SME
Organic Chemistry I · Problem type
ΔH of the propagation steps
Use this for “For CH₄ + Br₂, take C–H 439, H–Br 366, Br–Br 193 and C–Br 293 kJ/mol. Find ΔH of each propagation step and overall.”
Diagram
ΔH₁: R–H + X· + X₂ → R· + H–X + X₂, +7 kJ/mol
ΔH₂: R· + H–X + X₂ → R–X + H–X + X·, −108 kJ/mol
ΔH = ΔH₁ + ΔH₂ = 7 + (−108)
ΔH = −101 kJ/mol
Down the ladder: heat is given off (exothermic).
This is an example. Type your own values to start a new problem.
D(C–H)
Bond enthalpy of C–H
example · kJ/mol
D(H–X)
Bond enthalpy of H–X
example · kJ/mol
D(X–X)
Bond enthalpy of X–X
example · kJ/mol
D(C–X)
Bond enthalpy of C–X
example · kJ/mol
ΔH₁
Enthalpy change of step 1
calculated · kJ/mol
ΔH₂
Enthalpy change of step 2
calculated · kJ/mol
ΔH
Overall enthalpy change
calculated · kJ/mol
Formulas
ΔH₁ = D(C–H) − D(H–X)
7 = 439 − 432
ΔH₂ = D(X–X) − D(C–X)
−108 = 243 − 351
ΔH = ΔH₁ + ΔH₂
−101 = 7 + (−108)
Assumptions
•
Step 1: X· takes an H from R–H, breaking C–H and forming H–X. Step 2: R· takes X from X₂, breaking X–X and forming C–X.
•
Breaking a bond takes in its bond enthalpy and forming one gives the same amount back; all species are gases.
•
The two steps add to the overall reaction R–H + X₂ → R–X + H–X, so their ΔH values add too.
•
X–X and H–X take the four halogens’ values: F–F 159, Cl–Cl 243, Br–Br 193, I–I 151; H–F 570, H–Cl 432, H–Br 366, H–I 298 kJ/mol.
Step-by-step
WE KNOW
D(C–H) = 439 kJ/mol, D(H–X) = 432 kJ/mol, D(X–X) = 243 kJ/mol, D(C–X) = 351 kJ/mol
FIND
enthalpy change of step 1 (ΔH₁), enthalpy change of step 2 (ΔH₂), overall enthalpy change (ΔH)
Step 1 · Find enthalpy change of step 1 (ΔH₁)
Use ΔH₁ = D(C–H) − D(H–X)
Energy in to break C–H less energy out from forming H–X.
ΔH₁ = 439 − 432
ΔH₁ = 7 kJ/mol → ΔH₁ > 0: taking the H is uphill, so this step sets the rate; the more uphill, the more selective the halogen.
Step 2 · Find enthalpy change of step 2 (ΔH₂)
Use ΔH₂ = D(X–X) − D(C–X)
Energy in to break X–X less energy out from forming C–X.
ΔH₂ = 243 − 351
ΔH₂ = −108 kJ/mol
Step 3 · Find overall enthalpy change (ΔH)
Use ΔH = ΔH₁ + ΔH₂
The two steps add to the overall reaction, so add their ΔH values.
ΔH = 7 + (−108)
ΔH = −101 kJ/mol → ΔH < 0: the chain gives off heat overall, so it keeps running once started.
Check
Put the numbers back in. Both sides should match.
7 = 439 − 432 ✓
−108 = 243 − 351 ✓
−101 = 7 + (−108) ✓
Topic check
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