Home
Class 11
CHEMISTRY
Consider the reaction 2CO(g)+O(2)(g)hArr...

Consider the reaction `2CO(g)+O_(2)(g)hArr2CO_(2)(g)+Heat`
Under what conditions shift is undeterminable ?

A

Addition of `O_(2)` and decrease in volume

B

Addition of CO and removal of `CO_(2)` at constant volume

C

Increase in temperature and decrease in volume

D

Addition of CO and increase in temperature at constant volume

Text Solution

AI Generated Solution

The correct Answer is:
To determine the conditions under which the shift in the equilibrium of the reaction \[ 2CO(g) + O_2(g) \rightleftharpoons 2CO_2(g) + \text{Heat} \] is undeterminable, we will analyze the effects of various changes on the equilibrium position according to Le Chatelier's principle. ### Step-by-Step Solution: 1. **Understanding the Reaction**: - The reaction involves the conversion of carbon monoxide (CO) and oxygen (O₂) into carbon dioxide (CO₂) with the release of heat. This is an exothermic reaction. 2. **Identifying the Factors Affecting Equilibrium**: - According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift in a direction that counteracts the change. 3. **Analyzing Each Condition**: - **Condition 1**: Addition of oxygen and decrease in volume. - Adding oxygen increases the concentration of reactants, shifting the equilibrium to the right (towards products). - Decreasing volume shifts the equilibrium towards the side with fewer moles of gas (from 3 moles on the left to 2 moles on the right). - Both changes favor the formation of products, making the shift determinable. - **Condition 2**: Addition of carbon monoxide and removal of carbon dioxide at constant volume. - Adding CO increases reactant concentration, shifting equilibrium to the right. - Removing CO₂ decreases product concentration, shifting equilibrium to the right. - Both changes favor product formation, making the shift determinable. - **Condition 3**: Increase in temperature and decrease in volume. - Increasing temperature for an exothermic reaction shifts the equilibrium to the left (towards reactants). - Decreasing volume shifts the equilibrium to the right (towards products). - These changes are contradictory, making the shift undeterminable. - **Condition 4**: Addition of CO and increase in temperature at constant volume. - Adding CO shifts the equilibrium to the right (towards products). - Increasing temperature shifts the equilibrium to the left (towards reactants). - These changes are also contradictory, making the shift undeterminable. 4. **Conclusion**: - The conditions under which the shift is undeterminable are: - **Condition 3**: Increase in temperature and decrease in volume. - **Condition 4**: Addition of CO and increase in temperature at constant volume. ### Final Answer: The shift in equilibrium is undeterminable under the following conditions: - Increase in temperature and decrease in volume. - Addition of CO and increase in temperature at constant volume.
Promotional Banner

Similar Questions

Explore conceptually related problems

In the reaction: 2CO(g) +O_(2)(g) hArr 2CO(g) , entropy decreases.

K_(p)//K_(c) for the reaction CO(g)+1/2 O_(2)(g) hArr CO_(2)(g) is

K_(p)//K_(c) for the reaction CO(g)+1/2 O_(2)(g) hArr CO_(2)(g) is

For the reaction CO(g)+(1)/(2) O_(2)(g) hArr CO_(2)(g),K_(p)//K_(c) is

For the reaction, CO(g) +(1)/(2) O_2(g) hArr CO_2 (g), K_p//K_c is equal to

The energy profile diagram for the reaction: CO(g)+NO_(2)(g) hArr CO_(2)(g)+NO(g) is given below: the heat of the reaction is

The energy profile diagram for the reaction: CO(g)+NO_(2)(g) hArr CO_(2)(g)+NO(g) is given below: The activation energy of the forward reaction is

In the given reaction N_(2)(g)+O_(2)(g) hArr 2NO(g) , equilibrium means that

For the reaction CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g) at a given temperature, the equilibrium amount of CO_(2)(g) can be increased by

The energy profile diagram for the reaction: CO(g)+NO_(2)(g) hArr CO_(2)(g)+NO(g) is given below: The activation energy of the backward reaction is