I Review | Constants Periodic Table Use bond enthalpies in the table below to estimate AH for each of the following reactions. Vo AEO ? Average bond enthalpy (kJ/mol) AH = kJ C-H: 413 N-H: 391 0-H: 463 F-F: 155 Submit Previous Answers Request Answer C-C: 348 N-N: 163 0-O: 146 C=C: 614N-O: 201 0=0: 495 Cl-F: 253 X Incorrect; Try Again; 5 attempts remaining C-N: 293 N-F: 272 O-F: 190 Cl-–Cl: 242 C-O: 358 N-Cl: 200 0-Cl: 203 Br-F: 237 234 Br-Cl: 218 <> C=0: 799 N-Br: 243 0-I: Part B C-F: 485 H-H: 436 Br-Br: 193 C-Cl: 328 H-F: 567 H. C-Br: 276 H-Cl: 431 I-Cl: 208 H-C-H + 2 0=0– 0=C=0 + 2 H-O-H C-I: 240 H-Br: 366 H-I: 299 I-Br: 175 I-I: 151 H. CH4 plus 2 02 goes to CO2 plus 2 H20. Express your answer KIIUJUUTES to tree sigmncamm ngures. ινα ΑΣφ kJ AH = Submit Request Answer Next > Provide Feedback
I Review | Constants Periodic Table Use bond enthalpies in the table below to estimate AH for each of the following reactions. Vo AEO ? Average bond enthalpy (kJ/mol) AH = kJ C-H: 413 N-H: 391 0-H: 463 F-F: 155 Submit Previous Answers Request Answer C-C: 348 N-N: 163 0-O: 146 C=C: 614N-O: 201 0=0: 495 Cl-F: 253 X Incorrect; Try Again; 5 attempts remaining C-N: 293 N-F: 272 O-F: 190 Cl-–Cl: 242 C-O: 358 N-Cl: 200 0-Cl: 203 Br-F: 237 234 Br-Cl: 218 <> C=0: 799 N-Br: 243 0-I: Part B C-F: 485 H-H: 436 Br-Br: 193 C-Cl: 328 H-F: 567 H. C-Br: 276 H-Cl: 431 I-Cl: 208 H-C-H + 2 0=0– 0=C=0 + 2 H-O-H C-I: 240 H-Br: 366 H-I: 299 I-Br: 175 I-I: 151 H. CH4 plus 2 02 goes to CO2 plus 2 H20. Express your answer KIIUJUUTES to tree sigmncamm ngures. ινα ΑΣφ kJ AH = Submit Request Answer Next > Provide Feedback
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Thermochemistry
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Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
Question
![### Determining Enthalpy Change using Bond Enthalpies
#### Bond Enthalpy Table
Use the bond enthalpies in the table below to estimate ΔH for each of the following reactions:
| Bond | Bond Enthalpy (kJ/mol) |
|----------|------------------------|
| C–H | 413 |
| C–C | 348 |
| C=C | 614 |
| C≡C | 839 |
| C–N | 293 |
| C=N | 615 |
| C≡N | 891 |
| C–O | 358 |
| C=O | 799 |
| C–F | 485 |
| C–Cl | 328 |
| C–Br | 276 |
| C–I | 240 |
| H–H | 436 |
| H–F | 567 |
| H–Cl | 431 |
| H–Br | 366 |
| H–I | 299 |
| N–N | 163 |
| N=N | 418 |
| N≡N | 941 |
| N–O | 201 |
| O–O | 146 |
| O=O | 495 |
| O–Cl | 243 |
| O–I | 201 |
| Cl–Cl | 242 |
| Cl–F | 253 |
| Cl–Br | 218 |
| Br–Br | 193 |
| I–I | 151 |
| I–Cl | 208 |
| I–Br | 175 |
#### Part B: Reaction Enthalpy Calculation
For the reaction:
\[
\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
\]
To solve for ΔH, follow these steps:
1. Identify bonds broken and formed.
2. Use the bond enthalpies table to calculate the energy required to break bonds (positive value).
3. Calculate the energy released from forming bonds (negative value).
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Transcribed Image Text:### Determining Enthalpy Change using Bond Enthalpies
#### Bond Enthalpy Table
Use the bond enthalpies in the table below to estimate ΔH for each of the following reactions:
| Bond | Bond Enthalpy (kJ/mol) |
|----------|------------------------|
| C–H | 413 |
| C–C | 348 |
| C=C | 614 |
| C≡C | 839 |
| C–N | 293 |
| C=N | 615 |
| C≡N | 891 |
| C–O | 358 |
| C=O | 799 |
| C–F | 485 |
| C–Cl | 328 |
| C–Br | 276 |
| C–I | 240 |
| H–H | 436 |
| H–F | 567 |
| H–Cl | 431 |
| H–Br | 366 |
| H–I | 299 |
| N–N | 163 |
| N=N | 418 |
| N≡N | 941 |
| N–O | 201 |
| O–O | 146 |
| O=O | 495 |
| O–Cl | 243 |
| O–I | 201 |
| Cl–Cl | 242 |
| Cl–F | 253 |
| Cl–Br | 218 |
| Br–Br | 193 |
| I–I | 151 |
| I–Cl | 208 |
| I–Br | 175 |
#### Part B: Reaction Enthalpy Calculation
For the reaction:
\[
\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
\]
To solve for ΔH, follow these steps:
1. Identify bonds broken and formed.
2. Use the bond enthalpies table to calculate the energy required to break bonds (positive value).
3. Calculate the energy released from forming bonds (negative value).
4.
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