ChemGem

How to Use Hess’s Law (Step-by-Step, with Practice)

ΔH belongs to a reaction exactly as written. Scale the equation and ΔH scales; reverse it and ΔH flips sign. Hess’s law (adding reactions to reach a target) then lets you add known reactions to reach one you need.

Spot it, solve it, practice it: 4 steps, 1 worked example, 8 practice problems. Free, no account.

ChemGemUpdated Section 7.3

The route · 4 steps

  1. Match each supplied reaction to the target.
  2. Reverse or scale whole equations; adjust ΔH the same way.
  3. Cancel intermediates and add the ΔH values.
  4. Check the sum matches the target.
Try one now See a worked example
Spot it

How to recognize Hess’s law questions

Look for given reactions with ΔH values that you must reverse, multiply, or add to reach a target reaction.

Example wording, with the clue words marked:

A → B has ΔH = −100 kJ. What is ΔH for B → A?

Not sure it’s this type? Compare the neighbors:

Step-by-step

Worked example: Hess’s law

The idea

Hess’s law: add the steps, add their enthalpies
ΔHoverall = ΔHstep 1 + ΔHstep 2 + …
It works because ΔH depends only on the start and end, not on the path.
Two paths, same total
Enthalpy, HABC−30 kJ+10 kJ−20 kJdirect
A → B → C and the direct A → C start and end at the same levels, so both give −20 kJ.
Change the equation, change ΔH the same way
What you do to the equationWhat happens to ΔHExample
Use it as writtenKeep ΔH2H2 + O2 → 2H2O   ΔH = −571.6 kJ
Multiply every coefficient by a factorMultiply ΔH by that factor× ½: H2 + ½O2 → H2O   ΔH = −285.8 kJ
Reverse itFlip the sign2H2O → 2H2 + O2   ΔH = +571.6 kJ
Add equationsAdd their ΔH valuesCancel any substance on both sides first
ΔH is for the reaction as written, under the stated conditions. Water values are for liquid water.

On the periodic table: H2, O2 and H2O are different substances made from the same elements; keep each formula intact when you scale or reverse.

On the periodic table: The periodic table gives H as atomic number 1, average mass 1.008 u (an average, not a mass number; use supplied rounded masses when given). Reaction enthalpies come from supplied energy data, not from the table.

Worked example · 3 steps

A → B has ΔH = −30 kJ. B → C has ΔH = +10 kJ. Find ΔH for A → C, then C → A.

A, B and C are placeholder substances, not element symbols.

  1. 1
    Add the reaction steps

    B is made in the first step and used in the second, so it cancels.

    A → B
    B → C
    A → C
  2. 2
    Add their enthalpies

    Keep each sign. The larger negative step keeps the sum negative.

    ΔH = (−30 kJ) + (+10 kJ) = −20 kJ
  3. 3
    Reverse for the second target

    Reversing the reaction reverses the sign.

    ReactionΔH
    A → C−20 kJ
    C → A (reversed)+20 kJ
  4. ✓
    Answer

    A → C: ΔH = −20 kJ; C → A: ΔH = +20 kJ

Your turn

Hess’s law practice problems

Answer first, then check. Hints open one at a time; the answer and full solution open after you try.

Problem 1 of 8 · warm-up

A → B has ΔH = −100 kJ. What is ΔH for B → A?

Hint 1 · the clue

The same reaction is reversed.

Hint 2 · the next move

The same reaction written backward is asked: reversing a reaction flips the sign of ΔH.

Your path:

  1. See B → A is A → B reversed
  2. Keep the size of ΔH
  3. Flip its sign
Answer

Answer: +100 kJ

Full worked solution

What you know: A → B has ΔH = −100 kJ. What you want: ΔH for B → A.

  1. Notice that B → A is the same reaction written backward. The starting and ending states swap.
  2. Enthalpy depends only on the starting and ending states. Swapping them reverses the sign of ΔH and keeps its size.
  3. So ΔH = −(−100 kJ) = +100 kJ.

Check: Go A → B and then back B → A: −100 kJ + 100 kJ = 0. You end where you began, so the total change must be zero.

Your route: Reverse reaction → Reverse ΔH sign

Answer: +100 kJ

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Problem 2 of 8

A → B has ΔH = −110 kJ. What is ΔH for B → A?

Hint 1 · the clue

The same reaction is reversed.

