 # Hess’s Law Homework Help ## Be open-minded! Give us a chance to serve you. ## What is Hess’s Law?

Hess’s law states that irrespective of the multiple steps or the number of intermediates occurring in a reaction, the total of all the individual reaction is equivalent to the total enthalpy change. Enthalpy change comes from the enthalpy which is known to be the central factor in thermodynamics. Enthalpy can be defined as the overall heat content of a system whereas the enthalpy change can be as the heat passing in and out of the system during a reaction. Hess’s can also be defined as the conservation of the energy law. Here we can determine the enthalpy change for ABC = A+B+C by calculating the enthalpy change caused by the actual reactions. In other words, we can constitute A+B = AB and AB+C = ABC to determine the enthalpy change.

## What is Hess’s Law Equation?

In our guide to Hess’s law practice problems help, you can understand the Hess’s law equation in the easiest of manner with the help of our expert tutors in the topic. You can check the Hess’s law equation:

∆Hnet = ∑∆Hr

Here, ∆Hnet = Net change in Enthalpy

∑∆Hr = Sum Change in Enthalpy reactions

By using the above equations, you can solve Hess’s law problems.

## What does Hess’s Law signify?

Hess’s law is very important and the significance of it lies in the fact that it’s not only powerful but also allows us to combine multiple equations to generate new chemical reactions of which enthalpy could be calculated instead of measuring it directly. The entire enthalpy calculation is accomplished by performing basic algebraic operations. These operations are based on the chemical equations of reactions which were previously determined values for enthalpy of formation.

## Hess’s Law Example

Here are some of Hess’s law example which will help you in understanding how to solve Hess’s law related problems.

Example #1: Calculate the enthalpy for this reaction:

2C(s) + H2(g) —> C2H2(g)

ΔH° = ??? kJ

Given the following thermochemical equations:

C2H2(g) + 5⁄2O2(g) —> 2CO2(g) + H2O(ℓ)

ΔH° = −1299.5 kJ

C(s) + O2(g) —> CO2(g)

ΔH° = −393.5 kJ

H2(g) + 1⁄2O2(g) —> H2O(ℓ)

ΔH° = −285.8 kJ

Solution:

1) Determine what we must do to the three given equations to get our target equation:

1. a) first eq: flip it to put C2H2 on the product side
2. b) second eq: multiply it by two to get 2C
3. c) third eq: do nothing. We need one H2 on the reactant side and that’s what we have.

2) Rewrite all three equations with changes applied:

2CO2(g) + H2O(ℓ) —> C2H2(g) + 5⁄2O2(g)

ΔH° = +1299.5 kJ

2C(s) + 2O2(g) —> 2CO2(g)

ΔH° = −787 kJ

H2(g) + 1⁄2O2(g) —> H2O(ℓ)

ΔH° = −285.8 kJ

Notice that the ΔH values changed as well.

3) Examine what cancels:  2CO2 first & second equation

H2O first & third equation

5⁄2O2 first & sum of the second and third equation

4) Add up ΔH values for our answer:  +1299.5 kJ + (−787 kJ) + (−285.8 kJ) = +226.7 kJ

Example #2: Calculate the enthalpy of the following chemical reaction:

CS2(ℓ) + 3O2(g) —> CO2(g) + 2SO2(g)

Given:  C(s) + O2(g) —> CO2(g)

ΔH = −393.5 kJ/mol

S(s) + O2(g) —> SO2(g)

ΔH = −296.8 kJ/mol

C(s) + 2S(s) —> CS2(ℓ)

ΔH = +87.9 kJ/mol

Solution:

1) What to do to the data equations:

leave eq 1 untouched (want CO2 as a product)

multiply second eq by 2 (want to cancel 2S, also want 2SO2 on product side)

flip 3rd equation (want CS2 as a reactant)

2) The result:

C(s) + O2(g) —> CO2(g)

ΔH = −393.5 kJ/mol

2S(s) + 2O2(g) —> 2SO2(g)

ΔH = −593.6 kJ/mol <— note multiply by 2 on the ΔH

CS2(ℓ) —> C(s) + 2S(s)

ΔH = −87.9 kJ/mol <— note sign change on the ΔH

3) Add the three revised equations. C and 2S will cancel.

