Consider the Haber synthesis of gaseous NH3 (∆H°f = -46.1 kJ/mol;...

Question

Consider the Haber synthesis of gaseous NH3 (∆H°f = -46.1 kJ/mol; ∆G°f = -16.5 kJ/mol:  
(d) What are the equilibrium constants Kp and Kc for the reaction at 350 K? Assume that ∆H° and ∆S° are independent of temperature.

Accepted Answer

Hello. Everyone in this video, we're talking about the nitrogen dioxide which can be prepared by using the combustion of nitrogen with oxygen. We want to cure calculate for the equilibrium constants of K. P. And K. S. For the formation of nitrogen dioxide, we're assuming that temperature does not affect delta H. And delta S. Alright, so for my equation here we have half a mole of N two gas reacting with one mole of 02 gas. And this yields one mole of N. 02. So we know that the formula for delta G of reaction is equal to the delta age of reaction. My T. For temperature multiplied by the delta A delta S. Of my reaction. And this delta G of reaction has another formula which is negative R. Times T. Multiplied by the natural log of K. P. So first we'll go ahead and start off with calculating from the delta H of reaction that's equal to one, multiplied by negative 33.2 kg joules per mole minus. Let's see we have zero plus zero. So this delta age of reaction is equal to negative 33. kg jewels per mole. As for my delta S of reaction is equal to one times 2 40.1 jules per mole times kelvin's minus this with health Times 1 9. 1.6 Joules per mole Times Calvin's. And we're adding this with again one multiplied by two of 5.2 jewels per more times Calvin. The value that we get from. This is equal to negative 60.9 units being jewels for a mole times kelvin's. Okay so now plugging in my formula for a delta G of reaction, this is then equal to -3 3. killer joules per mole minus my tea, which is 500 kelvin's multiplied by negative 60.9 jules per mole times kelvin. Of course we want to go ahead. Change our jewels unit in to kill a jewel. So it matches my delta H. So we'll do a direct conversion for everyone. Kill a jewel that we have. We have 10223 or 1000 jewels. Alright so once I put everything into my calculator I get that my delta g of reaction is equal to negative 2.75 killer jewels per mole. Alright again we have my delta G of reaction equaling to the second formula over here we go ahead, utilize this to get our K. P. So again I'll just rewrite that formula. So a delta G. Of reaction is equal to negative R. Times T. Multiplied by the natural log of K. P. So we can go ahead and isolate four the natural log of K. P. So that is that this equals two delta G. Of reaction over negative R. Of T. So I'm gonna go ahead and plug in those values. So we have Negative negative 2.75 killer jewels, thermal. We want to convert our killer jewels into jewels. So for every 1000 jewels we have one killer jewel gonna go ahead and divide this of course are are and that is 8. joules per mole times kelvin. And our temperature were established to be 500 kelvin's. So once you put all that into the calculator we get the value of 0.66153 to each. Go to the natural log of K. P. How we can solve for K. P. Then is taking the E. Here. So we're doing E raised to the power of both sides. So of course on the left side here the natural law would just be canceled. But we have to take the E. To the power of zero point 66153 for the right side. And once we do so we get that my K. P value here is equal to 1.93775. And of course running it to the correct number of significant figures, which is just three is 1.94. So here we have my first answer for this problem. Now doing it with our K. C. Let's go ahead and scroll down a little bit for more space. All right, so how can relate K. P. And K. C. So K P is equal to K. C. Multiplied by R R. T. Race to power of delta N from delta and here we're doing the most initially or final minus initial. So we have one minus half plus one. That gives us a delta and value of negative 0.5. So now isolating our K. C. We get K. P. Multiplied by R. T. Raised the power of negative delta N. Not plugging in my values. We have 1.93775. Multiplied by 0.8 to 06. Soon as B A. T. M. Times leader, divide by mole times Calvin's, then go ahead and multiply this by team. So that's temperature and that is 500 Calvin's is raised to the negative negative 0.5. Which of course they just canceled out Just be positive. And once you put this into the calculator we can see that K. C. is equal to 12.41219. I guess Writing this to the correct number of significant figures, which is just three. That gives us 12.4. So this is going to be my K. c. value for this problem.

Consider the Haber synthesis of gaseous NH3 (∆H°f = -46.1 kJ/mol; ∆G°f = -16.5 kJ/mol: (d) What are the equilibrium constants Kp and Kc for the reaction at 350 K? Assume that ∆H° and ∆S° are independent of temperature.

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