### How To Do Hess’S Law Problems?

- Marvin Harvey
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Hess’ Law – using three equations and their enthalpies Germain Henri Hess, in 1840, discovered a very useful principle which is named for him: The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps.

2C(s) + H 2 (g) -> C 2 H 2 (g) | ΔH° = ??? kJ |

Given the following thermochemical equations:

C 2 H 2 (g) + 5 ⁄ 2 O 2 (g) -> 2CO 2 (g) + H 2 O(ℓ) | ΔH° = −1299.5 kJ |

C(s) + O 2 (g) -> CO 2 (g) | ΔH° = −393.5 kJ |

H 2 (g) + 1 ⁄ 2 O 2 (g) -> H 2 O(ℓ) | ΔH° = −285.8 kJ |

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a) first eq: flip it so as to put C 2 H 2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing. We need one H 2 on the reactant side and that’s what we have.2) Rewrite all three equations with changes applied:

2CO 2 (g) + H 2 O(ℓ) -> C 2 H 2 (g) + 5 ⁄ 2 O 2 (g) | ΔH° = +1299.5 kJ |

2C(s) + 2O 2 (g) -> 2CO 2 (g) | ΔH° = −787 kJ |

H 2 (g) + 1 ⁄ 2 O 2 (g) -> H 2 O(ℓ) | ΔH° = −285.8 kJ |

Notice that the ΔH values changed as well.3) Examine what cancels: 2CO 2 ⇒ first & second equation H 2 O ⇒ first & third equation 5 ⁄ 2 O 2 ⇒ first & sum of 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: CS 2 (ℓ) + 3O 2 (g) -> CO 2 (g) + 2SO 2 (g) Given:

C(s) + O 2 (g) -> CO 2 (g) | ΔH = −393.5 kJ/mol |

S(s) + O 2 (g) -> SO 2 (g) | ΔH = −296.8 kJ/mol |

C(s) + 2S(s) -> CS 2 (ℓ) | ΔH = +87.9 kJ/mol |

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leave eq 1 untouched (want CO 2 as a product) multiply second eq by 2 (want to cancel 2S, also want 2SO 2 on product side) flip 3rd equation (want CS 2 as a reactant) 2) The result:

C(s) + O 2 (g) -> CO 2 (g) | ΔH = −393.5 kJ/mol | ||

2S(s) + 2O 2 (g) -> 2SO 2 (g) | ΔH = −593.6 kJ/mol | CS 2 (ℓ) -> C(s) + 2S(s) | ΔH = −87.9 kJ/mol |

3) Add the three revised equations. C and 2S will cancel.4) Add the three enthalpies for the final answer. Example #3: Given the following data:

SrO(s) + CO 2 (g) -> SrCO 3 (s) | ΔH = −234 kJ |

2SrO(s) -> 2Sr(s) + O 2 (g) | ΔH = +1184 kJ |

2SrCO 3 (s) -> 2Sr(s) + 2C(s, gr) + 3O 2 (g) | ΔH = +2440 kJ |

Find the ΔH of the following reaction: C(s, gr) + O 2 (g) -> CO 2 (g)

- Solution:
- 1) Analyze what must happen to each equation:

a) first eq -> flip it (this put the CO 2 on the right-hand side, where we want it) 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) c) third equation -> flip it (to put the SrCO 3 on the other side so we can cancel it), divide by two (since we need to cancel only one SrCO 3 ) Notice that what we did to the third equation also sets up the Sr to be cancelled. Why not also multiply first equation by two (to get 2SrO for canceling)? Because we only want one CO 2 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):

SrCO 3 (s) -> SrO(s) + CO 2 (g) | ΔH = +234 kJ |

SrO(s) -> Sr(s) + 1 ⁄ 2 O 2 (g) | ΔH = +592 kJ |

Sr(s) + C(s, gr) + 3 ⁄ 2 O 2 (g) -> SrCO 3 (s) | ΔH = −1220 kJ |

3) Here is a list of what is eliminated when everything is added: SrCO 3, SrO, Sr, 1 ⁄ 2 O 2 The last one comes from 3 ⁄ 2 O 2 on the left in the third equation and 1 ⁄ 2 O 2 on the right in the second equation.4) Add the equations and the ΔH values: +234 + (+592) + (−1220) = −394

