Hello everyone, It is funny that I thought I understood dynamic equilibrium really well but now I have gone blank again. I can do all the calculations and I understand Le chatelier principle well. Here are the problems I have. 2A+B <----> C * If forward reaction equals backward reaction wouldn't those two balance out and be zero. Is the reason why this doesn't happen is because both reactions are continuosly occuring. But please elaborate if you can. *I checked so many sites on this topic and none of them really talks much about what happens to the moles in dynamic equilibrium. Can anyone tell me what happens to the moles in dynamic equilibrium. I understand that if the reaction occurs according to stoichiometric ratios there would be one way reaction. So roughly when we react 2 moles of A and 1 mole of B do we get like 0.8 moles of C.I'm just asking does something like this happen. *If in equilibrium the reactants and products don't act according to their stoichiometric ratios why do we assume so in calculations. Do they always provide us with ICL data in equilibrium questions. If anyone can take their time and at least answer one of these questions I would be very greatful. Thank you
The thing with reactions is, you need to see where the imbalance, or asymmetry is. Either a product is formed and no longer participates (by being removed somehow from the remaining reactants), or a product is converted back into reacting species. It all depends on endo- and exothermics, concentrations, the physical separation of product species, etc. Or a reaction proceeds to an equilibrium, which depends on initial conditions - concentrations, the presence or absence of a solvent medium and so on. So a reaction is something like a system with a resonant state - which is the equilibrium - that it transforms or changes around. Any help at all?
In a dynamic equilibrium, the reaction goes to both directions, i.e. reversible. This means, if you add the amount of the species in one side, let say you increase the amount of A and B, the reaction will shift to the right side, means C will increase; and, if you add the amount of C, the reaction will shift to the left, i.e. A and B will increase until both sides reach equilibrium. This is not the case in non equilibrium, i.e. irreversible reaction. Let say: 2A+B ----> C in which the arrow only goes to the right direction, adding A and B will increase C, but adding C will not increase A and B, because it is irreversible. The moles in the dynamic equilibrium I assume will be constant in ratio, and so does the total weight of each side. Here for example: CO + 2 H2 ⇌ CH3OH 1 mole of CO plus 2 moles of hydrogen will result in 1 mole of methanol. The molecular weight of CO = 12 + 16 = 28 gram/mole The molecular weight of H2 = 1x2 = 2 gram/mole The molecular weight of CH3OH = 12 + 3 + 16 + 1 = 32 gram/mole. On the left side, there are 1 mole of CO, and hence the weight is: 1 mole x 28 gram/mole = 28 gram On the left side, there are also 2 moles of H2, and hence the weight is: 2 mole x 2 gram/mole = 4 gram Total weight on the left side = 28 + 4 = 32 gram. On the right side, there are 1 mole of CH3OH, and the weight is: 1 mole x 32 gram/mole = 32 gram. The total weight of each sides are the same, i.e. 32 gram. The moles ratio of the equation is: (1+2)/1 = 3 Now, if you add another mole of CH3OH, the reaction will shift to the left and be: 2CO + 4 H2 ⇌ 2CH3OH there are now 2 moles of CO and 4 moles of H2, but the mole ratio of left and right side stays the same: (2+4)/2 = 3 For irreversible reaction, I suppose you can find the example in oxidation. Reacting something with oxygen normally irreversible. For example, you burn fuel, e.g. methane (CH4), i.e. you react CH4 with O2. The equation will be: CH4 + 2O2 → CO2 + 2H2O The more you add the fuel (the methane) and the oxygen, the more you will get CO2 and H2O. But, adding CO2 and H2O will not result in the increase of methane and oxygen.
