হেসের তাপ সমষ্টি সূত্র

(Video) Hess's Law and Heats of Formation

by Professor Dave Explains

Hess's law allows us to manipulate coefficients or reverse the direction of a reaction when we need to. The only catch is that we must change the $\Delta H$ in the same way. We can then add or subtract all of the energy changes to find out the exact $\Delta H$ for the reaction we are looking at.

(Video) Hess's Law Common Test Question

by Melissa Maribel

Have a test coming up? Melissa will help you prepare by walking you step-by-step through a problem using Hess's law.

We've learned about enthalpy. It gives us a way to measure heat transfer at constant pressures. It's also a state function. So, it only depends on the system's initial and final states. The benefit is our path doesn't matter! For example, let's say I have a cup of liquid water at 25.0 celsius. I could heat it directly to 30.0 celsius. But, I could also heat it to 35.0 and then cool it to 30.0. Either way, our final state is the same. In both cases, the change in enthalpy would be equal.

But what about chemical reactions? Again, the path doesn't matter. Chemists have experimentally determined the reaction enthalpies for tons of chemical reactions. We can use the predetermined ones to construct other reactions. Then, we can add their enthalpy changes. We call this Hess's law.

Hess's Law Rules

The definition of Hess's law is, "for a reaction done in a series of steps, the overall enthalpy change is the sum of the steps."

Hess's law has a few rules.

1. The enthalpy change of a reversed reaction has the opposite sign.

For example, let's say we had a reaction, $A + B \rightarrow C, \Delta H= 100.0 J$. The reverse reaction, $C \rightarrow A + B$ would have $\Delta H=-100.0$

1. Any time we alter the mole ratios, we change the enthalpy by the same amount.

For example, again let's say we had a reaction, $A + B \rightarrow C, \Delta H= 100.0 J$. If we needed 2C, our reaction becomes $2A + 2B \rightarrow 2C, \Delta H= 200.0 J$.

1. Similar to a reaction mechanism, our steps add to our overall reactions.

2. We always add the enthalpies of the steps.

Steps in a Hess's Law Question

For Hess's law problems, you'll be given an overall reaction. Your goal is to find the enthalpy change for this reaction. You'll also be given a list of other chemical reactions. It may have only a couple reactions that you need to manipulate. Or your teacher could give you a whole page filled with reactions. Either way, the steps are the same.

1. Find a reaction step that contains each reactant and product in the overall reaction. A single step might contain multiple reactants or products. But, you may have to find a reaction for each one.

2. If you need to reverse any reactions, do so. Remember to change the sign on the enthalpy.

3. Adjust the mole ratios via multiplication. Remember to multiply the enthalpy too.

4. Add the step reaction to achieve the overall reaction. Cancel anything that is on both a product and reactant.

5. Add the enthalpies changes for the step reactions. Now you have the enthalpy change for your overall reaction.

Hess's Law Examples

Let's start with an easy example. We want to convert graphite to diamond. They are both allotropes of solid carbon. We are given a combustion reaction for each one. We'll use the combustion data to find the overall change in enthalpy.

Goal:

• $C_{(s)graphite} \rightarrow C_{(s)diamond}$

Given:

• $C_{(s)diamond} + O_{2(g)} \rightarrow CO_{2(g)}$ $\Delta H= -395.0 KJ$

• $C_{(s)graphite} + O_{2(g)} \rightarrow CO_{2(g)}$ $\Delta H= -394.0 KJ$

Step 1: We'll need both reactions.

• $C_{(s)diamond} + O_{2(g)} \rightarrow CO_{2(g)}$ $\Delta H= -395.0 KJ$

• $C_{(s)graphite} + O_{2(g)} \rightarrow CO_{2(g)}$ $\Delta H= -394.0 KJ$

Step 2: We'll need to flip the combustion of diamond.

• $CO_{2(g)} \rightarrow C_{(s)diamond} + O_{2(g)}$ $\Delta H= +395.0 KJ$
• $C_{(s)graphite} + O_{2(g)} \rightarrow CO_{2(g)}$ $\Delta H= -394.0 KJ$

Step 3: We don't need to change the mole ratios.

Step 4: Add the two reactions. Cancel anything that is both a product and reactant.

• $CO_{2(g)} \rightarrow C_{(s)diamond} + O_{2(g)}$ $\Delta H= +395.0 KJ$

• $C_{(s)graphite} + O_{2(g)} \rightarrow CO_{2(g)}$ $\Delta H= -394.0 KJ$

• $Overall: \cancel{CO_{2(g)}} +C_{(s)graphite} + \cancel{O_{2(g)}} \rightarrow C_{(s)diamond} + \cancel{O_{2(g)}} + \cancel{CO_{2(g)}}$

• $C_{(s)graphite} \rightarrow C_{(s)diamond}$

Step 5: Add the enthalpies.

• $\Delta H_{overall}= +395.0 KJ + -394.0 KJ= 1.0 KJ$

Now let's look at a more challenging example. In environmental chemistry, nitric oxide was a significant cause of ozone depletion. Before the catalytic converter, it was produced by car engines. Now, the catalytic converter catalyzes its reduction to nitrogen gas. Let's look at nitric oxide's interaction with ozone.

Calculate the enthalpy for $NO_{(g)} + O_{(g)} \rightarrow NO_{2(g)}$.

Given:

• $NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)}$ $\Delta H= -199.0 KJ$

• $O_{3(g)}\rightarrow\frac{3}{2}O_{2(g)}$ $\Delta H= -142.0 KJ$

• $O_{2(g)}\rightarrow 2O_{(g)}$ $\Delta H= 495.0 KJ$

Step 1: This question is unique. We can see that the first equation has both nitrogen oxide and nitrogen dioxide. But, it also has ozone and diatomic oxygen. We'll need to use the other equations to cancel them.

Step 2: We'll need to flip the third equation. We need singlet oxygen to be a reactant. It's harder to see but, we'll also need to reverse the second equation. The reason is that we'll need to cancel ozone from the first equation.

• $NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)}$ $\Delta H= -199.0 KJ$

• $\frac{3}{2}O_{2(g)} \rightarrow O_{3(g)}$ $\Delta H= +142.0 KJ$

• $2O_{(g)} \rightarrow O_{2(g)}$ $\Delta H= -495.0 KJ$

Step 3: Now, we multiply the third equation by one-half. That's because we only need one singlet oxygen as a reactant.

• $\frac{1}{2}(2O_{(g)} \rightarrow O_{2(g)})$ $\frac{1}{2}(\Delta H= -495.0 KJ)$ $\rightarrow$ $O_{(g)} \rightarrow \frac{1}{2}O_{2(g)}$ $\Delta H= -247.5 KJ$

Step 4: Now, we can add the three equations.

• $NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)}$

• $\frac{3}{2}O_{2(g)} \rightarrow O_{3(g)}$

• $O_{(g)} \rightarrow \frac{1}{2}O_{2(g)}$

• Overall: $NO_{(g)} + \cancel{O_{3(g)}}+\cancel{\frac{3}{2}O_{2(g)}} +O_{(g)} \rightarrow NO_{2(g)} + \cancel{O_{2(g)}} + \cancel{O_{3(g)}} + \cancel{\frac{1}{2}O_{2(g)}}$

Step 5: Finally, we add the enthalpies.

• $\Delta H= -199.0 KJ+142 KJ +-247.5 KJ= -304.5 KJ$.