# DQ5: Chemical Equilibrium**Available until Sunday, June 28, 2020 12:59 PM CDT** **Must post first.**This discussion fulfills the objective #6 of chapter 15.Apply Le Chatelier's principle and decide how (by altering temperature, pressure, etc.) you could increase the amount of the products in the following reactions:1. $2CO_2(g) \leftrightharpoons 2CO(g) + O_2(g) \quad \Delta H^{0} = 566 \, \text{kJ} $2. $2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g) $**[Start a New Thread]**_Filter by:_ All Threads _Sort by:_ Oldest Thread_You must start a thread before you can read and reply to other threads_ --- ### Explanation:**Le Chatelier's Principle:** Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.### Reaction 1:\[ 2CO_2(g) \leftrightharpoons 2CO(g) + O_2(g) \quad \Delta H^{0} = 566 \, \text{kJ} \]For this reaction:- **Increase Temperature:** Since the reaction is endothermic ($\Delta H > 0$), increasing the temperature will push the equilibrium towards the products (right).- **Changing Pressure:** The number of moles of gases increases from 2 to 3 from reactants to products. Decreasing pressure will favor the formation of more moles of gas (move to the right).### Reaction 2:\[ 2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g) \]For this reaction:- **Increase Pressure:** Since there are fewer moles of gas on the product side (3 to 2), increasing pressure will push the equilibrium towards the products (right).- **Temperature:** The effect of temperature is not given due to lack of $\Delta H$ information.Ensure you start a thread before reading and replying to other threads.

A reaction is said to be in equilibrium if the rate of the forward reaction is equal to the rate of…

Apply Le Chatelier's principle and decide how (by altering temperature, pressure, etc.) you could increase the amount of the products in the following reactions: 1) 2CO2 (g) ----> 2CO (g) + O2 (g) AHº=566 kJ 2) 2SO2 (g) + O2 (g) ----> 2SO3 (g)

Apply Le Chatelier's principle and decide how (by altering temperature, pressure, etc.) you could increase the amount of the products in the following reactions: 1) 2CO2 (g) ----> 2CO (g) + O2 (g) AHº=566 kJ 2) 2SO2 (g) + O2 (g) ----> 2SO3 (g)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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# DQ5: Chemical Equilibrium

**Available until Sunday, June 28, 2020 12:59 PM CDT**
**Must post first.**

This discussion fulfills the objective #6 of chapter 15.

Apply Le Chatelier's principle and decide how (by altering temperature, pressure, etc.) you could increase the amount of the products in the following reactions:

1. $2CO_2(g) \leftrightharpoons 2CO(g) + O_2(g) \quad \Delta H^{0} = 566 \, \text{kJ} $
2. $2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g) $

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---

### Explanation:

**Le Chatelier's Principle:**

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

### Reaction 1:
\[ 2CO_2(g) \leftrightharpoons 2CO(g) + O_2(g) \quad \Delta H^{0} = 566 \, \text{kJ} \]

For this reaction:
- **Increase Temperature:** Since the reaction is endothermic ($\Delta H > 0$), increasing the temperature will push the equilibrium towards the products (right).
- **Changing Pressure:** The number of moles of gases increases from 2 to 3 from reactants to products. Decreasing pressure will favor the formation of more moles of gas (move to the right).

### Reaction 2:
\[ 2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g) \]

For this reaction:
- **Increase Pressure:** Since there are fewer moles of gas on the product side (3 to 2), increasing pressure will push the equilibrium towards the products (right).
- **Temperature:** The effect of temperature is not given due to lack of $\Delta H$ information.

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