration of a solid is constant!

STEP 7

Therefore, removing some A will cause **no shift**.

STEP 8

**Doubling the volume** effectively **decreases the pressure**.The system will try to **increase the pressure** by shifting to the side with **more gas molecules**.

STEP 9

Since there are **four moles** of gas on the left and **one mole** on the right, doubling the volume will cause a **leftward shift**.

STEP 10

Just like removing A, *adding* more of a solid reactant won't change the equilibrium position.The concentration of a solid is constant.

STEP 11

So, adding more A will cause **no shift**.

STEP 12

* **Leftward shift:** Double the volume * **Rightward shift:** Halve the volume * **No shift:** Remove some A, Add more A

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Analyze the reaction
2. Halve the volume
3. Remove some A
4. Double the volume
5. Add more A

We're given the reaction: $A(s) + 4B(g) \rightleftharpoons C(g)$ .
This tells us that one mole of solid A reacts with four moles of gaseous B to produce one mole of gaseous C.

When we halve the volume, we're effectively increasing the pressure.
Le Châtelier's principle tells us the system will try to reduce this stress by shifting towards the side with fewer gas molecules.

There are four moles of gas on the left side and one mole of gas on the right side.
So, halving the volume will cause a rightward shift towards the side with fewer gas molecules.

Since A is a solid, changing its amount doesn't affect the equilibrium position.
Think of it like this: the concentration of a solid is constant!

Therefore, removing some A will cause no shift.

Doubling the volume effectively decreases the pressure.
The system will try to increase the pressure by shifting to the side with more gas molecules.

Since there are four moles of gas on the left and one mole on the right, doubling the volume will cause a leftward shift.

Just like removing A, adding more of a solid reactant won't change the equilibrium position.
The concentration of a solid is constant.

So, adding more A will cause no shift.

* Leftward shift: Double the volume * Rightward shift: Halve the volume * No shift: Remove some A, Add more A