Hint 2 · the next move

The same reaction written backward is asked: reversing a reaction flips the sign of ΔH.

Your path:

  1. See B → A is A → B reversed
  2. Keep the size of ΔH
  3. Flip its sign
Answer

Answer: +110 kJ

Full worked solution

What you know: A → B has ΔH = −110 kJ. What you want: ΔH for B → A.

  1. Notice that B → A is the same reaction written backward. The starting and ending states swap.
  2. Enthalpy depends only on the starting and ending states. Swapping them reverses the sign of ΔH and keeps its size.
  3. So ΔH = −(−110 kJ) = +110 kJ.

Check: Go A → B and then back B → A: −110 kJ + 110 kJ = 0. You end where you began, so the total change must be zero.

Your route: Reverse reaction → Reverse ΔH sign

Answer: +110 kJ

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Problem 3 of 8

For A → B, ΔH = −100 kJ. What is ΔH for 3A → 3B?

Hint 1 · the clue

The reaction amounts are multiplied.

Hint 2 · the next move

Every coefficient is multiplied by 3: scaling a reaction scales ΔH by the same factor, with no sign change.

Your path:

  1. Compare coefficients: all times 3
  2. Multiply ΔH by 3
  3. Keep the sign
Answer

Answer: −300 kJ

Full worked solution

What you know: A → B has ΔH = −100 kJ. What you want: ΔH for 3A → 3B.

  1. Compare the equations. Every coefficient in 3A → 3B is 3 times the original. The direction is unchanged.
  2. ΔH is for the reaction exactly as written, so three times the amount releases three times the heat. Multiply ΔH by 3.
  3. ΔH = 3 × (−100 kJ) = −300 kJ.

Check: The sign stays negative because the reaction still runs forward. Only the size changes.

Watch out: Scaling does not flip the sign. Only reversing the reaction does that.

Your route: Scale all coefficients → Scale ΔH equally

Answer: −300 kJ

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Problem 4 of 8

For A → B, ΔH = −110 kJ. What is ΔH for 3A → 3B?

Hint 1 · the clue

The reaction amounts are multiplied.

Hint 2 · the next move

Every coefficient is multiplied by 3: scaling a reaction scales ΔH by the same factor, with no sign change.

Your path:

  1. Compare coefficients: all times 3
  2. Multiply ΔH by 3
  3. Keep the sign
Answer

Answer: −330 kJ

Full worked solution

What you know: A → B has ΔH = −110 kJ. What you want: ΔH for 3A → 3B.

  1. Compare the equations. Every coefficient in 3A → 3B is 3 times the original. The direction is unchanged.
  2. ΔH is for the reaction exactly as written, so three times the amount releases three times the heat. Multiply ΔH by 3.
  3. ΔH = 3 × (−110 kJ) = −330 kJ.

Check: The sign stays negative because the reaction still runs forward. Only the size changes.

Watch out: Scaling does not flip the sign. Only reversing the reaction does that.

Your route: Scale all coefficients → Scale ΔH equally

Answer: −330 kJ

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Problem 5 of 8

A → B has ΔH = −100 kJ and B → C has ΔH = +15 kJ. What is ΔH for A → C?

Hint 1 · the clue

Several reaction steps lead to a target.

Hint 2 · the next move

Two step reactions share B, and A → C is asked: add the steps and their ΔH values (Hess's law).

Your path:

  1. Check both steps run toward A → C
  2. Add equations; B cancels
  3. Add ΔH values with signs
Answer

Answer: −85 kJ

Full worked solution

What you know: A → B, ΔH = −100 kJ. B → C, ΔH = +15 kJ. What you want: ΔH for A → C.

  1. Match the target. A is a reactant in step 1 and C is a product in step 2, just as in A → C. So use both steps as written.
  2. Add the equations: A + B → B + C. B is made in step 1 and used in step 2, so it cancels, leaving A → C.
  3. Add the enthalpies with their signs: −100 kJ + 15 kJ = −85 kJ.

Check: The large negative step outweighs the small positive one, so the total should be negative and a bit smaller in size than 100. It is.

Watch out: Keep the + sign on 15. Subtracting it would give −115 kJ.

Your route: Add equations → Cancel intermediate → Add ΔH

Answer: −85 kJ

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Problem 6 of 8

A → B has ΔH = −110 kJ and B → C has ΔH = +15 kJ. What is ΔH for A → C?

