SME
Material & Energy Balances · Problem type
Heat of reaction: Q̇ = ξ̇ΔH°ᵣ
Use this for “10 mol/s of CH₄ burns at 25 °C, ΔH°ᵣ = −802.3 kJ/mol. How much heat leaves?”
Diagram
ΔH°ᵣ: Reactants, 25 °C → Products, 25 °C, −802.3 kJ/mol
This is an example. Type your own values to start a new problem.
ξ̇ (mol/s)
=
ṅA (mol/s)
÷
νA
Q̇ (kW)
=
ξ̇ (mol/s)
×
(
ΔH°ᵣ (kJ/mol)
)
ṅA is how fast one reactant is used up and νA its coefficient in the balanced equation (1 for CH₄).
Formulas
ξ̇
=
ṅA
νA
10
=
10
1
Q̇
=
ξ̇
×
ΔH°ᵣ
−8,023 = 10 × (−802.3)
Assumptions
•
Feed and products are both at 25 °C, so only the reaction changes the enthalpy.
•
Steady open system, no shaft work; kinetic and potential energy changes are negligible.
•
The extent ξ̇ counts reaction events per second: a reactant used up over its coefficient.
•
Q̇ < 0: heat leaves the reactor. ΔH°ᵣ = −802.3 kJ/mol for CH₄ with water as vapor.
Step-by-step
WE KNOW
ṅA = 10 mol/s, νA = 1, ΔH°ᵣ = −802.3 kJ/mol
FIND
extent of reaction (ξ̇), heat added (Q̇)
Step 1 · Find extent of reaction (ξ̇)
Use ξ̇ = ṅA ÷ νA
Each reaction event uses νA molecules of the reactant, so divide what is used up by νA.
ξ̇
=
10
1
ξ̇
=
10
mol/s
Step 2 · Find heat added (Q̇)
Use Q̇ = ξ̇ × ΔH°ᵣ
The energy balance with feed and products at 25 °C: each mole of reaction gives ΔH°ᵣ.
Q̇
=
10
×
(−802.3)
Q̇
=
−8,023
kW
Check
Put the numbers back in. Both sides should match.
10
=
10
1
✓
−8,023 = 10 × (−802.3) ✓
Topic check
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