 The problem reads, use Hess's laws for the following reaction to determine whether it is exothermic or endothermic, to find the change in entropy of the reaction, and to determine whether the reaction is spontaneous or not at standard state. Check your answer to see using your results from A and B and Gibbs free energy equation. Here's our reaction, so 202 plus CH4 goes to 2H2O plus CO2, everything is a gas. So A, determine whether it's exothermic or endothermic. That means we want to know whether the change in enthalpy of the reaction, everything that's at standard state because it doesn't say, is positive or negative. So we need this form of Hess's law, which says that the change in enthalpy at standard state of the reaction is the sum of the changes in enthalpy for the formation of the products minus the sum of the changes in enthalpy at standard state for the formation of the reactants. So we need to find the change in enthalpy for the formation of the four constituents. Let's draw a table underneath this and then get our table for the values. So the change in enthalpy for the formation of an element is always just zero, but we'll check it on our table, CH4, so O2 gas, like we said, formation is zero, and then we're looking for CH4. CH4 is minus 74.8 kilojoules per mole, minus 74.8, and we'll write over here that it's kilojoules per mole. Now we're looking for H2O in gas, so we need to scroll down to get H2O. There it is. H2O is a gas, minus 241.8, minus 241.8, and then CO2. CO2 is a gas, is minus 393.5, minus 393.5. Okay, so what do we have? We have delta H. That standard state of the reaction is equal to the sum of the products. We have two moles of this, so 2 times minus 241.8 plus, because it's a sum, a minus 393.5 minus, and now this is a sum, 2 times zero plus, minus 74.8. So how much is that with our trusty calculator? So 2 times minus 241.8 minus 393.5, minus parentheses, zero we won't put in, and then minus 74.8, and the parentheses equals. So minus 802.3 kilojoules per mole. A minus 802.3 kilojoules per mole, because this is smaller than zero. It is exothermic, and with that we have the answer to question A. And now we're going to use the entropy form of Hess's law. It says that the change in entropy at standard state of the reaction is the sum of the entropy at standard state of the products minus the sum of the entropy at standard state of the reactants. Do notice that there is no delta and no F here, and that is also true in the table. So here's our table, we look at O2. S at standard state of O2 is 205.1 joules per mole Kelvin. G205.1, and here we have joules per mole Kelvin. CH4, CH4, 186.3, 186.3. And now H2O, we need to scroll down to H2O. There it is, H2O, 188.8, 188.8. And CO2, CO2 is 213.7, 213.7. Okay, so what do we have here? We have the change in entropy at standard state of the reaction is equal to, we have two of these, so two times 188.8 plus 213.7 minus, and then we have two times 205.1, and plus 186.3. So that is, okay, let's store this result in A. That was the change in enthalpy, and now we're at 2 times 188.8 plus 213.7 minus parentheses 2 times 205.1 plus 186.3. And the parentheses, enter, minus 5.2, so minus 5.2 joules per mole Kelvin. So minus 5.2 joules per mole Kelvin. That is the answer to part B. Determine whether it's spontaneous or not, so we need the change in the free energy of the reaction, and for that we're going to use third form of Hess's law that says the change in the free energy at standard state of the reaction is the sum of the change in the free energy of the formation of the products at standard state minus the sum in the change in the free energy of the formation at standard state of the reactants. So we need to look those four things up on our table. Hence the third row. As with the change in the enthalpy, the change in the free energy of an element because its formation is zero. We'll check it, but it should be zero. And so the first one we're looking at is O2. The change in the free energy under formation is in fact zero. Now CH4 is minus 50.7 minus 50.7. We need H2O. H2O. We need to scroll down. H2O. There it comes. H2O is minus 228.6. Minus 228.6. And then CO2. CO2 is minus 394.4. Minus 394.4. We didn't see the units, but I will tell you they are kilojoules per mole. So we have delta G at standard state of the reaction is equal to 2 times minus 228.6 plus minus 394.4 minus parenthesis 2 times zero and then plus that minus, so minus 50.7. Calculator, we're going to store this value in B. And now we're doing 2 times minus 228.6 minus 394.4 minus parenthesis zero minus 50.7 and the parenthesis enter. Minus 800.9. So minus 800.9 and that's kilojoules per mole. The important thing is it's less than zero, so it is spontaneous. So this is our answer to part C. And now we're supposed to use Gibbs free energy equation, which is this, to find out if we have calculated delta G theta of the reaction correctly. So we need to substitute what we got here for this. For T we substitute 298.15 and for delta S theta we substitute our answer to this. So let's check that. Minus 802.3 kilojoules per mole and then minus 298.15 Kelvin times minus times minus 5.2 joules per mole. So let's see how much that is. This was in A and this was in B. So we have second and then recall and then A entered and that says minus 298.15 times second recall B. And then this was in joules so we need to divide it by 1000 to get kilojoules 1000 and enter. Minus 800.75 minus 800.9 we are good. Minus 800.7 and that was in kilojoules per mole and that is very close to what we had. And that is very close to what we have so these are the same and we are good. And that is that.