C(s, gr) + O 2 (g) -> CO 2 (g) | ΔH f o = −394 kJ |

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2NO(g) + O 2 (g) -> 2NO 2 (g) | ΔH = −116 kJ |

2N 2 (g) + 5O 2 (g) + 2H 2 O(ℓ) -> 4HNO 3 (aq) | ΔH = −256 kJ |

N 2 (g) + O 2 (g) -> 2NO(g) | ΔH = +183 kJ |

Calculate the enthalpy change for the reaction below:

3NO 2 (g) + H 2 O(ℓ) -> 2HNO 3 (aq) + NO(g) | ΔH = ??? |

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a) first eq -> flip; multiply by 3 ⁄ 2 (this gives 3NO 2 as well as the 3NO which will be necessary to get one NO in the final answer) b) second eq -> divide by 2 (gives two nitric acid in the final answer) c) third eq -> flip (cancels 2NO as well as nitrogen) 2) Comment on the oxygens: a) step 1a above puts 3 ⁄ 2 O 2 on the right b) step 1b puts 5 ⁄ 2 O 2 on the left c) step 1c puts 2 ⁄ 2 O 2 on the right In addition, a and c give 5 ⁄ 2 O 2 on the right to cancel out the 5 ⁄ 2 O 2 on the left.3) Apply all the changes listed above:

3NO 2 (g) -> 3NO(g) + 3 ⁄ 2 O 2 (g) | ΔH = +174 kJ |

N 2 (g) + 5 ⁄ 2 O 2 (g) + H 2 O(ℓ) -> 2HNO 3 (aq) | ΔH = −128 kJ |

2NO(g) -> N 2 (g) + O 2 (g) | ΔH = −183 kJ |

4) Add the equations and the ΔH values: +174 + (−128) + (−183) = −137 kJ

3NO 2 (g) + H 2 O(ℓ) -> 2HNO 3 (aq) + NO(g) | ΔH = −137 kJ |

Example #5: Calculate ΔH for this reaction: CH 4 (g) + NH 3 (g) -> HCN(g) + 3H 2 (g) given:

N 2 (g) + 3H 2 (g) -> 2NH 3 (g) | ΔH = −91.8 kJ |

C(s) + 2H 2 (g) -> CH 4 (g) | ΔH = −74.9 kJ |

H 2 (g) + 2C(s) + N 2 (g) -> 2HCN(g) | ΔH = +270.3 kJ |

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a) first eq ⇒ flip and divide by 2 (puts one NH 3 on the reactant side) b) second eq ⇒ flip (puts one CH 4 on the reactant side) c) third eq ⇒ divide by 2 (puts one HCN on the product side) 2) Rewite all equations with the changes:

NH 3 (g) -> 1 ⁄ 2 N 2 (g) + 3 ⁄ 2 H 2 (g) | ΔH = +45.9 kJ | CH 4 (g) -> C(s) + 2 H 2 (g) | ΔH = +74.9 kJ | 1 ⁄ 2 H 2 (g) + C(s) + 1 ⁄ 2 N 2 (g) -> HCN(g) | ΔH = +135.15 kJ |

3) What cancels when you add the equations: 1 ⁄ 2 N 2 (g) ⇒ first and third equations C(s) ⇒ second and third equations 1 ⁄ 2 H 2 (g) on the left side of the third equation cancels out 1 ⁄ 2 H 2 (g) on the right, leaving a total of 3H 2 (g) on the right (which is what we want) 4) Calculate the ΔH for our reaction: (+45.9 kJ) + (+74.9 kJ) + (+135.15) = +255.95 kJ Rounded off to three sig figs gives +260. kJ (note use of explicit decimal point) Example #6: Determine the heat of reaction for the oxidation of iron: 2Fe(s) + 3 ⁄ 2 O 2 (g) -> Fe 2 O 3 (s) given the thermochemical equations:

