Consider the following reaction between oxides of nitrogen: NO21g...

Question

Consider the following reaction between oxides of nitrogen: NO21g2 + N2O1g2¡3 NO1g2
(a) Use data in Appendix C to predict how ΔG for the reaction varies with increasing temperature.

Accepted Answer

Hello everyone in this question, we kind of have three parts. So we want to first calculate for the delta H. And then the delta S. For the preparation of a city acid. And we're also being asked as the temperature increases. Will our delta G value for the reaction increase decrease or remain the same? So let's go ahead and take a look. So first things first I'm going to calculate for my delta H since that's the first part. So my delta H is equal to the delta age of my products, minus the delta age of my reactant. Again, we're just using the values given to us right over here. And of course also using this chemical equation to give us the socioeconomic trek ratios. So beginning with the delta age of my products, we have, let's see we have only one product. So we have one mold this And this corresponding delta h value is negative 487.0 units. For this is just killer joules per mole. I won't put this for now because there's only one unit. And then for my delta age of my reactant. So we have one mole of this and one mole of this. So let's go ahead and put that then. So for my first reactant, its corresponding delta h value is going to be -201.2. And again we have one mole of the second starting material and its corresponding down to age value is negative 100.5. So putting all these into my calculator? I get the numerical value of -175.3. And of course our units is just killer joules per mole. So that's going to be one of my answers for this problem. Just scrolling down to give us a little bit more space, we can go ahead and solve our change of entropy czar delta S. So delta S. Is equal to the entropy of our reactant minus the entropy of our products. So for delta S of course we're doing the same exact thing using the table given to us. So we have one mole multiplied with 159.8. The units here are just going to be jewels per mole times kelvin. I won't include this here just because we're just dealing with one consistent unit. Now for my products I have two products but they're each with one mole with different entropy values, First one being 237.6 and of course we're adding that with the second product. So one more Multiplied with 1 97.9. So putting everything into my calculator, I get the numerical value of negative 75.7 jewels. That's over kelvin, we want the delta us to be in units of killer jewels. So let's go ahead and do a direct conversion. So for everyone, Killer jewel that we have, we gives us 1000 jewels. We'll see here that the units of jewels are canceled nicely to give us the unit of killer joules per kelvin. So putting this calculation then into my calculator, I get the value of negative 0. units being again killer joules per kelvin. So that's going to be my delta S. Value. And my second answer for this problem finally solving for or to see if my delta G will increase, decrease or ruin the same as the temperature increases. We can go ahead and use the gibbs free energy equation which is as we call delta G. Equaling two delta H. Subtracted from our temperature. R. T. Multiplied by our delta S. So the negative T. Multiplied by delta S. Gives us. Let's see here. So we have calculated for delta S, which is our zero point 2757. And we see here that the two negative signs will go ahead and cancel. So we get a positive value again scrolling down. So final answer for this part then is as our delta key increases or the opposite. Actually, so as T increases. So as the temperature increases, our delta G value will also increase or become more and more positive. Alright, and this is going to be my third and final answer for this problem. Thank you all so much for watching

Consider the following reaction between oxides of nitrogen: NO21g2 + N2O1g2¡3 NO1g2 (a) Use data in Appendix C to predict how ΔG for the reaction varies with increasing temperature.

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