These reactions are important in catalytic converters in automobiles. Calculate ΔG° for each at 298 K. Predict the effect of increasing temperature on the magnitude of ΔG°.
a. 2 CO(g) + 2 NO(g) → N₂(g) + 2 CO₂(g)
b. 5 H₂(g) + 2 NO(g) → 2 NH₃(g) + 2 H₂O(g)
c. 2 H₂(g) + 2 NO(g) → N₂(g) + 2 H₂O(g)
d. 2 NH₃(g) + 2 O₂(g) → N₂O(g) + 3 H₂O(g)

Question

These reactions are important in catalytic converters in automobiles. Calculate ΔG° for each at 298 K. Predict the effect of increasing temperature on the magnitude of ΔG°.

a. 2 CO(g) + 2 NO(g) → N₂(g) + 2 CO₂(g)

b. 5 H₂(g) + 2 NO(g) → 2 NH₃(g) + 2 H₂O(g)

c. 2 H₂(g) + 2 NO(g) → N₂(g) + 2 H₂O(g)

d. 2 NH₃(g) + 2 O₂(g) → N₂O(g) + 3 H₂O(g)

Accepted Answer

Welcome back everyone in this example, we're told that at 2 98 kelvin, what is the change in gibbs free energy for the below reaction, we have one mole of silicon dioxide solid reacting with four moles of hydrogen fluoride gas to produce one mole of silicon touch of fluoride gas and two moles of water in liquid form. In part two of this question, we need to determine what's going to happen to the magnitude of our change in gibbs free energy or reaction if the temperature is increased. So for part one, we need to find our change in gibbs free energy of our reaction by recalling the following formula where we're going to take the standard free energy change of our products. Subtracted from the standard free energy change of our reactant according to our given equation here. So we're going to actually use the color red for the standard free energy change of our reactant as we fill in our equation to find our result for our change in Gibbs free energy of our reaction. So let's do the work for that. Below here we're going to say that delta G of our reaction is equal to first beginning with our free energy change of our products. We should recognize that on our product side as we read our equation, we have one mole of our silicon touch of fluoride gas. And in our textbooks or online we're going to look up the standard free energy change of our molecule silicon tetrachloride gas. And we would see that it's equal to a value of negative 1572.8 with units of kayla jewels per mole. So we also have a second product being our two moles of water. So we're going to add that into our calculation here. So we have again two moles of water multiplied by from our textbooks or online. The standard free energy change of water, which is a value equal to negative 237. with units again Kayla jewels per mole. So let's make these units clear. It's kayla jewels in the numerator. So this is going to complete the sum of our standard free energy change of our product side. And now we want to subtract this from the standard free energy change of our reactant side. So let's actually continue that below for enough room here. So beginning with our first reactant we have as we read from our equation, one mole of our silicon dioxide solid, which is going to be multiplied by silicon dioxides. Standard free energy change according to our text books as the value negative 856.64 units, kayla jewels per mole. And then for our second reactant we're gonna add onto that, our formals of our hydrogen fluoride gas, which is multiplied by hydrogen fluoride gas. Standard free energy change according to our text books of a value being negative 273.2 kg joules per mole. So this will complete the sum of our standard free energy change of are reacting side. And now we're just going to apply the math for this entire equation here in our calculators. So just to show the simplification We're going to get for our standard free energy change some of our products, the value negative 1098.54. And then this is going to be again subtracted from our some of standard free energy change of our reactant which gives us the value negative 1949.44. And we should keep our units kayla jules Permal consistent. So let's write that in on both sides and actually correction for our some in part one. That's actually going to gilda value of negative 2047.6 kg joules per mole. So now that we have simplified this enough, we can place this into our calculators, recognize that this here is going to be subtracting a negative number. So this will actually be addition. And we would be able to say that our change in gibbs free energy of our reaction is equal to a value of negative 97.62 kg joules per mole. So this is going to be our answer here for part one as our change in gibbs. Free energy of our reaction. Now because our free energy of our reaction is a negative value. We would say that therefore we have a spontaneous reaction and that is due to the fact that we recognize that our reactant, our favorite here because we have more free energy on our reactant side. And so we don't need any additional energy to get this reaction going now. Moving on to solve part two of our prompt here which is figuring out the magnitude of our Gibbs free energy change if temperature is increased, we want to recall our formula where we relate our change in gibbs free energy delta G to entropy and entropy. And we would recall that that's equal to our change in entropy subtracted from our temperature which is multiplied by our change in entropy of our reaction. So this is something that we recall from our textbooks and from our lessons and as we stated this is entropy and this is our entropy. So what we should recognize from this formula is that because we recall that entropy is our measure of the redistribution of thermal energy of our system here. We would recognize why by focusing on our gas particles here that we have only just one mole of our silicon touch of fluoride gas versus we have four moles of our hydrogen fluoride gas on the reactant side. And so we have less moles of gas that is produced. And because we have less moles of gas produced on our reactant side, we would say that our entropy is going to be therefore our change in entropy is going to be negative. And so writing out our relationship here, we would rather say that our change in Gibbs free energy of our reaction is equal to our change in entropy subtracted from our temperature of our reaction multiplied by our negative entropy change. Now from our part one of our answer, we have a negative magnitude of our change in Gibbs free energy of our reaction and part two is asking if our magnitude or what our magnitude will be if temperature is increased. So as we can see from this formula that we've written out, we have a direct relationship between our change in gibbs free energy and temperature because they're all in the numerator here and so that means that therefore as temperature increases, we would say our change in Gibbs free energy is also going to become more positive. I'm sorry, let's make sure that that's all visible. So this says that it's going to be more positive as temperature increases since they have a direct relationship and that's actually going to answer part two of this question. Our change in Gibbs free energy of our reaction is going to become more positive in magnitude as temperature increases. So everything that is highlighted in yellow represents our final answer to complete this example. So this is going to correspond to choice. See in our multiple choice, I hope everything I reviewed was clear if you have any questions, leave them down below and I'll see everyone in the next practice video

Verified Solution

Key Concepts

Gibbs Free Energy (ΔG)

Temperature and Reaction Spontaneity

Catalytic Converters and Reaction Mechanisms