Hess's Law - Online Tutor, Practice Problems & Exam Prep

Hess's law states that the overall enthalpy change (ΔH) of a reaction is the sum of the enthalpy changes of its individual steps. Rearranging thermochemical equations affects ΔH proportionally; multiplying or dividing coefficients alters ΔH similarly, while reversing the reaction changes the sign of ΔH. This principle allows for the calculation of enthalpy in complex reactions by combining partial reactions and canceling intermediates. Understanding these concepts is crucial for mastering thermodynamics and reaction mechanisms in chemistry.

Hess's Law

Now Hess's law has to do with the rearrangement of thermochemical equations to help us find the overall enthalpy of reaction. Recall that a thermochemical equation is a chemical equation that includes an enthalpy of reaction. We're going to say the thermochemical equation and its enthalpy of reaction, which is ΔH_rnxn, are directly proportional. This means that any change to the original equation will cause the same change within your enthalpy of reaction.

Let's say we have an original thermochemical equation, where we have 2 moles of magnesium as solid plus O₂ gas, and it produces 2 moles of magnesium oxide. The enthalpy associated with this is -1204 kilojoules. I can do different things, different rearrangements to this thermochemical equation that will affect it, but also affect my ΔH of reaction. So, let's say we multiply it. Here we have our original equation, and let's say we multiply it by 2. Now, if I multiply it by 2, the coefficients now become 4 for magnesium, 2 for oxygen, and 4 for magnesium oxide. If I multiply the equation by 2, then that means I also have to multiply my ΔH by 2. So this original value of -1204 kilojoules gets multiplied by 2, and now my new enthalpy of the reaction is -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 for magnesium solid, 0.5 for O₂, and 1 for magnesium oxide. Again, whatever I do to the equation, I must do the same thing to my ΔH. So -1204 gets divided by 2, and it becomes -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 MgO solid. And what was a reactant can now become products, so 2 Mg solid plus ½ O₂. Now when I reverse the reaction, what that does is it reverses the sign of ΔH. So, it was -1204, now it's going to become +1204. So just realize that if I reverse the reaction, I reverse the sign of ΔH. So, just remember, these are the 3 things I can do to a thermochemical equation, and they have a direct impact on my ΔH of reaction for that particular thermochemical equation.

Hess's Law Example 1

In this example question, it says, if the formation equation for boron trioxide is given as the following: 4 moles of boron (B) solid plus 3 moles of oxygen (O₂) gas produces 2 moles of boron trioxide (B₂O₃) solid, with 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 ΔH. 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 also multiplied by 2. Now remember, when you reverse the reaction, that's going to reverse the sign of ΔH, so now it's going to be positive. And whatever number I multiply my equation by, I multiply ΔH 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, and multiplying by a value means I multiply my ΔH value by that same number.

Calculate the ∆H_rxn for the following thermochemical equation:

S (s) + $\frac{3}{2}$ O₂ (g) → SO₃ (s) ΔH_rxn = ?

When given the following:

2 S (s) + 3 O₂ (g) → 2 SO₃ (s) ΔH_rxn = −296.8 kJ

• A. -296.8 kJ
• B. -148.4 kJ
• C. 148.4 kJ
• D. 296.8 kJ

Hess's Law

Many reactions cannot be carried out in a single step, but instead require multiple steps to get to the final product. We're going to 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 going to say here that this fluorine, which is 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 one and one on each, so they both cancel out each other entirely. What's left behind comes down. So this fluorine (F2) comes down to give me this reactant, and then this xenon trifluoride 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 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 determine the overall reaction and associated with it an overall enthalpy of reaction.