2Fe(s) + 6H 2 O(ℓ) -> 2Fe(OH) 3 (s) + 3H 2 (g) | ΔH = +322 kJ |

Fe 2 O 3 (s) + 3H 2 O(ℓ) -> 2Fe(OH) 3 (s) | ΔH = +289 kJ |

2H 2 (g) + O 2 (g) -> 2H 2 O(ℓ) | ΔH = –572 kJ |

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2Fe(s) + 6H 2 O(ℓ) -> 2Fe(OH) 3 (s) + 3H 2 (g) | ΔH = +322 kJ | 2Fe(OH) 3 (s) -> Fe 2 O 3 (s) + 3H 2 O(ℓ) | ΔH = −289 kJ | 3H 2 (g) + 3 ⁄ 2 O 2 (g) -> 3H 2 O(ℓ) | ΔH = –858 kJ | 3 ⁄ 2 |

2) Adding up the equations gives the target equation. Adding the enthalpies gives us our answer: 2Fe(s) + 3 ⁄ 2 O 2 (g) -> Fe 2 O 3 (s) ΔH = −825 kJ

- Note how the multiplying factor doesn’t have to be an integer value.
- Example #7: Using the following thermochemical equations, calculate the standard enthalpy of combustion for one mole of liquid acetone (C 3 H 6 O).

3C(s) + 3H 2 (g) + 1 ⁄ 2 O 2 (g) -> C 3 H 6 O(ℓ) | ΔH° = −285.0 kJ |

C(s) + O 2 (g) -> CO 2 (g) | ΔH° = −394.0 kJ |

H 2 (g) + 1 ⁄ 2 O 2 (g) -> H 2 O(ℓ) | ΔH° = −286.0 kJ |

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C 3 H 6 O(ℓ) + 4O 2 (g) -> 3CO 2 (g) + 3H 2 O(ℓ) 2) The first data equation needs to be reversed, so as to put acetone on the reactant side. Here are all three data equations with the first one changed:

C 3 H 6 O(ℓ) -> 3C(s) + 3H 2 (g) + 1 ⁄ 2 O 2 (g) | ΔH° = +285.0 kJ |

C(s) + O 2 (g) -> CO 2 (g) | ΔH° = −394.0 kJ |

H 2 (g) + 1 ⁄ 2 O 2 (g) -> H 2 O(ℓ) | ΔH° = −286.0 kJ |

Note the sign change in the enthalpy when the equation is reversed.3) The second data equation needs to be changed to create a situation where the 3C(s) will cancel when the equations are added together:

C 3 H 6 O(ℓ) -> 3C(s) + 3H 2 (g) + 1 ⁄ 2 O 2 (g) | ΔH° = +285.0 kJ |

3C(s) + 3O 2 (g) -> 3CO 2 (g) | ΔH° = −1182.0 kJ |

H 2 (g) + 1 ⁄ 2 O 2 (g) -> H 2 O(ℓ) | ΔH° = −286.0 kJ |

Note that the enthalpy was also multiplied by three.4) The 3H 2 (g) (present in the first data equation) also needs to be removed from the final answer. Another multiplication by 3 is used:

C 3 H 6 O(ℓ) -> 3C(s) + 3H 2 (g) + 1 ⁄ 2 O 2 (g) | ΔH° = +285.0 kJ |

3C(s) + 3O 2 (g) -> 3CO 2 (g) | ΔH° = −1182.0 kJ |

3H 2 (g) + 3 ⁄ 2 O 2 (g) -> 3H 2 O(ℓ) | ΔH° = −858.0 kJ |

Note that the enthalpy was also multiplied by three.5) Add the three data equations together to recover the target equation laid out in step 1. Note the following: 1 ⁄ 2 O 2 (g) will cancel from each side, leaving 4O 2 (g) on the left-hand side. The enthalpies are added together to obtain the final answer of −1755 kJ.

