To calculate the equilibrium temperature and

the range of temperatures for which this reaction is product favored, we will need the Gibbs

free energy equation. Delta G equals delta H minus T delta S. At

equilibrium delta G will be 0, and if we want to solve for T we’ll substitute in the values

for enthalpy and entropy, which are given. So delta H is 98.8 kilojoules per mole minus

T times the entropy, which is in joules per mole Kelvin – so I’m going to convert

that into kilojoules – 0.1415 kilojoules per mole Kelvin. I’m going to rearrange, move the 98.8 to

the other side and that will equal negative T times .1415. Of course, the negatives cancel, and solving

for T gives 698 Kelvin. This is the equilibrium temperature. The product-favored reaction is a spontaneous

reaction, and a spontaneous reaction will have delta G values less than zero. We know that when delta G is equal to 0, temperature

is equal to 698 Kelvin. So let’s look at what happens to G if T

is less than 698 Kelvin, and when T is greater than 698 Kelvin. Let’s pick a temperature less than 698 and

see what happens to G. Perhaps the easiest temperature to pick would be T equals 0 Kelvin. If T is 0, then the entropy term disappears

and we end up with delta G being positive. When you have a delta G greater than 0, that’s

a nonspontaneous reaction; nonspontaneous also means reactant favored. If the reaction is nonspontaneous at temperatures

below the equilibrium temperature, then it will be spontaneous at temperatures above

the equilibrium temperature. But for proof, let’s plug in a temperature

greater than 698 Kelvin. So I’m going to pick 1,000 because that

makes the math a little bit easier. If T is equal to 1,000 then when we multiply

we end up with 141.5 kilojoules per mole Kelvin – over here in the entropy term, which then

makes the delta G less than 0, which is spontaneous, or product favored. Therefore, we can say the reaction is spontaneous,

or product favored, at all temperatures greater than 698K.