 The problem reads, calculate the standard enthalpy and entropy changes at 25 degrees Celsius for the reaction. 2cu plus 1 half o two equals cu two o, given that the change in the free energy is equal to minus 40,500 minus 3.92 t log t plus 29.5 t calories. So what do we want to find? The standard enthalpy change, so that's delta h theta, and the standard entropy change, so that would be delta s theta. First we're going to find delta h theta using delta g theta and the Gibbs-Hemsoltz equation, and then we'll use delta h theta and delta g theta and use the Gibbs equation in standard state to find delta s theta. So delta h theta is going to be minus t squared times this, so we need to find this here. So first we need to divide delta g theta by t, so delta g theta divided by t would be minus 40,500 divided by t minus 3.92 log t. This t is cancelled, plus 29.5, that t is cancelled. Now we need to take the partial with that with respect to t. We're considering the pressure to be fixed, so delta g theta over t, so minus 1 over t, the derivative of minus 1 over t is plus 1 over t squared. So 40,500 over t squared, the derivative of log of t is 1 over t times 1 over ln of 10. This is a base of 10, so the derivative of log base 10 of t is 1 over t times 1 over ln of 10, and the derivative of this is 0. So we'll just write 0 there, so we know. So delta h theta equals minus t squared times this part here. So that would be how much? Equals? We would have minus 40,500 plus 3.92 divided by ln of 10 times t. And we're interested in calculating this for t equal to 25 degrees Celsius, but this t is in absolute, so that would be 29815 Kelvin. So delta h theta 29815 Kelvin is equal to minus 40,500 plus 3.92 over ln of 10 times 29815, which is so minus 40,500 plus 3.92 divided by ln of 10 and parentheses times 298.15. Minus 39,992 minus 39,992 and that would be calories. And that is the answer to the first part of the question. Now in the second part of the question we're looking for the standard entropy change, so that we said was delta s theta. So we change this and we get delta g theta under constant pressure is equal to delta h theta minus t times delta s theta. So we would have delta s theta is equal to minus 1 over t times delta g theta minus delta h theta. So we substitute everything we have at the temperature of 298.15 Kelvin and we will get delta s theta. So delta s theta minus 1 over t times minus 40,500 minus 392 t log t plus 29.5 t. This minus here times that minus there becomes plus 39,992. So this is going to be calories over Kelvin equals the sum of these two is how much? Minus 40,500 plus 39,992 minus 508 would be plus 508 over t minus 392 log t plus 392 minus 295. Now we need to substitute 298.15. So we get delta s theta at 298.15. K is equal to 508 divided by 298.15 plus 392 times log of 298.15 minus 295 equals hopefully we'll be able to squish it in here. So 508 divided by 298.15 plus 392 times log of 298.15 parenthesis minus 29.5. Enter minus 1809 minus 1810 minus 1810 and that's calories versus Kelvin is the answer to the second part of the problem.