If you worked through part A with me, you saw that the delta G for the reaction, with the amounts given in part A, gave us a delta G of 45.5 kJ/mol. This is a positive value, but it dropped down some from the standard delta G at 55.9. Now, we're ready to make those concentrations even lower and see how that affects the delta G of the reaction. We will start with the same equation that we did in part A. The non-standard delta G = standard delta G + R times T and then times the ln Q where Q = the Ag ion concentration times the Cl ion concentration. The reason we do not include the silver chloride is because it's a solid. The value for the standard delta G, I'm going to go ahead and use it in units of J/mol with 5.59 x 10 to the 4th J/mol. And I use it in Joules because our R is in Joules. Times mole kelvin on the denominator. Our K is 298, that's the temperature in kelvin and the ln. The silver ion concentration is 1.0 x 10 to the -8. The chloride concentration is also that value so that'd be squared. This is going to give for us 5.59 times 10 to the 4th J/mol- 9.13 times 10 to the 4th J/mol. And that will give us a value of delta G of a -3.54 times 10 to the 4th J/mol or -35.4 kJ/mol. So with these amounts, very low concentrations of silver and chloride, we end up with a negative sign for our delta G, which tells me that this reaction now is spontaneous. In other words it will proceed to the right spontaneously. So change in the concentration has taken us from a very positive delta G, if it was standard state, brought it down a little bit, if it were lower concentrations than one. But as we take those concentrations down really low from the product side, the reaction will eventually become spontaneous in the four direction.