too add to the questions in a equillbrium system say 3h2+n2-->2nh3 all in gas form if another gas like oxygen was added what would happen to equillbrium system
or Nacl(s)--> na^+(aq) + cl^-(aq)
what affect would changing pressure or conecntration or increasing temp have on this system because a solid is involved would it be different.
1.equillibrium constant only changes for change in temp? not for change in pressure etc.
2. rate of reaction can be measure by measuring how long it takes for products to form or how long it takes for reactants to be consumed? so is the rate of consumption of products = to rate of production of reactants
but the promblem is each indiviual reactant and product has a differnt rate of reaction?
3.when water vapour moleculed collide with liquid surface of water how is energy transfered and to what for the h2o(g) to turn h2o(l)
2 answers
1. yes. only T changes K.
2.I think the coefficients must be considered. If 2A + B ==> C, the production of C is only 1/2 that of A. right?
3. Look at it this way with an equation.
H2O(l) + energy ==> H2O(g). So if we are going form right to left, then energy is given up (it's an exothermic process) so the energy is given up, either to the liquid molecules or to the surrounding air.
Second post:
For N2 + 3H2 ==> 2NH3, if we add helium (so no reaction can take place--oxygen might do that), then
Kp = p(H2)^3*p(N2)/p(NH3). So by adding helium to the system, do you change the partial pressures of any of the reactants or products? Nope. So Kp can't change.
For the NaCl system, that is a strong electrolyte and it has no K. But even if it did, adding more solid to solid that's already there doesn't change the concn of the solide, It is still 1 by definition. Increasing T of NaCl will change the solubility. You know that.
2.I think the coefficients must be considered. If 2A + B ==> C, the production of C is only 1/2 that of A. right?
3. Look at it this way with an equation.
H2O(l) + energy ==> H2O(g). So if we are going form right to left, then energy is given up (it's an exothermic process) so the energy is given up, either to the liquid molecules or to the surrounding air.
Second post:
For N2 + 3H2 ==> 2NH3, if we add helium (so no reaction can take place--oxygen might do that), then
Kp = p(H2)^3*p(N2)/p(NH3). So by adding helium to the system, do you change the partial pressures of any of the reactants or products? Nope. So Kp can't change.
For the NaCl system, that is a strong electrolyte and it has no K. But even if it did, adding more solid to solid that's already there doesn't change the concn of the solide, It is still 1 by definition. Increasing T of NaCl will change the solubility. You know that.
11 answers