Use the data in Appendix B to calculate the equilibrium pressure ...

Question

Use the data in Appendix B to calculate the equilibrium pressure of CO2 in a closed 1 L vessel that contains each of the following samples: (a) 15 g of MgCO3 and 1.0 g of MgO at 25 °C Assume that ∆H° and ∆S° are independent of temperature.

Accepted Answer

Hello everyone. So in this video we're talking about this reaction right over here already being asked to use the given thermodynamic data to calculate the equilibrium pressure of oxygen gas and a 1.5 liter vessel that contains 10 g of H. T. O. At 350 degrees Celsius. We're under the assumption that the delta S. And delta H. Do not vary with temperature. Alright, so from my equation here we're using the delta G of reaction is equaling two Delta H. of reaction Maya's my T. For temperature multiplied by delta S. Of reaction. Another formula for delta G of reaction and that is negative R times T. Multiplied by the natural log of K. P. And we also know that K. P. Is equal to the pressure then of 02 is the first solving from a delta H. of reaction is equal to 0-plus zero -2, multiplied by the entire team of my H. G. O. Those negative 90.8 kg joules per mole. Once you put that into the calculator we get them I delta H of reaction is 81.6 kg jewels per mole as from a delta S of reaction. That is equal to two times 75.9 jewels per more times kelvin plus one times two oh five times two jewels per mole times Calvin. We're subtracting this with two times 70.3 joules per mole times kelvin's. Alright, so if you put all this into a calculator, we get that my delta S of reaction is equal to positive 216.4 jewels per mole times kelvin. So, using the equation from earlier to solve our K. P. Value, which is the delta G. Is equal to delta H. Of reaction minus T. Times delta S of reaction. Also equal to negative R times T. Natural log of K. P. All right, so again, we're just user realizing this right here, which will continue in green. So we have 1 81.6 kg joules per mole. That's my delta age of reaction Minour teeth or temperature and temperature were given to be 30 or 350°C. of course we want this in Kelvin. So we're adding to 73.15. They were multiplying this with delta S of reaction and that is to 16.4 joules per mole times kelvin, we should know that. We want our unit of jewels to be translated into killer jewels to do a direct conversion. So for every one kill a jewel that we have, we get 1000 jewels, then this is going to be equal to, I'll just continue in a different line in case we run out of space. So negative are so our value is just our ideal gas law constant is 8.314 joules per mole times kelvin. Again, we want our jewels unit to be in killer jewels. So for everyone kill a jewel that we have, we have 1000 jewels multiplied by T. Where he said that is 3 50 degrees Celsius plus to 73.15. Lastly multiplied by the natural log of K. P. So we're gonna go ahead and simplify this here. But we should get 46.75034 killer jewels per mole. Going to negative 5. killer jewels per mole multiplied by the natural log of K. P. now isolating the natural log of K. P. This gives us negative .75034 Killer Jewels per molt is divided by 5.18087 killer jewels for more. And this gives me that the natural log of K. P. Is equal to negative 9.2365. Now isolating R. K. P. We take the race to whatever it is on both sides. So for this if we take the E. Two part of natural log of K. P. Where basically canceling out our natural log. So this gives me that my K. P. Is equal to E. race to power of negative 9.02365. Again, we already know that key P. Is equal to the pressure of our auction gas. I'm putting this into my calculator here. This gives me 1.21 times 10 to the negative four units being an A. T. M. So this right here is going to be my final answer for this problem. Thank you all so much for watching.

Use the data in Appendix B to calculate the equilibrium pressure of CO2 in a closed 1 L vessel that contains each of the following samples: (a) 15 g of MgCO3 and 1.0 g of MgO at 25 °C Assume that ∆H° and ∆S° are independent of temperature.

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