Bond Energy Worksheet - Free Printable

Here are the step-by-step solutions for the problems on the worksheet.

To solve these, we use the formula provided in the example:
$\Delta H_{rxn}$ = [Sum of Energy to Break Bonds] – [Sum of Energy Released Forming Bonds]

* Breaking bonds (reactants) requires energy (positive value).
* Forming bonds (products) releases energy (we subtract this value).

We will use the values from the "Average Bond Energies" table in the image.

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Part 1: Estimate the enthalpy change ($\Delta H_{rxn}$)

1. $H-H + Cl-Cl \rightarrow H-Cl + H-Cl$

* Bonds Broken (Reactants):
* 1 mol H–H bond: $436 \text{ kJ/mol}$
* 1 mol Cl–Cl bond: $242 \text{ kJ/mol}$
* Total Energy In = $436 + 242 = 678 \text{ kJ}$
* Bonds Formed (Products):
* 2 mol H–Cl bonds: $2 \times 431 \text{ kJ/mol} = 862 \text{ kJ}$
* Total Energy Out = $862 \text{ kJ}$
* Calculation:
* $\Delta H = 678 - 862 = -184 \text{ kJ}$

2. Ethene ($C_2H_4$) + $F-F \rightarrow$ Difluoroethane ($C_2H_4F_2$)

* Bonds Broken (Reactants):
* 1 mol C=C double bond: $614 \text{ kJ/mol}$
* 4 mol C–H bonds: These stay intact in the product, so we can ignore them to save time, or count them on both sides. Let's count only what changes.
* 1 mol F–F bond: $155 \text{ kJ/mol}$
* *Note: The C-H bonds do not break.*
* Total Energy In = $614 + 155 = 769 \text{ kJ}$
* Bonds Formed (Products):
* 1 mol C–C single bond: $348 \text{ kJ/mol}$
* 2 mol C–F bonds: $2 \times 485 \text{ kJ/mol} = 970 \text{ kJ}$
* Total Energy Out = $348 + 970 = 1318 \text{ kJ}$
* Calculation:
* $\Delta H = 769 - 1318 = -549 \text{ kJ}$

3. Methanol ($CH_3OH$) + Methanol ($CH_3OH$) \rightarrow Dimethyl Ether ($CH_3OCH_3$) + Water ($H_2O$)

* Bonds Broken (Reactants):
* We break the O–H bond in the first methanol: $1 \times 467 \text{ kJ}$
* We break the O–H bond in the second methanol: $1 \times 467 \text{ kJ}$
* (The C-O and C-H bonds remain intact).
* Total Energy In = $467 + 467 = 934 \text{ kJ}$
* Bonds Formed (Products):
* We form a new C–O bond in the ether: $1 \times 358 \text{ kJ}$
* We form two O–H bonds in water: $2 \times 467 \text{ kJ} = 934 \text{ kJ}$
* Total Energy Out = $358 + 934 = 1292 \text{ kJ}$
* Calculation:
* $\Delta H = 934 - 1292 = -358 \text{ kJ}$

4. $CH_2Cl_2 + O \rightarrow COCl_2 + H-Cl$
*(Note: The reactant is Dichloromethane. The oxygen atom is written as 'O', implying atomic oxygen with no initial bonds to break).*

* Bonds Broken (Reactants):
* In $CH_2Cl_2$, we break 2 C–H bonds and 2 C–Cl bonds to rearrange into phosgene ($COCl_2$).
* 2 mol C–H bonds: $2 \times 413 = 826 \text{ kJ}$
* 2 mol C–Cl bonds: $2 \times 339 = 678 \text{ kJ}$
* Atomic Oxygen (O) has no bonds to break.
* Total Energy In = $826 + 678 = 1504 \text{ kJ}$
* Bonds Formed (Products):
* In $COCl_2$ (Phosgene), there is 1 C=O bond and 2 C–Cl bonds.
* 1 mol C=O bond: $799 \text{ kJ}$
* 2 mol C–Cl bonds: $2 \times 339 = 678 \text{ kJ}$
* In H–Cl, there is 1 H–Cl bond.
* 1 mol H–Cl bond: $431 \text{ kJ}$
* Total Energy Out = $799 + 678 + 431 = 1908 \text{ kJ}$
* Calculation:
* $\Delta H = 1504 - 1908 = -404 \text{ kJ}$

5. Acetaldehyde ($CH_3CHO$) + $O_2 \rightarrow$ Acetic Acid ($CH_3COOH$)

* Bonds Broken (Reactants):
* In Acetaldehyde, we break the C=O double bond and one C–H bond (on the carbonyl carbon) to insert the oxygen.
* 1 mol C=O: $799 \text{ kJ}$
* 1 mol C–H: $413 \text{ kJ}$
* In $O_2$, we break the O=O double bond.
* 1 mol O=O: $495 \text{ kJ}$
* Total Energy In = $799 + 413 + 495 = 1707 \text{ kJ}$
* Bonds Formed (Products):
* In Acetic Acid, we form a new C–O single bond and an O–H bond (creating the carboxylic acid group), and reform the C=O bond.
* 1 mol C=O: $799 \text{ kJ}$
* 1 mol C–O: $358 \text{ kJ}$
* 1 mol O–H: $467 \text{ kJ}$
* Total Energy Out = $799 + 358 + 467 = 1624 \text{ kJ}$
* Calculation:
* $\Delta H = 1707 - 1624 = +83 \text{ kJ}$

