For slightly soluble materials (Ksp)we call it Qsp and it is the ion product of the ions.In fact, in my school days we called it ion-product and I still consider that a better name. Then we compare Qsp with Ksp to know if there is a ppt or not.
For equilibrium constants it is usually called Qrxn (reaction quotient which I consider a misnomer) and it is the same form as Keq BUT you use the concentrations in the problem and not those at equilibrium. Again, one compares Qrxn with Keq to determine which way the reaction will go to reach equilibrium. I can elaborate on these if necessary.
what is the Q of a rxn?
3 answers
so it is during the reaction process
Not necessarily.
Suppose we have a ppt, AgCl, which has a value of Ksp = 1 x 10^-10 (that's a cloe number but I don't know the exact value).
Now we have a solution The is 0.1M in NaCl and we dump in AgNO3 that is 0.001M. Will a ppt form?
Qsp = (Ag^+)(Cl^-) = (0.001)(0.1) = 1E-4. This is larger than Ksp = 1E-10; therefore, a ppt of AgCl will occur.
On the other hand if NaCl = 1E-6M and AgNO3 = 1E-6M, then Qsp = (Ag^+)(Cl^-)= (1E-6)(1E-6) = 1E-12. That is smaller than Ksp = 1E-10 so no ppt of AgCl will form.
For a reaction of
2N2O5 ==> 2N2 + 5O2. I;ll make up a number for Keq of 1E-20 (a very small number).
Keq = (N2)^2(O2)^5/((N2O5)^2
N2O5 = 1M, N2 = 2M and O2 = 1M. Which way is equilibrium; i.e., will the reaction move to the left or to the right?
Qrxn = (N2)^2(O2)^5/(N2O5)^2
Qrxn = (2)^2((1)^5/(1)^2 = 4. That is larger than Keq = 1E-20 (by a bunch) so it will move to the left to reach equilibrium.
Suppose we have a ppt, AgCl, which has a value of Ksp = 1 x 10^-10 (that's a cloe number but I don't know the exact value).
Now we have a solution The is 0.1M in NaCl and we dump in AgNO3 that is 0.001M. Will a ppt form?
Qsp = (Ag^+)(Cl^-) = (0.001)(0.1) = 1E-4. This is larger than Ksp = 1E-10; therefore, a ppt of AgCl will occur.
On the other hand if NaCl = 1E-6M and AgNO3 = 1E-6M, then Qsp = (Ag^+)(Cl^-)= (1E-6)(1E-6) = 1E-12. That is smaller than Ksp = 1E-10 so no ppt of AgCl will form.
For a reaction of
2N2O5 ==> 2N2 + 5O2. I;ll make up a number for Keq of 1E-20 (a very small number).
Keq = (N2)^2(O2)^5/((N2O5)^2
N2O5 = 1M, N2 = 2M and O2 = 1M. Which way is equilibrium; i.e., will the reaction move to the left or to the right?
Qrxn = (N2)^2(O2)^5/(N2O5)^2
Qrxn = (2)^2((1)^5/(1)^2 = 4. That is larger than Keq = 1E-20 (by a bunch) so it will move to the left to reach equilibrium.