At equilibrium, what happens if CO2 is added to the reaction mixture at constant temperature and volume? (Select all that apply.)
The reaction absorbs energy.
The reaction releases energy.
[H2O] increases.
[H2O] decreases.
[CO] increases.
[CO] remains constant.
[CO2] decreases.
[H2] decreases.
[H2] remains constant.
H2O(g) + CO(g) equilibrium reaction arrow CO2(g) + H2(g) The forward reaction above is exothermic.?
2016-02-25 3:02 pm
回答 (1)
2016-02-25 4:46 pm
H₂O(l) + CO(g) ⇌ CO₂(g) + H₂(g) .... ΔH = -ve
CO₂ is on the right hand side of the equation.
When CO₂ is added to the system at equilibrium, according to Le Chatelier's principle, the equilibrium position shifts to the left.
As the backward reaction is endothermic.
• The reaction absorbs energy.
Because the equilibrium position shits to the left, the concentrations of the components on the left increase, and that of the components on the right decrease, i.e.
• [H2O] increases. (H₂O is the component on the left.)
• [CO] increases. (CO is the component on the left.)
• [CO₂] decreases. (CO₂ is the component on the right.)
• [H₂] decreases. (H₂ is the component on the right.)
CO₂ is on the right hand side of the equation.
When CO₂ is added to the system at equilibrium, according to Le Chatelier's principle, the equilibrium position shifts to the left.
As the backward reaction is endothermic.
• The reaction absorbs energy.
Because the equilibrium position shits to the left, the concentrations of the components on the left increase, and that of the components on the right decrease, i.e.
• [H2O] increases. (H₂O is the component on the left.)
• [CO] increases. (CO is the component on the left.)
• [CO₂] decreases. (CO₂ is the component on the right.)
• [H₂] decreases. (H₂ is the component on the right.)
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