Example #3: Given the following data:

SrO(s) + CO2(g) —> SrCO3(s)

ΔH = −234 kJ

2SrO(s) —> 2Sr(s) + O2(g)

ΔH = +1184 kJ

2SrCO3(s) —> 2Sr(s) + 2C(s, gr) + 3O2(g)

ΔH = +2440 kJ

Find the ΔH of the following reaction:

C(s, gr) + O2(g) —> CO2(g)

Solution:

1) Analyze what must happen to each equation:

1. a) first eq —> flip it (this put the CO2 on the right-hand side, where we want it)
2. b) second eq —> do not flip it, divide through by two (no flip because we need to cancel the SrO, divide by two because we only need to cancel one SrO)
3. c) third equation —> flip it (to put the SrCO3 on the other side so we can cancel it), divide by two (since we need to cancel only one SrCO3)

Notice that what we did to the third equation also sets up the Sr to be canceled. Why not also multiply the first equation by two (to get 2SrO for canceling)? Because we only want one CO2 in the final answer, not two. Notice also that I ignored the oxygen. If everything is right, the oxygen will take care of itself.

2) Apply all the above changes (notice what happens to the ΔH values):

SrCO3(s) —> SrO(s) + CO2(g)

ΔH = +234 kJ

SrO(s) —> Sr(s) + 1⁄2O2(g)

ΔH = +592 kJ

Sr(s) + C(s, gr) + 3⁄2O2(g) —> SrCO3(s)

ΔH = −1220 kJ

3) Here is a list of what is eliminated when everything is added:

SrCO3, SrO, Sr, 1⁄2O2

The last one comes from 3⁄2O2 on the left in the third equation and 1⁄2O2 on the right in the second equation.

4) Add the equations and the ΔH values:

+234 + (+592) + (−1220) = −394

C(s, gr) + O2(g) —> CO2(g)

ΔH fo = −394 kJ

Notice the subscripted f. This is the formation reaction for CO2 and its value can be looked up, either in your textbook or online.

Example #4: Given the following information:

2NO(g) + O2(g) —> 2NO2(g)

ΔH = −116 kJ

2N2(g) + 5O2(g) + 2H2O(ℓ) —> 4HNO3(aq)

ΔH = −256 kJ

N2(g) + O2(g) —> 2NO(g)

ΔH = +183 kJ

Calculate the enthalpy change for the reaction below:

3NO2(g) + H2O(ℓ) —> 2HNO3(aq) + NO(g)

ΔH = ???

Solution:

1) Analyze what must happen to each equation:

1. a) first eq —> flip; multiply by 3⁄2 (this gives 3NO2 as well as the 3NO which will be necessary to get one NO in the final answer)
2. b) second eq —> divide by 2 (gives two nitric acids in the final answer)
3. c) third eq —> flip (cancels 2NO as well as nitrogen)

2) Comment on the oxygens:

1. a) step 1a above puts 3⁄2O2 on the right
2. b) step 1b puts 5⁄2O2 on the left
3. c) step 1c puts 2⁄2O2 on the right

Besides, a and c give 5⁄2O2 on the right to cancel out the 5⁄2O2 on the left.

3) Apply all the changes listed above:

3NO2(g) —> 3NO(g) + 3⁄2O2(g)

ΔH = +174 kJ

N2(g) + 5⁄2O2(g) + H2O(ℓ) —> 2HNO3(aq)

ΔH = −128 kJ

2NO(g) —> N2(g) + O2(g)

ΔH = −183 kJ

4) Add the equations and the ΔH values:

+174 + (−128) + (−183) = −137 kJ

3NO2(g) + H2O(ℓ) —> 2HNO3(aq) + NO(g)

ΔH = −137 kJ

These are some fundamental examples of demonstrating Hess’s law problem. To know more avail our service to learn from our experts who can unstuck you from any Hess’s law related problem. Also, we provide materials with more examples and Hess’s law practice problems help to ace your Hess’s law test. Moreover, our services are available 24×7.

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