The fuel in high-efficiency natural-gas vehicles consists primari...

Question

The fuel in high-efficiency natural-gas vehicles consists primarily of methane 1CH42. (a) How much heat is produced in burning 1 mol of CH41g2 under standard conditions if reactants and products are brought to 298 K and H2O1l2 is formed?

Accepted Answer

Hey everyone today, we're being asked to calculate the heat that is produced when two molds of butane gas or C four H 10 combust with oxygen gas to produce C 02 and H 20 under standard conditions. So as I mentioned, which it isn't stated in the problem, but a combustion reaction occurs when a compound reacts with excess oxygen gas or 02 present in the surroundings to form carbon dioxide and water as main products. So to work with this, we first need to find the balanced reaction equation for this reaction and this will make sense once we actually get to calculating the heat. So let's do that. First. We have C four H 10, C four H 10, which is butane plus 02, both of which are gasses which will form carbon dioxide gas and water, which could either be a gas or a liquid. But we'll just say it will also be a gas. However, the question states that we have two Molds of butane gas. So let's write that in. We have two moles of butane. So since we have two molds of butane, we can go ahead and leave the oxygen for now. We can leave the oxygen alone will come back to it. But let's balance out the equation using the carbon and hydrogen is present in butane. So we have C four times two. So we have eight total carbons on this side, which means in order to get eight carbons on the right side. On the product side, we need to have eight carbon dioxides. Similarly, we have 10 hydrogen in butane Multiply that by two. That brings us to 20. So to have 20 hydrogen on this side, we would need 10 water molecules. So that makes sense. However, we're still left with an unbalanced oxygen here and we can find out what needs to be the coefficient by finding out how many oxygen's are on the right side in the product side of the reaction. Let's do that. We have eight times two 02, which leaves us with 16 oxygen's Plus 10 oxygen's from here, which leaves us a total of 26 oxygen molecules. So if we want to have 26 oxygen molecules on the right, then this means we need to have 1302 molecules on the left 13 moles of 02 to produce 26 moles of oxygen on the right side of the equation. So this here is our balanced equation right here. So with this in mind, we can actually go ahead and find the uh heat produced when this reaction occurs. Now to find the heat, we're actually, we're actually just looking for the change in entropy and to find the change in entropy of a reaction, we need to use a formula. The change in entropy of the reaction is equal to the difference in the standard entropy of formation of the products, products and the change in standard entropy of formation of the reactant. So let's go ahead and write these values out the standard entropy of formation. Our experimental values that are calculated and then put in tables that you can either find on google or on the internet or in your textbook. But and let's write the reactant in red for the reactions. The standard entropy of formation of carbon dioxide is negative 93.5 kila jules Permal for water. The standard entropy of formation is negative -2 85. joules per more for the reactant. The standard entropy of formation for butane C four H is -1 25. 1 25.7 killer joules per mole. And finally, for oxygen gas as it is a natural element oh to exist naturally and is the most stable state. The standard entropy information will be zero killer joules per mole. So with all this in mind, we can go ahead and put this back into our reaction formula. Now we need to keep in mind that we need to utilize the molar ratios as that's why we balance the equation. Let's write that out for the products, we will have eight mole c. 0 times -3 93.5 killer whales Permal added to The standard entropy information for water, which is 10 moles times -2 85. killer jules Permal. And this value will be uh or from this value, we will subtract the standard entropy of formation of the product, which is simply two more Butane C4 H times -1 25.7. And will be done because the standard entropy of formation for oxygen gas is just zero. So that term will be zero. Let me tidy this up a bit before moving on. So we will subtract the Senate and the formation of the react ints and we'll write this in green to keep up with the color coding 1 25.7. So adding all of this up, we eventually reach a final value of negative 5,754. killer jewels for two moles, two moles of butane. Therefore, the heat produced when two moles of butane gas combusts is in fact negative 5754.9 kg jewels. I hope this helps. And I look forward to seeing y'all in the next one.

The fuel in high-efficiency natural-gas vehicles consists primarily of methane 1CH42. (a) How much heat is produced in burning 1 mol of CH41g2 under standard conditions if reactants and products are brought to 298 K and H2O1l2 is formed?

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