C 3 H 8 (g) | ΔH 1 = −2219.9 kJ |

C(s, gr) | ΔH 2 = −393.5 kJ |

H 2 (g) | ΔH 3 = −285.8 kJ |

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3C(s, gr) + 4H 2 (g) -> C 3 H 8 (g) ΔH = ??? 2) Write the chemical equations for combustion of the three chemical species given:

C 3 H 8 (g) + 5O 2 (g) -> 3CO 2 (g) + 4H 2 O(ℓ) | ΔH 1 |

C(s, gr) + O 2 -> CO 2 (g) | ΔH 2 |

H 2 + 1 ⁄ 2 O 2 (g) -> H 2 O(ℓ) | ΔH 3 |

3) Modify the three data equations so as to reproduce the target equation:

3CO 2 (g) + 4H 2 O(ℓ) -> C 3 H 8 (g) + 5O 2 (g) | −ΔH 1 (reversed equation) |

3C(s, gr) + 3O 2 -> 3CO 2 (g) | 3ΔH 2 (multiplied by 3) |

4H 2 + 2O 2 (g) -> 4H 2 O(ℓ) | 4ΔH 3 (multiplied by 4) |

4) Add the enthalpies for the answer: −ΔH 1 + 3ΔH 2 + 4ΔH 3 −(−2219.9) + (3) (−393.5) + (4) (−285.8) = −103.8 kJ Answer may be verified, Example #9: Determine the standard enthalpy of formation for butane, using the following data:

C 4 H 10 (g) + 13 ⁄ 2 O 2 (g) -> 4CO 2 (g) + 5H 2 O(g) | ΔH 1 = −2657.4 kJ |

C(s, gr) + O 2 (g) -> CO 2 (g) | ΔH 2 = −393.5 kJ |

2H 2 (g) + O 2 (g) -> 2H 2 O(g) | ΔH 3 = −483.6 kJ |

Comment: note that the first and third equations are not standard combustion equations. The water in each equation is as a gas. In standard combustion equations, water is a liquid (its standard state). Solution: 1) The equation for the formation of butane is as follows: 4C(s, gr) + 5H 2 (g) -> C 4 H 10 (g) 2) The three data equations are modified as follows:

4CO 2 (g) + 5H 2 O(g) -> C 4 H 10 (g) + 13 ⁄ 2 O 2 (g) | −ΔH 1 |

4C(s, gr) + 8 ⁄ 2 O 2 (g) -> 4CO 2 (g) | 4ΔH 2 |

5H 2 (g) + 5 ⁄ 2 O 2 (g) -> 5H 2 O(g) | 2.5ΔH 3 |

4) The enthalpies are added together: −ΔH 1 + 4ΔH 2 + 2.5ΔH 3 −(−2657.4) + (−1574) + (−1209) = −125.6 kJ Is it correct?, Example #10: Calculate the enthalpy of formation for acetylene (C 2 H 2 ), given the following data:

C(s, gr) + O 2 (g) -> CO 2 (g) | ΔH = −393.5 kJ |

H 2 (g) + 1 ⁄ 2 O 2 (g) -> H 2 O(ℓ) | ΔH = −285.8 kJ |

2C 2 H 2 (g) + 5O 2 (g) -> 4CO 2 (g) + 2H 2 O(ℓ) | ΔH = −2598 kJ |

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2C(s, gr) + H 2 (g) -> C 2 H 2 (g) Remember, a formation reaction has all substances in their standard states and only one mole of product is produced.2) Manipulate the data equations: eq 1 -> do not flip, multiply by 2 (gets the 2C we need on the reactant side) eq 2 -> leave untouched (keeps H 2 on the reactant side and in the desired amount) eq 3 -> flip, divide by 2 (puts C 2 H 2 on the product side in the desired amount) 3) The result:

2C(s, gr) + 2O 2 (g) -> 2CO 2 (g) | ΔH = −787.0 kJ |

H 2 (g) + 1 ⁄ 2 O 2 (g) -> H 2 O(ℓ) | ΔH = −285.8 kJ |

2CO 2 (g) + H 2 O(ℓ) -> C 2 H 2 (g) + 5 ⁄ 2 O 2 (g) | ΔH = +1299 kJ |

4) Adding the three reactions together yields the desired equation. Adding the three enthalpies yields the enthalpy of formation for acetylene:

2C(s, gr) + H 2 (ℓ) -> C 2 H 2 (g) | ΔH = +226.2 kJ |

Bonus Example: Given:

