Hess's Law Calculations (SL IB Chemistry)

Calculate the enthalpy of reaction for:

 2N₂ (g) + 6H₂ (g) → 4NH₃ (g)

Given the data:

4NH₃ (g) + 3O₂ (g) → 2N₂ (g) + 6H₂O (l),           ΔH₁ = -1530 kJ mol^-1

H₂ (g) + ½O₂ (g )→  H₂O (l),                               ΔH₂ = -288 kJ mol^-1

Answer:

• Step 1: Begin by writing the target enthalpy change at the top of your diagram from left to right:

• Step 2: Next, write the alternative route at the bottom of your cycle and connect the top and bottom with arrows pointing in the correct directions:

• Step 3: Add the enthalpy data and adjust, as necessary, for different molar amounts

• Step 4:Write the Hess's Law calculation out:

ΔH_r = +6ΔH₂ – ΔH₁ = + (-288 x 6) - ( -1530 ) = -198 kJ

• • Two important rules:
    • If you follow the direction of the arrow you ADD the quantity
    • If you go against the arrow you SUBTRACT the quantity

What is the enthalpy change, in kJ, for the reaction below?

4FeO (s) + O₂ (g) → 2Fe₂O₃ (s)

Given the data:

2Fe(s) + O₂ (g) → 2FeO (s)                  ∆H = – 544 kJ

   4Fe (s) + 3O₂ (g) → 2Fe₂O₃ (s)             ∆H = –1648 kJ

Answer:

• Step 1: Draw the Hess cycle and add the known values

• Step 2: Write the Hess's Law calculation out:

Follow the alternative route and the process the calculation

ΔH_r = - ( - 544 x 2) + (- 1648) = - 560 kJ

Consider the following reactions.

   N₂ (g) + O₂ (g) → 2NO (g)         ∆H = +180 kJ

   2NO₂ (g) → 2NO (g) + O₂ (g)    ∆H = +112 kJ

What is the ∆H value, in kJ, for the following reaction?

   N₂ (g) + 2O₂ (g) → 2NO₂ (g)

Answer:

• Step 1: Identify which given equation contains the product you want

   This equation contains the desired product on the left side:

  2NO₂ (g) → 2NO (g) + O₂ (g)                             ∆H = +112 kJ

• Step 2: Adjust the equation if necessary, to give the same product. If you reverse it, reverse the ΔH value

   Reverse it and reverse the sign

2NO (g) + O₂ (g)  →   2NO₂ (g)                         ∆H = -112 kJ

• Step 3:  Adjust the equation if necessary, to give the same number of moles of product

   The equation contains the same number of moles as in the question, so no need to adjust the moles

• Step 4: Identify which given equation contains your reactant

   This equation contains the reactant

N₂ (g) + O₂ (g) → 2NO (g)         ∆H = +180 kJ

• Step 5: Adjust the equation if necessary, to give the same reactant. If you reverse it, reverse the ΔH value

   No need to reverse it as the reactant is already on the left side

• Step 6: Adjust the equation if necessary, to give the same number of moles of reactant

 

• Step 7: Add the two equations together

N₂ (g) + O₂ (g) → 2NO (g)                                             ∆H = +180 kJ

2NO (g) + O₂ (g) → 2NO₂ (g)                                        ∆H  = -112 kJ

• Step 8: Cancel the common items

N₂ (g) + O₂ (g) + 2NO (g) + O₂ (g) → 2NO (g) + 2NO₂ (g)

• Step 9: Add the two ΔH values together to get the one you want

N₂ (g) + 2O₂ (g) → 2NO₂ (g)                         ∆H  = +180-112 = +68 kJ

The enthalpy changes for two reactions are given.

Br₂ (l) + F₂ (g) → 2BrF (g)         ΔH = x kJ

Br₂ (l) + 3F₂ (g) → 2BrF₃ (g)      ΔH = y kJ

What is the enthalpy change for the following reaction?

BrF (g) + F₂ (g) → BrF₃ (g)

A. x – y

B. y - x

C. ½ (–x + y)

D. ½ (x – y)

Answer:

• The correct option is C.
  • The second equation contains the desired product, but it needs to be halved to make 1 mole

Br₂ (l) + 3F₂ (g) → 2BrF₃ (g)    ΔH = y   becomes

 ½Br₂ (l) + 1½F₂ (g) → BrF₃ (g)   ½ΔH = ½y

• • The first equation contains the reactant, but it needs to be reversed and halved:

Br₂ (l) + F₂ (g) → 2BrF (g)       ΔH = x   becomes

BrF (g)  → ½Br₂ (l) + ½F₂ (g)   ½ΔH = -½x

• • Combine the two equations and cancel the common terms:

½Br₂ (l) + 1½F₂ (g) → BrF₃ (g)      ½ ΔH = y kJ

BrF (g)  → ½Br₂ (l) + ½F₂ (g)      ½  ΔH = -x kJ

BrF (g) + F₂ (g) → BrF₃ (g)    ΔH = ½y  +  -½x  =  ½(-x + y)