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Part 2: Draw Lewis structures and Estimate $\Delta H_{rxn}$

6. $H_2(g) + CO_2(g) \rightarrow H_2O(g) + CO(g)$

* Bonds Broken:
* 1 mol H–H: $436 \text{ kJ}$
* 2 mol C=O (in $CO_2$): $2 \times 799 = 1598 \text{ kJ}$
* Total In = $436 + 1598 = 2034 \text{ kJ}$
* Bonds Formed:
* 2 mol O–H (in $H_2O$): $2 \times 467 = 934 \text{ kJ}$
* 1 mol C≡O (in $CO$, triple bond): $1072 \text{ kJ}$
* Total Out = $934 + 1072 = 2006 \text{ kJ}$
* Calculation:
* $\Delta H = 2034 - 2006 = +28 \text{ kJ}$

7. $2H_2O(g) \rightarrow 2H_2(g) + O_2(g)$

* Bonds Broken:
* 4 mol O–H bonds (in 2 water molecules): $4 \times 467 = 1868 \text{ kJ}$
* Total In = $1868 \text{ kJ}$
* Bonds Formed:
* 2 mol H–H bonds: $2 \times 436 = 872 \text{ kJ}$
* 1 mol O=O bond: $495 \text{ kJ}$
* Total Out = $872 + 495 = 1367 \text{ kJ}$
* Calculation:
* $\Delta H = 1868 - 1367 = +501 \text{ kJ}$

8. $CO(g) + 2H_2(g) \rightarrow CH_3OH(g)$

* Bonds Broken:
* 1 mol C≡O (in $CO$): $1072 \text{ kJ}$
* 2 mol H–H: $2 \times 436 = 872 \text{ kJ}$
* Total In = $1072 + 872 = 1944 \text{ kJ}$
* Bonds Formed:
* 3 mol C–H: $3 \times 413 = 1239 \text{ kJ}$
* 1 mol C–O: $358 \text{ kJ}$
* 1 mol O–H: $467 \text{ kJ}$
* Total Out = $1239 + 358 + 467 = 2064 \text{ kJ}$
* Calculation:
* $\Delta H = 1944 - 2064 = -120 \text{ kJ}$

9. $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

* Bonds Broken:
* 1 mol N≡N: $941 \text{ kJ}$
* 3 mol H–H: $3 \times 436 = 1308 \text{ kJ}$
* Total In = $941 + 1308 = 2249 \text{ kJ}$
* Bonds Formed:
* 6 mol N–H (2 molecules of $NH_3$, each has 3 bonds): $6 \times 391 = 2346 \text{ kJ}$
* Total Out = $2346 \text{ kJ}$
* Calculation:
* $\Delta H = 2249 - 2346 = -97 \text{ kJ}$

10. $H_2(g) + C_2H_4(g) \rightarrow C_2H_6(g)$

* Bonds Broken:
* 1 mol H–H: $436 \text{ kJ}$
* 1 mol C=C: $614 \text{ kJ}$
* (4 C-H bonds remain unchanged)
* Total In = $436 + 614 = 1050 \text{ kJ}$
* Bonds Formed:
* 1 mol C–C: $348 \text{ kJ}$
* 2 mol C–H (the new ones added): $2 \times 413 = 826 \text{ kJ}$
* Total Out = $348 + 826 = 1174 \text{ kJ}$
* Calculation:
* $\Delta H = 1050 - 1174 = -124 \text{ kJ}$

11. $C_3H_8(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)$
*(Note: The equation in the image appears to have a typo in the balancing or formula, usually Propane $C_3H_8$ reacts with $5O_2$. However, I will solve strictly based on the bonds present in the written equation: Reactants $C_3H_8 + 7O_2$, Products $4CO_2 + 6H_2O$. This implies the reactant might actually be something else or the coefficients are wrong, but let's look at the atoms. Left: 3 C, 8 H, 14 O. Right: 4 C, 12 H, 14 O. The equation is unbalanced. Assuming standard combustion of Propane $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$ is the intended question, but the text says $C_3H_8 + 7O_2 \rightarrow 4CO_2 + 6H_2O$. Let's assume the text meant Butane ($C_4H_{10}$) or similar? No, it clearly writes $C_3H_8$. Let's calculate based on the bonds listed in the products vs reactants as written, assuming the student should just sum the bonds shown.)*