2C 2 H 6 + 7O 2 -> 4CO 2 + 6H 2 O | ΔH = −3119.7 kJ | (1) |

2H 2 + O 2 -> 2H 2 O | ΔH = −478.84 kJ | (2) |

2CO + O 2 -> 2CO 2 | ΔH = −565.98 kJ | (3) |

please calculate delta H for the following reaction: C 2 H 6 + O 2 -> 3H 2 + 2CO Solution: Comment: this is not the usual ChemTeam manner of solving Hess’ Law problems. Which is why I coped it, so as to allow you to analyze how another brain approaches these problems. Pay close attention to the reasoning going on in step 4.1) Multiply equation (2) by 3 and designate as equation (4):

6H 2 + 3O 2 -> 6H 2 O | ΔH = −1436.52 kJ | (4) |

2) Multiply equation (3) by 2 and designate as equation (5):

4CO + 2O 2 -> 4CO 2 | ΔH = −1131.96 kJ | (5) |

3) Add equation (4) and equation (5) and designate as equation (6):

6H 2 + 4CO + 5O 2 -> 6H 2 O + 4CO 2 | ΔH = −2568.48 kJ | (6) |

4) Subtract equation (6) from equation (1) and designate as equation (7):

2C 2 H 6 + 7O 2 − (6H 2 + 4CO + 5O 2 ) -> 4CO 2 + 6H 2 O − (6H 2 O + 4CO 2 ) | ΔH = −551.22 kJ | |

2C 2 H 6 + 2O 2 − 6H 2 − 4CO -> 4CO 2 − 4CO 2 + 6H 2 O − 6H 2 O | ΔH = −551.22 kJ | |

2C 2 H 6 + 2O 2 -> 6H 2 + 4CO | ΔH = −551.22 kJ | (7) |

5) Dividing equation (7) by 2: C 2 H 6 + O 2 -> 3H 2 + 2CO ΔH = −275.61 kJ : Hess’ Law – using three equations and their enthalpies

### What is Hess’s law example problem?

Hess’s Law Example Problem – Question: Find the enthalpy change for the reaction CS 2 (l) + 3 O 2 (g) → CO 2 (g) + 2 SO 2 (g) when: C(s) + O 2 (g) → CO 2 (g); ΔH f = -393.5 kJ/mol S(s) + O 2 (g) → SO 2 (g); ΔH f = -296.8 kJ/mol C(s) + 2 S(s) → CS 2 (l); ΔH f = 87.9 kJ/mol Solution: Hess’s Law problems can take a little trial and error to get started.

One of the best places to begin is with a reaction with only one mole of reactant or product in the reaction. Our reaction needs one CO 2 in the product and the first reaction also has one CO 2 product. C(s) + O 2 (g) → CO 2 (g) ΔH f = -393.5 kJ/mol This reaction gives us the CO 2 needed on the product side and one of the O 2 needed on the reactant side.

The other two O 2 can be found in the second reaction. S(s) + O 2 (g) → SO 2 (g) ΔH f = -296.8 kJ/mol Since only one O 2 is in the reaction, multiply the reaction by two to get the second O 2, This doubles the ΔH f value.2 S(s) + 2 O 2 (g) → 2 SO 2 (g) ΔH f = -593.6 kJ/mol Combining these equations gives 2 S(s) + C(s) + 3 O 2 (g) → CO 2 (g) + SO 2 (g) The enthalpy change is the sum of the two reactions: ΔH f = -393.5 kJ/mol + -593.6 kJ/mol = -987.1 kJ/mol This equation has the product side needed in the problem but contains an extra two S and one C atom on the reactant side.

Fortunately, the third equation has the same atoms. If the reaction is reversed, these atoms are on the product side. When the reaction is reversed, the sign of the change in enthalpy is reversed. CS 2 (l) → C(s) + 2 S(s); ΔH f = -87.9 kJ/mol Add these two reactions together and the extra S and C atoms cancel out.

The remaining reaction is the reaction needed in the question. Since the reactions were added together, their ΔH f values are added together. ΔH f = -987.1 kJ/mol + -87.9 kJ/mol ΔH f = -1075 kJ/mol Answer: The change in enthalpy for the reaction CS 2 (l) + 3 O 2 (g) → CO 2 (g) + 2 SO 2 (g) is ΔH f = -1075 kJ/mol.

