Here it says for the reaction where one mole of nitrogen gas reacts with three moles of hydrogen gas to produce 2 moles of ammonia gas, our ΔH value is -92.4 kilojoules. ΔS_{0} equals -198 Joules per Kelvin. Is the reaction spontaneous under standard conditions? If not, at which temperature will it be spontaneous?

All right, so here we can follow the following steps in order to find our answer. For step one we say using the Gibbs free energy formula, set ΔG_{0} equal to 0. So here we're going to set it equal to 0. Plug in given values of ΔH and ΔS and solve for temperature. The foul temperature corresponds to equilibrium.

All right, so we're going to do what it says here. We're going to say here that we have ΔG = ΔH - TΔS. We set ΔG equal to 0 here. We're going to plug in the values here, so we're going to have -92.4 kilojoules for ΔH. We're solving for temperatures with temperatures unknown South. Entropy is in joules. Dividing that by 1000 will get us kilojoules. So now we have that filled in.

We're going to add 92.4 kilojoules to both sides, so when I do that I'm going to get 92.4 kilojoules equals. Now we have a negative times a negative. We're going to have equal to T * 198 kilojoules per Kelvin. Divide both sides here by the ΔS value and we'll isolate our temperature. So here kilojoules will cancel out and we'll be left with Kelvin. So here our temperature equals 466.67 Kelvin.

Now this is just the temperature to make ΔG equal to 0, which means this is the temperature for us to be at equilibrium. So found temperature corresponds equilibrium. So what do we do? Well, we go to Step 2. Here we say predict spontaneity by using the signs of ΔH and ΔS and that's where our Punnett square comes into play. Here on the left side we have +ΔH -ΔH. At the top we have +ΔS -ΔS.

The way we fill this in corresponds to our original equation, and it's set in a way where we're trying to manipulate temperature to help give us a ΔG that's less than 0 to make it spontaneous. Here, if ΔH and ΔS are both positive, then we'd be spontaneous at high temperatures, and the higher the temperature goes, the more spontaneous will become. So if spontaneous at high temperatures, reaction will become spontaneous above our calculated temperature.

If ΔH is positive and ΔS is negative, there will always be non spontaneous. So it doesn't matter what temperature we have. When ΔS is negative and ΔS is positive, then you're spontaneous. No matter the temperature, your Gibbs free energy will be less than 0. And then here when they're both negative, we're going to say we're only spontaneous at low temperatures is spontaneous at low temperatures, reaction will be spontaneous below our calculated temperature.

So let's go back to the original question. Here We have ΔS as negative and we have ΔS as negative, which means we fall. Here we're spontaneous at low temperatures O. That means here's our calculated temperature. It would mean that it's spontaneous below this temperature O any temperature below 466.67 Kelvin would give us a spontaneous reaction. So that would be the answer to this question.