How would increasing the volume of the reaction vessel affect these equilibria. You don't have to tell me the answers, but can you please explain to me how I get the answer.
a) NH4Cl(solid) <===> NH3(gas) + HCL(gas)
b) N2(gas) + O2(gas) <===> 2NO(gas)
If the volume increases, the pressure must decrease AND the concn must decrease (moles/greater volume = smaller concn).
If Kc = (NH3)(HCl). Then making (NH3) and (HCl) smaller means the product becomes less than Kc and the reaction must shift to increase it to the original value of Kc. The only way it can do that is to shift the equilibrium to the right. That's the long approach. The shorter approach is to say (for a gas reaction) that increasing P will shift the equilibrium to the side with the smaller number of moles. So increase P and we see a shift to left. Decrease P and it goes to the right. Same answer but shorter.
For b), changing the pressure has no effect BECAUSE the moles of gas is the same on both sides. But you can get the same answer if we want to go the concn route. Kc = (NO)^2/(N2)(O2)
If we increase the volume, then concn is moles/larger V = smaller number.
(moles NO/V)(moles NO/V)/(moles N2/V)(moles O2/V) and you notice V cancels so it makes no difference what we choose for a volume (or pressure).
To determine how increasing the volume of the reaction vessel would affect these equilibrium reactions, you need to consider the principles of Le Chatelier's principle. This principle states that when a system at equilibrium is subject to a change, it will adjust in such a way as to counteract the change and restore equilibrium.
In the case of the equilibria provided:
a) NH4Cl(s) ⇌ NH3(g) + HCl(g)
When the volume of the reaction vessel is increased, the total pressure of the system decreases. Since pressure is related to the number of gas molecules, reducing the pressure will shift the equilibrium in the direction that produces more gas molecules. In this case, according to Le Chatelier's principle, the equilibrium will shift towards the right, favoring the formation of NH3(g) and HCl(g).
b) N2(g) + O2(g) ⇌ 2NO(g)
Increasing the volume of the reaction vessel affects the pressure of the system similarly as in the previous case. However, in this reaction, the total number of gas molecules remains constant on both sides of the equation. Therefore, changing the volume will not directly affect the equilibrium position. Instead, it will only affect the density of gas molecules in the reaction vessel.
In both cases, increasing the volume of the reaction vessel will result in a decrease in the total pressure of the system. The effect on the equilibrium position depends on how the different species are distributed in the balanced equation and whether the number of gas molecules changes upon reaction. By considering these factors according to Le Chatelier's principle, you can determine how the equilibria will be affected by the change in volume.