#### What are the requirements for Hess’s law?

What is Hess’s Law? – Russian Chemist and Physicist Germain Hess developed the concepts of thermochemistry and physical chemistry. He introduced the concept known as Hess’s Law of Constant Heat of Summation or Hess’s Law for short. This law has to do with net enthalpy in a reaction.

- Overall, it states that the total enthalpy change of a reaction is the sum of all the changes, no matter the number of steps or stages in the reaction (i.e.
- Net enthalpy and the number of steps in a reaction are independent of each other).
- The ideas of this law are seen throughout science, such as in the principle of conservation of energy, or the first law of thermodynamics, and the statement that enthalpy is a state function,

There are some requirements that the reaction has to follow in order to use Hess’s Law. For example, if there are multiple steps to the reactions, each equation must be correctly balanced. Also, all the steps of the reaction must start and end at constant temperatures and pressures in order to keep reaction conditions constant.

#### How do I find the correct path for Hess’s law?

Solution – Hess’s Law says the total enthalpy change does not rely on the path taken from beginning to end. Enthalpy can be calculated in one grand step or multiple smaller steps. To solve this type of problem, organize the given chemical reactions where the total effect yields the reaction needed. There are a few rules that you must follow when manipulating a reaction.

- The reaction can be reversed. This will change the sign of ΔH f,
- The reaction can be multiplied by a constant. The value of ΔH f must be multiplied by the same constant.
- Any combination of the first two rules may be used.

Finding a correct path is different for each Hess’s Law problem and may require some trial and error. A good place to start is to find one of the reactants or products where there is only one mole in the reaction. You need one CO 2, and the first reaction has one CO 2 on the product side.

- C(s) + O 2 (g) → CO 2 (g), ΔH f = -393.5 kJ/mol This gives you the CO 2 you need on the product side and one of the O 2 moles you need on the reactant side.
- To get two more O 2 moles, use the second equation and multiply it by two.
- Remember to multiply the ΔH f by two as well.2 S(s) + 2 O 2 (g) → 2 SO 2 (g), ΔH f = 2(-326.8 kJ/mol) Now you have two extra S’s and one extra C molecule on the reactant side that you don’t need.

The third reaction also has two S’s and one C on the reactant side. Reverse this reaction to bring the molecules to the product side. Remember to change the sign on ΔH f, CS 2 (l) → C(s) + 2 S(s), ΔH f = -87.9 kJ/mol When all three reactions are added, the extra two sulfur and one extra carbon atoms are canceled out, leaving the target reaction.

### What is Hess’ law in a chemical reaction?

In a chemical reaction, Hess’ law states that the change of enthalpy (it means, the heat of reaction under constant pressure) is independent of direction between the states of final and original. Also, this law requires the change in enthalpy (Δ H) for a reaction to be determined, even though it can not be measured directly.4.

## How is Hess’s law used in real life?

- Heats of formation of unstable intermediates like CO (g) and NO (g).
- Heat changes in phase transitions and allotropic transitions.
- Lattice energies of ionic substances by constructing Born-Haber cycles if the electron affinity to form the anion is known,or
- Electron affinities using a Born-Haber cycle with a theoretical lattice energy.

### What does Hess’s law say about the enthalpy of a reaction?

Introduction – Hess’s Law is named after Russian Chemist and Doctor Germain Hess. Hess helped formulate the early principles of thermochemistry. His most famous paper, which was published in 1840, included his law on thermochemistry. Hess’s law is due to enthalpy being a state function, which allows us to calculate the overall change in enthalpy by simply summing up the changes for each step of the way, until product is formed.

#### Why is Hess’ law useful?

Uses of Hess’s Law – Hess’s law of constant heat summation can be useful to determine the enthalpies of the following.

- Heats of unstable intermediates formation such as NO(g) and CO(g).
- The ionic substances’ lattice energies by constructing the Born-Haber cycles, if the electron affinity is known to form the,
- Heat changes in allotropic transitions and phase transitions.
- Electron affinities with a Born-Haber cycle using theoretical lattice energy.

### How to use Hess’s law for enthalpy change?

- Enthalpy change for R to U is endothermic We can draw an arrow from R to U directly.
- Enthalpy change for S to T is+226 kJ mol-1 We can draw an arrow from S to T directly.
- Enthalpy change for S to U is+151 kJ mol-1