Hess's Law involves the use of partial reactions in the determination of the overall enthalpy of reaction.
Hess's Law
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concept
Hess's Law
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Now Hess's law has to do with the rearrangement of thermochemical equations to help us find the overall enthalpy of reaction. Now recall a thermochemical equation is a chemical equation that involves or includes an enthalpy of reaction. Now we're gonna say the thermochemical equation and its enthalpy of reaction, which is delta h r x n are directly proportional. This means that any change to the original equation will cause the same change within your enthalpy of reaction. So let's say we have an original equation here, original thermochemical equation, where we have 2 moles of magnesium as solid plus o 2 gas, and it produces 2 moles of magnesium oxide. The enthalpy associated with this is negative 1204 kilojoules. Now I can do different things, different rearrangements to this thermochemical equation that'll affect it, but also affect my delta h of reaction. So the things I can do is I can multiply this equation by a value, I can divide it by a number, or I can reverse the reaction. So let's say we multiply it. Here we have our original equation and let's say multiply it by 2. Now if I multiply it by 2, the coefficients now become 4, 2, and 4. If I multiply the equation by 2, then that means I also have to multiply my delta h by 2. So this original value of negative 1204 kilojoules gets multiplied by 2, and now my new enthalpy of the reaction is negative 2408 kilojoules. Now let's say instead of multiplying I divide it. So my original equation again, now I divide it by 2. Dividing it by 2 changes my coefficients now to, 1 to 1 for magnesium solid, to 1 half for 2, and one for magnesium oxide. Again, whatever I do to the equation I must do the same thing to my delta H. So negative 1204 gets divided by 2, so it becomes negative 602 kilojoules. Finally, the last thing I can do to my thermochemical equation is I can reverse it. What was a product can become a reactant, so that's 2 m g o solid. And what was a reactant can now become products. So 2 mg solid plus 102. Now when I reverse the reaction, what that does is it reverses the sign of delta h. So it was a negative 1204, now it's going to become a positive 1204. So just realize that if I reverse the reaction, I reverse the sign of delta h. So just remember, these are the 3 things I can do to a thermal chemical equation and they have a direct impact on my delta h of reaction for that particular thermal chemical equation.
Multiplication of the original reaction results in multiplication of ∆Hrxn.
Division results in the division of ∆Hrxn..
Reversing the original reaction results in reversing the sign of ∆Hrxn.
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example
Hess's Law Example 1
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In this example question it says, if the formation equation for boron trioxide is given as the following, 4 moles of b solid plus 3 moles of o 2 gas produces 2 moles of boron trioxide solid, has an enthalpy of negative 2547 kilojoules. What will be the new enthalpy value when it's rearranged? Alright. So we have to see what are the changes that the original equation underwent to determine the new delta h. Well, we can see one thing that happened is what was a product is now a reactant, and what was a reactant are now products, so we reversed it. The next thing we should notice are the coefficients. Here it was a 2, but now it's a 4. This was a 4, but now it's an 8. This was a 3, but now it's a 6. So we didn't only reverse the reaction, but we multiplied by 2. Now remember, when you reverse the reaction that's gonna reverse the sign of delta h, so now it's gonna be positive. And whatever number I multiply my equation by, I multiply delta h by that same number. So you multiply this by 2. When you do that it gives you 5,094 kilojoules. So for my new equation it would be a positive 5,094 kilojoules. Reversing the reaction reverses the sign, multiplying by a value means I multiply my delta h value by that same number.
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Problem
Problem
Calculate the ∆Hrxn for the following thermochemical equation:
When given the following:
A
-296.8 kJ
B
-148.4 kJ
C
148.4 kJ
D
296.8 kJ
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concept
Hess's Law
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Many reactions cannot be carried out in a single step, but instead require multiple steps to get to the final product. We're gonna say Hess's law states that the enthalpy of reaction of an overall reaction is the sum of the enthalpies, the standard enthalpy values of these multiple steps. So here we have a partial reaction 1, and a partial reaction 2. By combining them together they help to give me my overall reaction down here. Now what we do is this xenon difluoride is a reactant, so it comes down. Then we're gonna say here that this fluorine which is a solid, I mean not not solid, a product, cancels out with one of these fluorines which is a reactant. Remember, they can't exist on both sides, okay, this causes an imbalance. These are called reaction intermediates and they will cancel one another out. So this cancels out with one of these leaving us with one behind. This xenon here is a product, and this one here is a reactant. Again can't exist as both, there's 1 and 1 on each, so they both cancel out each other entirely. What's left behind comes down. So this f 2 comes down to give me this reactant, and then this xenon trifold comes down to give me this final product. To find the overall enthalpy of reaction for this overall reaction, you add up each of these partial standard enthalpy values. So when you add 123 and you subtract it, subtract 262, that's how we come up with this new delta h of reaction that's a negative 139 kilojoules. This in essence is what Hess's law tries to do. It takes partial reactions and from them helps us to determine the overall reaction and associated with it an overall enthalpy of reaction.
