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Problem 111

Synthesis gas, a mixture that includes the fuels …

Problem 111

Synthesis gas, a mixture that includes the fuels CO and $\mathrm{H}_{2}$, is used to produce liquid hydrocarbons and methanol. It is made at pressures up to 100 atm by oxidation of methane followed by the steam reforming and water-gas shift reactions. Because the process is exothermic, temperatures reach $950-1100^{\circ} \mathrm{C}$, and the conditions are such that the amounts of $\mathrm{H}_{2}, \mathrm{CO}, \mathrm{CO}_{2}, \mathrm{CH}_{4}$, and $\mathrm{H}_{2} \mathrm{O}$ leaving the reactor are close to the equilibrium amounts for the steam re-forming and water-gas shift reactions:
$$\begin{aligned} \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) & \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \quad \text { (steam re-forming) } \\ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) & \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \quad \text { (water-gas shift) } \end{aligned}$$
(a) At $1000 .$ ', what are $\Delta G^{\circ}$ and $\Delta H^{\circ}$ for the steam re-forming
reaction and for the water-gas shift reaction?
(b) By doubling the steam re-forming step and adding it to the water-gas shift step, we obtain the following combined reaction:
$$2 \mathrm{CH}_{4}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{CO}(g)+7 \mathrm{H}_{2}(g)$$
Is this reaction spontaneous at $1000 .^{\circ} \mathrm{C}$ in the standard state?
(c) Is it spontaneous at 98 atm and $50 . \%$ conversion (when $50 . \%$ of the starting materials have reacted)?
(d) Is it spontaneous at 98 atm and $90 . \%$ conversion?

Problem 110

Acetylene is produced commercially by the partial oxidation of methane. At $1500^{\circ} \mathrm{C}$and pressures of 1–10 bar, the yield of acetylene is about 20%. The major side product is carbon monoxide, and some soot and carbon dioxide also form.
(a) At what temperature is the desired reaction spontaneous:
$$2 \mathrm{CH}_{4}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{2}+2 \mathrm{H}_{2}+\mathrm{H}_{2} \mathrm{O}$$
(b) Acetylene can also be made by the reaction of its elements, carbon (graphite) and hydrogen. At what temperature is this formation reaction spontaneous?
(c) Why must this reaction mixture be immediately cooled?

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Video Transcript

All right, so we're asked to find the temperature at which this reaction will be spontaneous. So the best way to do that is to set our equation. Delta G, um, equals don't h minus T Delta s 20 So what we're gonna have to do is find Delta h and Delta s of reaction for our reaction here. So let's start off of finding the Delta h of reaction again. We're gonna use our equation of products minus reactant, and I always like to write out a little skeleton here, so I don't get confused. So we're gonna have to Delta f information or the Delta H of formation of C two h two plus the adult age information of age two plus the Delta H information reached 20 minus. These are all these are all our X and then minus or react insists the Delta H information of ch four and the Delta H of formation of Oh, to remember when we have coefficients in our equation, we always need to put them in our products minus reactant. That is, on a coefficient. Our products minus reaction equation. So we are going to put a two here, so two times developed age information of H two. And then we're going to have to hear times two and times, one half Great. So you can find the Delta H informations most like being the appendix of your textbook. You can also google them. Um, I have filled in the little skeleton the years, so these values correspond to the built age formation of these molecules. Up here again, we're gonna want to add in our coefficients do that again. It's always good practice on half. Great. So if you want to pause the video and do that calculation really quickly, you should get a Delta age of reaction of, ah, 134 0.914 killer jewels. And since we know we're going to put this in an equation with Delta s, which is in jewels, were gonna want to change our Delta H into jewels so that we have the same units throughout. So again, we're just gonna multiply by 1000 and will get 134 914 Jules Great. So we're going to do the exact same thing. We're Goto s reaction now. So again, I wrote on our little skeleton and let's fill in our I'll just do this. Um, our coefficients times two times. One half great again. I have provided the standard entropy use for all of these, But again, they should be in your, um, in textbook where you can find them online. So fill in our coefficients times two half. If you want to pause the video and do that calculation really quick, you should get a standard entropy. Uh, 176 0.7 jewels. Great. So now we're gonna again use the equation above Delta G equals Don't age minus t Built s. So again, we're cell setting Delta G 20 We're gonna fill in our delta age and jewels. So 1000 134,000 914 jewels minus X is temperature. That's what we're solving for. And 176.7 jewels. Great. And that should come out to be a temperature of 775 Kellman. So above this temperature, the reaction will be spontaneous. All right. So be asks us to find the temperature at which the formation of a pseudo lean will be spontaneous again. This is from its component parts. So we're gonna have C plus H two. And remember, we always have to balance our equations. Never assume they're balanced. So that is just going to be putting a to see here. So we have to seize on this side to H is on the side. To seize to H is perfect. Now we're gonna do the exact same thing again. We're wanted gonna want to get our Delta H in Delta s so we can set our Delta G equals Delta H minus T. Delta s Sorry. That's a little messy. We're gonna set Delta G 20 again. So we're gonna want to find our Delta h of reaction and our delta acid reaction. So let's start with our Delta H again. It's always products minus reactions. You want to drill that into your brain products minus reactant have my little skeleton here and again. Always remember to fill in the coefficient. So again, we have a two there, so we're gonna put it into our equation. So I have not provided the standard h of formation or the standard age. Yeah, the standard entropy of formation. But again, you should be able to find them in your textbook. And if you want to pause the video, do that calculation. They should come out to me. 227. Kill it, Jules. And just like last time, we're gonna want to change this into jewels. Since we know that Delta s is usually given in jewels, so that is going to come out to be 200 27,000, Jules. Great. And now we're gonna do the X. The same exact thing with our standard enter peace. So again, remember, to code are coefficients, and I haven't provided the standard entropy use. But again, they should be in your textbook, or you can google them. So pause video and do that quick calculation, and you should get 58.8 jewels. All right, You know where we're going from here? We're gonna plug it in to our equation up here again. So again, we're setting Delta G 20 So when you're trying to find the temperature that a reaction becomes spontaneous, you're always going to sell to set your Delta G 20 out. So we have 227,000 kill it, er, jewels on this for a delta H and then X is going to be our temperature we're solving for. And then we have 58.8 jewels over there and then solving for X, we should get a T equals 385. Excuse me. 3855 point for Calvin and above this temperature of the reaction will be spontaneous. All right, So for the last question, we know that it's asking us about this initial equation up here. This is how they make a settling, um, in a lab. I think it says yes. Okay, so we know it. We're told that this process happens at 15 100 degrees Celsius, which is about adding 273 to changes into Calvin Ah, 170 our 1000. Gosh, I'm having a rough day 1773 Calvin. And from the first part of this question, we know that above 775 we Kelvin, the reaction will be spontaneous. So we know that at this temperature the reaction will be spontaneous. That means that it's just going to continue going. It doesn't need any input of energy. So we want to cool the reaction reaction. So we know that the reaction will stop. Because by cooling this down, once we get below 775 Calvin, the reaction is no longer going to be spontaneous, so we're able to stop it.