Hi ya'll! Part a of this problem states:
"Consider the equilibrium CO + H2O <--> CO2 + H2. a) If the partial pressure of CO2 is increased, what happens to the partial pressure of H2?"
I know that if we are increasing pressure, the reaction will shift towards the side that has fewer moles. But here, the moles are equal on both sides so the reaction won't shift towards any side. How do I solve this problem? Is increasing/decreasing partial pressure different from increasing/decreasing pressure?
Textbook 5J.1
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Re: Textbook 5J.1
Think of Le Chatlier's. Increasing partial pressure of CO2 is another way of saying CO2 is added. Since CO2 is on the right side of the reaction, adding CO2 causes the right side to overpower the left side, thus the reaction will shift left to compensate for the addition. When the reaction shifts left, H2 decreases.
"If we are increasing pressure, the reaction will shift towards the side that has fewer moles" pertains to when pressure is affected by increasing/decreasing the volume of the vessel.
"If we are increasing pressure, the reaction will shift towards the side that has fewer moles" pertains to when pressure is affected by increasing/decreasing the volume of the vessel.
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Re: Textbook 5J.1
Hi, increasing the partial pressure of CO2 would cause a shift in the equilibrium reaction that would lead more reactants to be formed. As a result, H2 partial pressure would decrease, as the reaction shifts left.
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Re: Textbook 5J.1
In this case, the description of partial pressure of CO2 increases indicates that the CO2 concentration increases. Thus, the reaction shifts to the right.
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Re: Textbook 5J.1
Sine our concentration of CO2 is increasing the reaction must shift to the left given this is an equilibrium reaction. Thus a shift to the left would denote a decrease in the right side, or a decrease in H2 and CO2.
Re: Textbook 5J.1
Hi! The moles concept results when the pressure of where the reaction is taking place changes. Since the problem mentions partial pressure, especially of a product, we don't compare moles here. When the partial pressure of CO2 is increased, this results in a increase in reactants as the reaction shifts left, which causes a decrease in the partial pressure of H2.
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Re: Textbook 5J.1
Hi Narhayne!
When CO2 is added, Q>K because there are more products than reactants. As a result, the reaction progresses toward reactants to reach equilibrium again (Q=K). Since H2 is also a product, its concentration decreases in order to form more reactants. The shortcut you're talking about applies more to compression, and the exact details are in 5J.1-5J.2 in the textbook!
When CO2 is added, Q>K because there are more products than reactants. As a result, the reaction progresses toward reactants to reach equilibrium again (Q=K). Since H2 is also a product, its concentration decreases in order to form more reactants. The shortcut you're talking about applies more to compression, and the exact details are in 5J.1-5J.2 in the textbook!
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Re: Textbook 5J.1
If you increase the pressure on one side of the equation, the reaction will proceed so that the pressure increases on the other side of the equation. For this problem specifically, the reaction is CO + H2O --> CO2 + H2. Increasing P(CO2) will cause the reaction to proceed towards the reactants, thus decreasing P(H2). Decreasing P(CO) also causes the reaction to proceed towards the reactants, lowering P(CO2). The equilibrium constant stays the same because the temperature is not being manipulated at all.
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Re: Textbook 5J.1
Increasing the partial pressure of a specific substance means that you are adding more of it and increasing its molar concentration. Thus, in this example, the equilibrium will be shifted towards the reactants since we increased the amount of products.
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