C(s) + H2O(g) ⇌ CO(g) + H2 ΔH>0
For the system above at equilibrium, which changes will increase the amount of H2(g)?
I. Adding C(s)
II. Increasing the volume of the container
III. Increasing the temperature
The answer is II and III. I understand everything except why II is true. Can someone please explain why increasing the volume will make more H2?!?!
I hope I can. Volume goes up so pressure goes down. Increase in P make the reaction shift on the side with fewer mols GAS (solids don't count). There are two mols gas on the right, 1 mol on the left so increase P makes it go to the left. Decreasing P makes it go to the right which increases H2.Hope that helps.
OH, thank you soo much. I forgot about the solid thing!
Increasing the volume of the container will actually shift the equilibrium towards the side with more moles of gas in order to minimize the pressure. In this reaction, the forward reaction produces one mole of gas (H2) while the reverse reaction produces two moles of gas (H2O and CO).
Therefore, by increasing the volume, the system adjusts by shifting the equilibrium towards the side with more moles of gas to reduce the pressure. In this case, it will shift towards the right, favoring the forward reaction to produce more H2 gas.
Keep in mind that this explanation holds true only if the total pressure of the system remains constant during the volume change, such as in a closed container.
To understand why increasing the volume of the container will increase the amount of H2(g) in the given system at equilibrium, we need to apply Le Chatelier's principle.
In this system, the forward reaction (C(s) + H2O(g) ⇌ CO(g) + H2) is endothermic (ΔH > 0), indicating that it absorbs heat. According to Le Chatelier's principle, if a stress is applied to a system at equilibrium, the system will shift in a direction that reduces the impact of that stress.
Now, let's consider the effect of increasing the volume of the container. When the volume is increased, the pressure decreases as per Boyle's Law (P1V1 = P2V2). In this case, since the number of moles of gas on the left side of the equation (C(s) + H2O(g)) is greater than the number of moles of gas on the right side (CO(g) + H2), increasing the volume reduces the pressure of the system.
According to Le Chatelier's principle, if the pressure decreases, the system will shift in the direction that produces more moles of gas, in order to counteract the decrease in pressure. In this case, the system will shift to the right (forward reaction) to produce more H2(g) molecules because it is the side with more moles of gas.
Therefore, increasing the volume of the container will result in the system shifting to the right, increasing the amount of H2(g).
In summary, increasing the volume of the container will decrease the pressure, causing the system to shift to the side with more moles of gas, which in this case is the forward reaction (C(s) + H2O(g) → CO(g) + H2), resulting in an increase in the amount of H2(g).