Hint 1 · the clue

Several reaction steps lead to a target.

Hint 2 · the next move

Two step reactions share B, and A → C is asked: add the steps and their ΔH values (Hess's law).

Your path:

  1. Check both steps run toward A → C
  2. Add equations; B cancels
  3. Add ΔH values with signs
Answer

Answer: −95 kJ

Full worked solution

What you know: A → B, ΔH = −110 kJ. B → C, ΔH = +15 kJ. What you want: ΔH for A → C.

  1. Match the target. A is a reactant in step 1 and C is a product in step 2, just as in A → C. So use both steps as written.
  2. Add the equations: A + B → B + C. B is made in step 1 and used in step 2, so it cancels, leaving A → C.
  3. Add the enthalpies with their signs: −110 kJ + 15 kJ = −95 kJ.

Check: The large negative step outweighs the small positive one, so the total should be negative and a bit smaller in size than 110. It is.

Watch out: Keep the + sign on 15. Subtracting it would give −125 kJ.

Your route: Add equations → Cancel intermediate → Add ΔH

Answer: −95 kJ

YOUR PERIODIC-TABLE REFERENCE

1HHydrogen1.008Group 1 · Period 1

Atomic number identifies proton count. The decimal mass is an average, not an isotope’s mass number. Use supplied rounded masses when a problem gives them.

Problem 7 of 8

A reaction has ΔH = −30 kJ. Reverse it and double its coefficients. New ΔH?

Hint 1 · the clue

Handle one change at a time. Reversing a reaction flips the sign of ΔH; multiplying its coefficients multiplies ΔH by the same factor.

Hint 2 · the next move

The reaction is reversed and its coefficients doubled: reversing flips the sign, doubling doubles the size.

Your path:

  1. Reverse → flip the sign of ΔH
  2. Double → multiply ΔH by 2
Answer

+60 kJ: reversing changes the sign; doubling scales the magnitude.

Full worked solution

What you know: The original reaction has ΔH = −30 kJ. What you want: ΔH after reversing it and doubling its coefficients.

  1. Reverse the reaction. Reversing swaps the starting and ending states, so the sign flips: −30 kJ becomes +30 kJ.
  2. Double the coefficients. Twice the amount means twice the heat: 2 × (+30 kJ) = +60 kJ.
  3. So the new ΔH is +60 kJ: reversing changes the sign; doubling scales the magnitude.

Check: Do the steps in the other order: 2 × (−30 kJ) = −60 kJ, then reverse to get +60 kJ. The order does not matter.

Watch out: Doubling never changes the sign. Only reversing does.

Answer: +60 kJ: reversing changes the sign; doubling scales the magnitude.

Problem 8 of 8

Given (1) C(s) + O2(g) → CO2(g), ΔH = −394 kJ, and (2) CO(g) + ½O2(g) → CO2(g), ΔH = −283 kJ. Find ΔH for C(s) + ½O2(g) → CO(g).

Hint 1 · the clue

CO has to end up as a product. Which reaction should you reverse?

Hint 2 · the next move

Two given reactions with ΔH must be combined to make a target: that signals Hess's law.

Your path:

  1. Keep (1): C is a reactant
  2. Reverse (2) so CO is product; flip sign
  3. Add equations; CO2 cancels
  4. Add the ΔH values
Answer

−111 kJ

Full worked solution

What you know: (1) C(s) + O2(g) → CO2(g), ΔH = −394 kJ. (2) CO(g) + ½O2(g) → CO2(g), ΔH = −283 kJ. What you want: ΔH for C(s) + ½O2(g) → CO(g).

  1. Match the target. C is a reactant in (1) and in the target, so keep (1) as written: ΔH = −394 kJ.
  2. CO must be a product, but it is a reactant in (2). Reverse (2): CO2(g) → CO(g) + ½O2(g). Reversing flips the sign: ΔH = +283 kJ.
  3. Add the equations. CO2 appears on both sides and cancels. O2: 1 on the left minus ½ on the right leaves ½O2 on the left. Result: C(s) + ½O2(g) → CO(g), the target.
  4. Add the enthalpies: −394 kJ + 283 kJ = −111 kJ.

Check: The final equation matches the target exactly, so the sum is the answer.

Watch out: Reverse the step's sign when you reverse its equation. Using −283 would give −677 kJ.

Answer: −111 kJ

Watch out

Common Hess’s law mistakes