Hess's Law Example 2

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. The compound from a 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 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. 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 means I'm going to have to reverse the reaction. O₂ also becomes a reactant, O₂ also becomes a reactant, and CO₂ becomes a product. Now remember when I reverse the reaction that's going to reverse the sign for ΔH. 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 a reactant 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 NO gas. I need it to be one mole of it, and I need it to be a reactant. Here I see it, 2 issues. First, 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 going to 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 NO gas. And then when I reverse and divide by 2, this becomes half N₂ plus half O₂. 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 ΔH by 2, so this is negative 90.3 kilojoules. So now this NO matches up exactly with the overall equation. Keep going. We need 1 mole of CO₂ as a product, which we have here. We need half a mole of N₂ as a 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 O₂ here as a reactant, which will cancel out with this half O₂ 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 90.3, 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.

Calculate the ∆H_rxn for

S (s) + $\frac32$ O₂ (g) → SO₃ (g) ΔH_rxn = ?

Given the following set of reactions:

$\frac12$ S (s) + $\frac12$ O₂ (g) → $\frac12$ SO₂ (g) ΔH° = −296.8 kJ

2 SO₃ (g) → 2 SO₂ (g) + O₂ (g) ΔH° = 198.4 kJ

• A. -494.4 kJ
• B. -692.8 kJ
• C. 494.4 kJ
• D. -346.4 kJ

Calculate the ∆H_rxn for

ClF (g) + F₂ (g) → ClF₃ (g) ΔH_rxn = ?

Given the following reactions:

Cl₂O (g) + F₂O (g) → 2 ClF (g) + O₂ (g) ΔH° = −167.4 kJ

4 ClF₃ (g) + 4 O₂ (g) → 2 Cl₂O (g) + 6 F₂O (g) ΔH° = 682.8 kJ

2 F₂ (g) + O₂ (g) → 2 F₂O (g) ΔH_rxn = −181.7 kJ

• A. -177.85 kJ
• B. -355.73 kJ
• C. -258.36 kJ
• D. -487.27 kJ

Here’s what students ask on this topic:

What is Hess's Law and how is it used in thermochemistry?

Hess's Law states that the total enthalpy change (ΔH) of a reaction is the sum of the enthalpy changes of its individual steps. This principle is crucial in thermochemistry because it allows us to determine the overall enthalpy change for complex reactions by breaking them down into simpler steps. By rearranging thermochemical equations—multiplying, dividing, or reversing them—we can calculate the ΔH for each step and sum them to find the overall ΔH. This method is particularly useful when a reaction cannot be carried out in a single step.

How does multiplying or dividing a thermochemical equation affect the enthalpy change (ΔH)?

When you multiply or divide a thermochemical equation by a coefficient, the enthalpy change (ΔH) is affected proportionally. For example, if you multiply the entire equation by 2, you must also multiply the ΔH by 2. Conversely, if you divide the equation by 2, you must divide the ΔH by 2. This is because the enthalpy change is directly proportional to the quantities of reactants and products involved in the reaction.

What happens to the enthalpy change (ΔH) when a reaction is reversed?

When a thermochemical reaction is reversed, the sign of the enthalpy change (ΔH) is also reversed. For instance, if the ΔH of the forward reaction is -1204 kJ, then the ΔH of the reverse reaction will be +1204 kJ. This is because reversing the reaction changes the direction of heat flow, converting an exothermic process into an endothermic one, or vice versa.

How do you use Hess's Law to find the enthalpy change of an overall reaction from multiple steps?

To use Hess's Law to find the enthalpy change (ΔH) of an overall reaction, you sum the enthalpy changes of the individual steps. First, write down the partial reactions and their respective ΔH values. Next, rearrange and combine these reactions so that intermediates cancel out, leaving you with the overall reaction. Finally, add the ΔH values of the partial reactions to obtain the ΔH for the overall reaction. This method allows you to calculate the enthalpy change for complex reactions that occur in multiple steps.

What are reaction intermediates and how do they affect the application of Hess's Law?

Reaction intermediates are species that appear in the steps of a reaction mechanism but not in the overall balanced equation. When applying Hess's Law, these intermediates must cancel out. This is done by ensuring that intermediates appearing as reactants in one step are also present as products in another step. By canceling out intermediates, you can combine the partial reactions to form the overall reaction, allowing you to sum their enthalpy changes (ΔH) to find the total ΔH for the overall reaction.