September 23, 2021, 02:08:31 AM
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Topic: If system 2CO + O2 ----> 2 CO2 came to equilibrium, and more CO was added...  (Read 19877 times)

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Offline petstar21

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 If system 2CO + O2 ----> 2 CO2 came to equilibrium, and more CO was added, why does CO2 increase and O2 decrease??

If pressure on the equilibrium system above increases, does the quantity of CO2 increase, decraese or stay the same and why?? (how does pressure affect it)

Offline LQ43

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Offline Gerard

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If system 2CO + O2 ----> 2 CO2 came to equilibrium, and more CO was added, why does CO2 increase and O2 decrease??

If pressure on the equilibrium system above increases, does the quantity of CO2 increase, decraese or stay the same and why?? (how does pressure affect it)
-more products are formed since you put more reactants to the system, recall that partial pressure of a gas is related to its molar percenatge by the equation: Pa=Pt x Ya
there for if we increase the partial pressure of the CO (reactant) more CO2 (products) are produced and more O2 is consumed..
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Offline wilson

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If system 2CO + O2 ----> 2 CO2 came to equilibrium, and more CO was added, why does CO2 increase and O2 decrease??

If pressure on the equilibrium system above increases, does the quantity of CO2 increase, decraese or stay the same and why?? (how does pressure affect it)
If more (moles of) CO was added, the concentration of CO also increases, if the volume remains a constant. By Le Chatelier's Principle, to oppose the change/stress, i.e. we want to decrease the concentration of CO, the equilibrium shifts to the right side. Thus, more products (CO2) will be produced.

If you think about it logically at the molecular level, if you increase the concentration of CO, the probability of successful collision between CO and O2 is more likely than 2 molecules of CO2 colliding.

For the second part of the question, just apply the same principle. In this case, it's change of pressure. Which side will the chemical equilibrium lie to in order to decrease the pressure? Look at the number of moles of gas on each side. Recall that the number of moles of gas is directly proportional to its volume.

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