 The problem reads for the reaction CaCO3 solid to CaO solid plus CO2 gas at 25 degrees Celsius, delta G theta at the standard state is 130.4 kilojoules per mole. The change in enthalpy at the standard state is 178.3 kilojoules per mole and the change in enthalpy is not a function of temperature. We are to calculate the equilibrium pressure of CO2 at 298.15 kel, which is of course 25 degrees Celsius and at 1115 kelvin. So there's two problems here. The point that we can do this problem here is that these two are solids and we're left with just the gas here. What does that mean? We know that the free energy change for a mole of reaction is delta G equal to delta G theta standard state plus RT times ln of K, where K is the equilibrium constant. And at equilibrium delta G is zero, this is zero, so we get delta G theta is equal to minus RT ln of Kp and for ideal gases Kp is the equilibrium constant of the reaction. So in the numerator we have the product of the free energy of the products at 1 atm divided by the product of the free energy of the reactants at 1 atm. And the convention is that P of a solid is 1 divided by atm. So what do we have here for Kp? So we see that we have delta G theta, we know R, we know T is 25 degrees, this is in kelvin so it's 298. And so then our hope is that we can get the partial pressure at equilibrium for CO2 and the reason we can do that is because these are both 1. So what we have is Kp is equal to, all the powers are 1 because all the stoimetric coefficients are 1. So on the right side we have P of CaO, that's going to be a solid to the first degree times P of CO2 to the first degree divided by P of, on the left side we only have CaCO3 but it's also a solid. So what we have is that this is 1 over atm, 1 over atm. So they cancel and we're left with P of CO2. And this is actually what we want multiplied by 1 atm, which is what we want. It's the equilibrium pressure of CO2, we have to multiply this by 1 atm. Okay, so we are substituting and going to solve for this. So what do we have? So g theta is 130.4 kilojoules per mole equals minus 831146 joules mole. Kelvin times 298.15 Kelvin times ln of P of CO2. Now we need to solve this for P of CO2. So we have ln of, we're just going to write P here equals 130400 joules and everything cancels divided by minus 831146 times 29815 calculator. So 130400 divided by parenthesis minus 8.31146 times 298.15 and the parenthesis intern. So we get ln of P is minus 52 622 ln of P is minus 52 622. That means that this has a base of E, so P equals E to the minus 52 622. So we need E, so that's right here, second function E and then we need second function our answer from there and then enter. So we have 140 E to the minus 23. So P equals 140 times 10 to the minus 23. And so the partial pressure of CO2 at 25 degrees Celsius is 140 times 10 to the minus 23 and we multiply it by a standard atmosphere. So this is our answer to the first part of the problem. Now for the second part of the problem we have to calculate the equilibrium pressure of carbon dioxide at 1100.15 Kelvin. Notice that we have not yet used the second part of the information here and we're going to use that now in a standard way that is we're going to find delta G theta at 1100 by using this information here. We do that by using our standard Gibbs-Hemmeltz equation and remember that we can transform this and write it as delta G theta equals and then we're going to put this T on the other side. So minus T integral of delta H theta over T squared dt. Now the nice thing is because we've been told that delta H theta is not a function of temperature that means it's constant with respect to this integral. So we can take it out. Many times it was not constant and we had had the function inside with all its Ts. So right now we can say that delta G theta is equal to minus T times delta H theta times the integral of dt over T squared which is equal to minus T times delta H theta times the integral of this is minus 1 over T plus we have an arbitrary constant of integration. We can write this as dH theta times minus T and 1 over T is 1 and so we have minus T times C here and we're looking for C using the information we have and then we're going to calculate delta G theta at this temperature once we have that formula. So first we need to find this constant of integration. So delta G theta at 29815 Kelvin is equal to delta H theta at that temperature it really doesn't matter because it says it's temperature independent times 1 minus 298.15 Kelvin times RC. So here we have 130.4 kilojoules per mole here we have 178.3 kilojoules per mole and here we have 1 minus 298.15 Kelvin times so we have 130.4 divided by 178.3 is equal to 1 minus 298.15 times that constant that we're looking for. In our calculator we have 130.4 divided by 178.3 so that's 0.73 equals 1 minus 298.15 times C so 298.15 times C equals 1 minus 0.73 and C equals 1 minus 0.73 divided by 298.15 which we will now find so we need 1 minus our answer that's 0.268 and then we need to divide that by 298.15 and we get 9.01 10 to the minus 4 so 9.01 times 10 to the minus 4. So at 1100.15 Kelvin we want to calculate our new delta G theta delta H theta is the same and we have 1 minus our temperature is here and our C is here so how much is delta G theta so delta G theta at 1100.15 Kelvin is equal to 178.3 that's this times 1 minus 1115 times 9.01 times 10 to the minus 4 so let's calculate that so 178.3 times parentheses 1 minus 1100.15 times 9.01 and we need that ee there so second ee and we need minus 4 and end of parentheses and enter 1.563 so delta G theta is 1.563 at 1100.15 Kelvin 1.563 kilojoules per mole the same unit as it was so now we need to put this into that other equation and get our partial pressure so we have 1.563 kilojoules per mole equals minus 8.31146 joules per mole Kelvin times 1100.15 Kelvin times ln and we remember of P of carbon dioxide so ln of P equals 1.563 joules divided by minus 8.31146 joules times 1115 so calculating that 1.563 divided by minus 8.31146 times 1100.15 close parentheses equals so that's minus 0.17 minus 0.17 so P equals 8 to the minus 0.17 and let's get that so remember we did second e and then second answer and we can end the parentheses not enter 0.843 0.843 so our partial pressure of CO2 1100.15 Kelvin is 0.843 atmospheres and that is the answer to the second part of that promo