A 79.2 g chunk of dry ice (solid CO2) and 30.0 g of graphite (car...

Question

A 79.2 g chunk of dry ice (solid CO2) and 30.0 g of graphite (carbon) were placed in an empty 5.00-L container, and the mixture was heated to achieve equilibrium. The reaction is
CO 1g2 + C s ∆ 2 CO g
(b) What is the value of Kp at 1100 K if the gas density at 1100 K is 16.9 g/L?

Accepted Answer

Hello everyone today, we are being given the following problem, A mixture of 1.89 g of solid sulfur and 1.38 g of oxygen gas were placed in a 2.5 liter vessel it was heated up to reach equilibrium. The reaction involved was sulfur as follows their oxygen gas to make that sulfur dioxide. If the gas density at 1150 kelvin is 6. g per earlier, what is the value of K. P. At this temperature? So the first thing we need to do is we need to calculate the initial moles of oxygen. And to do that we have to take our given value Which was 1.38 grams of our oxygen gas. And then we have to multiply by our molar mass which according to the periodic table so that one mole of oxygen gas is equal to 0.0 g. Our units of g for oxygen will cancel out. We'll be left with 0.043 moles of oxygen gas. Next we're gonna calculate equilibrium concentrations using an I. C. E. Table. So we have our reaction here and recall that we are not going to use liquids or solids in our reaction expression. So we have our additional change in our final or equilibrium. So like I said, I mentioned earlier, we're not gonna use solids in this table. Are initial number of moles for oxygen gas. As we calculated before it was 0.43 bowls. Of course we didn't have any of our products. Since we're taking away from our reacting and adding to our product, our change will be minus X. For the reactant and positive X. For the product. Thus, R. E. Is going to be the first two rows added together. So we're gonna have 0.43 minus X. And then we're gonna have X. For the products. Next, we're going to find the total mass of gas in the flask. And what we're gonna do is we're gonna take our density. That we had density, which is six 0.7. Where to find the gas, the mass of the gas in the flask, we're gonna take our 1.66 g per liter times are 1.5 liter that we have, or 2.5 later Give us 2.49 g. Now, we need to solve for X. By using the total mass of gas in the flask and the molar mass of gas. So this is how we're gonna set it up. We're going to have the total mass of the gas, Which is 2.49 grams equal to the moles of oxygen gas had equilibrium. And that's gonna be represented by 0.43 minus X times the molar mass of oxygen, which is 32 g per mole plus our moles of sulfur dioxide gas at equilibrium represented by X times its molar mass, or 64 point oh seven g per mole Solving for X. We're gonna arrive at 0.035. So With this we can see the moles of oxygen gas at equilibrium by taking our moles of oxygen 0.043 And subtracting by X 0.035, giving us 0.0085 moles of oxygen to gas. Alright, initial there and then for S. 02, that will be our value there. And so now we have to calculate our partial pressures for both of them. So our partial pressure for oxygen is going to be the moles times the gas constant, times our temperature divided by our volume. So our moles of oxygen was 0. 85. The gas constant can be found in the reference text. It is 0.8 to 06 liters times atmospheres over moles, times kelvin. And then our degrees in temperature was 150 Kelvin. We're gonna invite them by our volume, which was 1.5 or 2.5 liters. This is gonna yield us 0.54 atmospheres and we're gonna do the same thing for our sulfur dioxide. Most times gas costs at times temperature divided by our volume are moles was 0.35. R. Gas constant was 0.8206 Leaders times atmospheres most times kelvin And then our temperature as before was 1150 Kelvin. We're gonna divide that by our volume And we're gonna get 2. atmospheres from these values. We can calculate our k p r k p is going to be our partial pressure for sulfur dioxide, divided by a partial pressure for oxygen. We're gonna have to .2 divided by 0.54 to give us 4. as our final answer. And with that we've answered the question overall. I hope this helped hand until next time.

A 79.2 g chunk of dry ice (solid CO2) and 30.0 g of graphite (carbon) were placed in an empty 5.00-L container, and the mixture was heated to achieve equilibrium. The reaction is CO 1g2 + C s ∆ 2 CO g (b) What is the value of Kp at 1100 K if the gas density at 1100 K is 16.9 g/L?

Watch next