Let's assume the question meant Propane Combustion: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$.
* Bonds Broken:
* Propane ($C_3H_8$): 2 C-C bonds, 8 C-H bonds.
* $2(348) + 8(413) = 696 + 3304 = 4000 \text{ kJ}$
* Oxygen ($5 O_2$): 5 O=O bonds.
* $5(495) = 2475 \text{ kJ}$
* Total In = $6475 \text{ kJ}$
* Bonds Formed:
* $CO_2$ ($3 \times 2$ C=O bonds): 6 C=O bonds.
* $6(799) = 4794 \text{ kJ}$
* $H_2O$ ($4 \times 2$ O-H bonds): 8 O-H bonds.
* $8(467) = 3736 \text{ kJ}$
* Total Out = $8530 \text{ kJ}$
* Calculation:
* $\Delta H = 6475 - 8530 = -2055 \text{ kJ}$

*(If we strictly follow the typo in #11: $C_3H_8 + 7O_2 \rightarrow 4CO_2 + 6H_2O$)*
* Broken: $C_3H_8$ (4000 kJ) + $7 O_2$ ($7 \times 495 = 3465$) = $7465 \text{ kJ}$
* Formed: $4 CO_2$ ($8 \times 799 = 6392$) + $6 H_2O$ ($12 \times 467 = 5604$) = $11996 \text{ kJ}$
* Result: $7465 - 11996 = -4531 \text{ kJ}$
*(Given the likely typo, the standard propane answer is -2055 kJ, but based strictly on the printed numbers, it is -4531 kJ. I will provide the calculation for the printed numbers as that is what is visually there).*

12. $CH_4(g) + 3Cl_2(g) \rightarrow CHCl_3(g) + 3HCl(g)$

* Bonds Broken:
* Methane ($CH_4$): Break 3 C-H bonds (one remains). $3 \times 413 = 1239 \text{ kJ}$
* Chlorine ($3 Cl_2$): Break 3 Cl-Cl bonds. $3 \times 242 = 726 \text{ kJ}$
* Total In = $1239 + 726 = 1965 \text{ kJ}$
* Bonds Formed:
* Chloroform ($CHCl_3$): Form 3 C-Cl bonds. $3 \times 339 = 1017 \text{ kJ}$
* HCl ($3 HCl$): Form 3 H-Cl bonds. $3 \times 431 = 1293 \text{ kJ}$
* Total Out = $1017 + 1293 = 2310 \text{ kJ}$
* Calculation:
* $\Delta H = 1965 - 2310 = -345 \text{ kJ}$

13. $HCN(g) + 2H_2(g) \rightarrow CH_3NH_2(g)$

* Bonds Broken:
* HCN: Break 1 C-H and 1 C≡N.
* C-H: $413 \text{ kJ}$
* C≡N: $891 \text{ kJ}$
* Hydrogen ($2 H_2$): Break 2 H-H.
* $2 \times 436 = 872 \text{ kJ}$
* Total In = $413 + 891 + 872 = 2176 \text{ kJ}$
* Bonds Formed:
* Methylamine ($CH_3NH_2$):
* 3 C-H bonds: $3 \times 413 = 1239 \text{ kJ}$
* 1 C-N bond: $305 \text{ kJ}$ (Using C-N value from table, note: table lists C-N as 305? No, table lists C-N as 305 is not there, it lists C-N 305? Wait, looking at table: C-N is 305 kJ/mol? Let me re-read the crop. Ah, the table says C-N 305 kJ/mol is NOT listed. It lists C-N 305? No, it lists C-N 305 is missing. It lists C-N 305? Let's look closer. Row 4, Col 2: C-N 305 kJ/mol. Okay, I see it now. Wait, row 4 col 2 is C-N. The value is 305? No, looking at the image:
* C-N is listed as 305 kJ/mol? Actually, standard value is ~305. The table shows C-N 305 kJ/mol.
* Let's check the table again carefully.
* Row 1: H-H, C-H, C=C
* Row 2: H-Cl, C-C, C≡C
* Row 3: H-F, C-N, C=O
* Wait, the table in the image has:
* C-N: 305 kJ/mol (It looks like 305 or 358? Let's check C-O. C-O is 358. C-N is above it. It looks like 305).
* Actually, let's look at the third column.
* C-N is listed as 305 kJ/mol.
* N-H is listed as 391 kJ/mol.
* So, forming $CH_3NH_2$:
* 3 C-H bonds: $3 \times 413 = 1239 \text{ kJ}$
* 1 C-N bond: $305 \text{ kJ}$
* 2 N-H bonds: $2 \times 391 = 782 \text{ kJ}$
* Total Out = $1239 + 305 + 782 = 2326 \text{ kJ}$
* Calculation:
* $\Delta H = 2176 - 2326 = -150 \text{ kJ}$

Final Answer:
1. -184 kJ
2. -549 kJ
3. -358 kJ
4. -404 kJ
5. +83 kJ
6. +28 kJ
7. +501 kJ
8. -120 kJ
9. -97 kJ
10. -124 kJ
11. -4531 kJ (Based strictly on the printed unbalanced equation) OR -2055 kJ (If corrected for Propane combustion)
12. -345 kJ
13. -150 kJ