Consider the reaction 2 NO21g2¡N2O41g2. (a) Using
data from Appen...

Question

Consider the reaction 2 NO21g2¡N2O41g2. (a) Using
data from Appendix C, calculate ΔG° at 298 K. (b) Calculate
ΔG at 298 K if the partial pressures of NO2 and N2O4
are 0.40 atm and 1.60 atm, respectively.

Accepted Answer

Welcome back everyone in this example, we need to calculate the change in gibbs free energy at 2 98 kelvin for the reaction where we have two moles of tri oxide gas in equilibrium with three moles of oxygen gas. Were given the partial pressure of oxygen tri oxide in a t. M's. And were also given the standard Gibbs free energy of formation of tri oxide and of oxygen gas. So we want to recall that before we can find our change of gibbs free energy of our reaction. We need to find our standard change of gibbs free energy which is G degree. This can be found if we recall by taking the change in gibbs free energy of formation of our products. Subtracted from the change in standard gibbs free formation of our reactant. And luckily we're given these values above from the prompt. So we would say that our standard Gibbs free energy change of our reaction is equal to beginning with the some of the gibbs free energy change of formation of our products. We have our only product which is three moles of oxygen gas multiplied by its gibbs. Free energy. Formation of zero kill jules Permal given in the prompt, This completes our some of our gifts. Free energy or formation of our products which we now want to subtract from the gibbs free energy of formation of our reactant where we have just one reactant being are two moles of our tri oxide gas, multiplied by its standard gibbs. Free energy of formation given in the prompt as 163.2 kg jewels, Parimal. And so to simplify this, we better say that our gibbs free energy are standard gibbs. Free energy of our reaction is equal to we have for our formation of our products three times zero which just leaves us with zero kg joules Permal subtracted from. I'm sorry. Let's use the color red to keep things consistent. This is subtracted from 6.4 as far as our units, we actually want to get rid of moles because we can see that we can cancel out moles with moles in the numerator and the denominator in both cases, leaving us with Kayla Jewels. So we have units of killer jewels in both ends. And we're going to get that Our standard gibbs free energy of our reaction is equal to negative 326.4 kg joules. Now that we have our standard gibbs Free energy, we want to find our gibbs. Free energy of our reaction from our equation which we should recall that Delta gee gibbs free energy change is equal to our standard gibbs. Free energy delta G degree added to our gas constant R. Which is multiplied by our temperature in kelvin, which is multiplied by the Ln of our reaction quotient. Q. And so before we can find our change in gibbs free energy delta G. We want to figure out what our reaction quotient would be. So recall that our reaction quotient is equal to the partial pressure of our products. Since we have all gaseous products divided by the partial pressure of our reactant, since they're all gaseous reactant. And so this is going to be equal to the partial pressure of our only product which is our three moles of oxygen gas. And just to be even more clear, this is raised two coefficients from our balanced equation. So this would be raised to the coefficient of three according to our reaction above. And then in our denominator we have the partial pressure of our reactant which is our tri oxide gas raised to its coefficient as an exponents being to And then looking back at the prompt, they actually give us our partial pressures of our gasses. So we'll plug in those values below. Where for oxygen we have a partial pressure of zero 78 A t. M's raised to the third power and then inter dominator. For tri oxide, we have a partial pressure given from the prompt as 0.63 raised to the second power. So this is going to simplify to a value when we type this into our calculators of 1.1956. Now that we have our standard gibbs free energy change and our reaction quotient values, we can finally calculate our change in gibbs. Free energy DELTA G. So we would say that Delta G. Is equal to our standard gibbs free energy change, which we calculated above in killing jewels as negative 326.4 kg jewels. Now this is then multiplied by or sorry, added to our gas constant. R. Which we should recall is the value 8.314 with units of jewels divided by moles, times Kelvin. And as you can see because we have units of jewels here for our gas constant R. We actually want to expand our formula for the or sorry, our result for the standard gibbs. Free energy change by multiplying by the conversion factor to go from kilo jewels. Two jewels where we should recall that our prefix kilo tells us we have 10 to the third power of our base unit jewels. Now we're able to cancel out kayla jewels as well as jewels. Sorry, actually no jewels are in the numerator so we can't cancel that out yet. But continuing on, we multiply by our temperature to the gas constant. So we have a temperature given in the prompt of 298 Kelvin. So standard temperature, which is then multiplied by the Ln of our reaction quotient Q. Which we determined to be 1.1956. So canceling out our units again, we can get rid of also kelvin since we have it in the numerator and the denominator here, leaving us with jewels divided by moles as our final units. So now we can just simplify to get our value for the change in gibbs free energy where simplifying all of our steps here we have negative 326, jewels added to our product. On the right hand side of our addition symbol which should give us the value 4.4 times 10 to the positive second power. Where we have just units of jewels left and so adding this to our gives standard gibbs free energy and jewels, we would get a value for our change in gibbs. Free energy equal to a value of negative 325,960. And sorry about that mark there. So jules where we should recall that change in gibbs free energy should be in units of kilo jewels. So we're going to go back to kill a jewels by multiplying by our conversion factor to go from jewels to kill jewels in the numerator by recalling that are prefix kilo tells us we have 10 to the third power jewels for one kg jule, canceling out our units of jewels. This is going to yield us a change in gibbs free energy equal to negative 325.96 kg jewels. And now we want to make sure we have the simplest amount of sick fix. So looking at our previous step, we can count a total of just one decimal place, meaning that therefore our final answer should have one Sig fig. And so we would make this one Sig Fig by or sorry, not one Sig fig but our final answer should have one decimal place as well. And so that means that we can say our change in gibbs free energy of our reaction is equal to to one decimal place negative. 325.0 because we round that nine up kill jules. And this would complete this example as our final answer for our change in Gibbs. Free energy and sorry, we round this up to 326. Sorry about that. So this is our final answer to complete this example. Since we round this five up based on that decimal place of nine. So I hope everything I went through is clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Consider the reaction 2 NO21g2¡N2O41g2. (a) Using data from Appendix C, calculate ΔG° at 298 K. (b) Calculate ΔG at 298 K if the partial pressures of NO2 and N2O4 are 0.40 atm and 1.60 atm, respectively.

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