Addition of ∆Hrxn values of multiple steps gives us the ∆Hrxn of the overall reaction.
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example
Hess's Law Example 2
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Here it says to calculate the enthalpy of reaction for the overall reaction. So we have carbon monoxide plus nitrogen monoxide producing carbon dioxide, and half a mole of nitrogen gas. Now here we're given these partial reactions below with their own known standard enthalpies of reaction. Now what do we do? Well, let's follow the rules. Step 1 says that I need to start with the first compound in the overall equation, and locate it in the set of partial reactions. Compound from partial reaction must match in terms of number and type. We'll see what that means with the one from the overall equation. This may require you to reverse, multiply, or divide the partial reaction, which will also affect your standard enthalpy value. Step 2 says keep moving on to the next compound in the overall equation until you locate all compounds in the partial reaction. So let's do steps 1 and 2 initially. Alright. So we have 2. So if we go back to step 1 it says that we need to locate the compound must we must locate, the first compound in the overall equation, and locate it in the set of partial reactions. Alright. So in my overall equation I have carbon monoxide, gas as a reactant, and there's one of it, one mole of it. So if I look in the partial reactions we see carbon monoxide is right here. But we have an issue. Although there's 1 mole of it like I want, it is a product. We need it to match up with the overall equation. We need it to be a reactant. To do that that means I'm gonna have to reverse the reaction. O 2 also becomes a reactant, and o 2 also becomes a reactant, and c o 2 becomes a product. Now remember when I reverse the reaction that's gonna reverse the sign for delta h. So now my carbon monoxide, there's one mole of it, so the number is good, and the type is also good because it's also reactive now just like in my overall equation. This equation is gone now because it's been reversed and is represented by this bottom one. Next, I need to find n l gas. I need it to be one mole of it, and I need it to be a reactant. Here I see it, 2 issues. 1st, it has a 2 in front of it. I need it to have a coefficient of 1, just like in my overall equation. Also, I need it to be a reactant. So what I'm gonna do here is I'm going to divide by 2 and reverse. So here I'm going to divide by 2 and reverse it, so that's going to become n o, gas. And then when I reverse and when I divide by 2, this becomes half n 2 plus half o 2. When I reverse the equation, remember that's going to change the sign so this becomes negative, And when I divide by 2, I have to divide delta h by 2, so this is negative 90.3 kilojoules. So now this n l matches up exactly with the overall equation. Keep going. We need 1 mole of c o two as product, which we have here. We need half a mole of n 2 as product, which we have here. So all the compounds I need match up with the overall equation. So now we go to step 3. Combine the partial reactions and cross out reaction intermediates if present. Reaction intermediates are compounds that look the same with one as a reactant and the other as a product. Once we do that, we go to step 4 where we add all the standard enthalpy values of my partial reactions to obtain the enthalpy of reaction of the overall reaction. Alright. So who are intermediates? Things that look the same, but one's a product and one's a reactant. We have half o two here as a reactant, which will cancel out with this half o two that's a product. Those are our only reaction intermediates. Everything else comes down if you look. When everything comes down it should give you back the overall equation you were looking for in the first place. It doesn't matter the order that you write them in, all that matters is that the coefficients match the overall equation, and that they're on the right side. Reactants here should match up with the reactants up in the overall. Products here should match up with the products up here. All we do now is we add up these enthalpies of reaction. So when I add up negative 283 plus this negative 90 0.3, that gives me a total enthalpy of reaction equal to negative 373.3 kilojoules. So this would be our Hess's law question, where we solved for the overall enthalpy of the reaction for the overall