We are given a standard delta G for the reaction, and it's a positive value. This tells me that if we were to start with one molar of the silver ion and one molar of the chloride ion, the reaction would not be at equilibrium, and it would proceed to the left. Both of these things, the Ksp and the standard delta G let me know that information. A very small, less than one value of Ksp tells me the equilibrium lies far to the left. A positive delta G means it's going to proceed to the left in order to get to equilibrium. But what we want to know here is what will a delta G be if we don't have one atmosphere or one molar of those solution, ions and solution. But we have 0.1 for the silver ion and 0.15 for the chloride ion. We will use this equation that denote delta G is equal to the standard delta G+RxT times the natural log of Q. Now for the reaction above, Q would be equal to the concentration of silver ion initially, and the chloride ion initially, of which these are the numbers we would put in. So we put in our delta G standard. And I'm going to go ahead and convert it to joules per mole, because my second term will have 8.314, and that has units of joules per mole times Kelvin. I'll put in my temperature of 298 Kelvin and I'll put the natural log of Q. And Q would be the silver concentration, which is 0.1 times the chloride concentration, which is 0.15. So it's the natural log of that product. 55,900 J/mol plus the product of all this. And the product of all this is actually a -1 value. So it's a -10,400 joules per mole. And this will give me 4.55x10 to 4th joules per mole. Or 45.5 kilo joules per mole. So the delta G standard was 55.9. By dropping down the concentration of these products of the reaction, the reaction will have a lower delta G.