In this problem we are asked to calculate a standard delta G for the reaction. There are many ways to obtain a standard delta G. So we look at the data provided for us and we see that K value is given. With that information, we will use this equation. Standard delta G is equal to minus RT natural log of K. And since this problem is the dissolution of silver sulfate, the K we want to use is the Ksp provided. Let's plug in the values that we know. R is 8.314 joules per mole kelvin. So we have our minus in our R value. When need or temperature in kelvin. It was provided for us in Celsius, so we convert that to kelvin. And we have 298 kelvin. And then the natural log of Ksp which is 1.4 x 10 to the -5. Before we proceed let's think about the sign delta G should be. A very negative exponent here tells me that this is a very small Ksp and tells me that the equilibrium lies far to the left. So we would expect, if we were to start out with one molar here and one molar here, that this reaction would proceed to the left. And if it proceeds to the left, that's a positive delta G. Plugging the values in for our R, our T, and our natural log K gives us a value of 2.77 times 10 to the e4. And that would be units of joules per mole, because the kelvins cancel. We could convert that to kilojoules. That would be 27.7 kilojoules per mole. So this a positive delta G for this reaction. Meaning if we started in our standard state conditions, this reaction would proceed to the left and provide for us 27.7 kilojoules of energy.