Standard Enthalpy Of Formation, Combustion And Bond Dissociation

Edited By Shivani Poonia | Updated on Jul 02, 2025 06:00 PM IST

Enthalpy of combustion: It is the heat developed when one mole of a substance is completely burnt in oxygen under standard conditions. This quantity becomes very important while assessing the energy content of fuels. Enthalpy of dissociation, ΔH_d, is the energy required to break one mole of a specific type of bond in a molecule into individual atoms in the gas phase. It provides a measure of the strength and stability of chemical bonds. Enthalpy of atomization, ΔH_a, is the energy required to transform one mole of a given compound into its constituent atoms in the gas phase. It, therefore, represents the energy required to break all bonds in a molecule to yield individual atoms.

This Story also Contains

Heat of Combustion
Enthalpy Of Dissociation Or Ionization
Some Solved Examples
Summary

Standard Enthalpy Of Formation, Combustion And Bond Dissociation

Heat of Combustion

1. It is a change in enthalpy when one mole of a substance is completely oxidized combusted, or burnt.

2. $\Delta \mathrm{H}$ is - ve here as heat is always evolved here that is, exothermic process.

3. Heat of combustion is useful in calculating the calorific value of food and fuels.

4. It is also useful in confirming the structure of organic molecules having C, H, O, N, etc.

5. Enthalpy change by combustion of 1 gm solid,1 gm liquid, or 1 cc gas is called calorific value.

$\begin{aligned} & \text { calorific value }=\frac{\text { Heat of combustion }}{\text { Molecular wt. }} \\ & \Delta \mathrm{H} \text { (heat of reaction) }=-\Sigma \Delta \mathrm{H}_{\mathrm{P}}^{\circ}-\Sigma \mathrm{H}_{\mathrm{R}}^{\circ}\end{aligned}$

transition from liquid to gas.

Enthalpy Of Dissociation Or Ionization

It is defined as, "The quantity of heat absorbed when one mole of a substance is completely dissociated into its ions". Example,

$\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-} \quad \Delta \mathrm{H}=13.7 \mathrm{Kcal}$

Heat Of Atomization

It is the enthalpy change (heat required) when bonds of one mole of a substance are broken down completely to obtain atoms in the gaseous phase (isolated) or it is the enthalpy change when one mole of atoms in the gas phase is formed from the corresponding element in its standard state. In the case of diatomic molecules, it is also called bond dissociation enthalpy.

It is denoted by $\Delta \mathrm{H}_{\mathrm{a}}$ or $\Delta \mathrm{H}^*$.
Example,
$
\begin{aligned}
& \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g})-435 \mathrm{~kJ} \\
& \Delta \mathrm{H}=+435 \mathrm{~kJ} / \mathrm{mol} \\
& \mathrm{CH}_4(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g})+1665 \mathrm{~kJ} \\
& \Delta \mathrm{H}=+1665 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$

Phase Transition And Transition Energy

The change of matter from one state (solid, liquid, or gas) to another state is called Phase Transition.
Such changes occur at definite temperatures such as melting point (solid to liquid), boiling point (liquid to vapors), etc., and are accompanied by absorption or evolution of heat. The enthalpy change during such phase transitions is called heat of transition or transition energy.

Example, C (diamond) $\rightarrow \mathrm{C}$ (amorphous) $\Delta \mathrm{H}=3.3 \mathrm{Kcal}$

Some Solved Examples

Example 1: Calculate the standard enthalpy of combustion (kJ/mol) of glucose

Given, $\Delta H_f^0 \mathrm{CO}_2=-394 \mathrm{~kJ} / \mathrm{mol} \Delta H_f^0 \mathrm{H}_2 \mathrm{O}=-286 \mathrm{~kJ} / \mathrm{mol} \Delta H_f^0 C_6 H_{12} O_6=-1275 \mathrm{~kJ} / \mathrm{mol}$,

1)2805

2)-2805

3)1275

4)-1275

Solution

$\Delta H_f^o O_2$ is equal to Zero because O₂ is in the standard state.

First, write the balanced equation of the reaction,

$\begin{aligned} & \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O} \\ & \Delta H_c=\left(6 * \Delta H_f^0 \mathrm{H}_2 0+6 * \Delta H_f^0 \mathrm{CO}_2\right)-\left(\Delta H_f^0 C_6 H_{12} O_6+6 * \Delta H_f^0 O_2\right) \\ & \Delta H_c=6 \times(-394)+6 \times(-286)-(-1275)-6 \times(0) \\ & \Delta H_c=-2805 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$

Example 2: At 25⁰C 1 atm pressure, the enthalpies of combustion are as given below:

Substance H₂ C (graphite) C₂H₆(g) $\frac{\Delta_c \mathrm{H}^{\ominus}}{\mathrm{kJ} \mathrm{mol}^{-1}}$ -286.0 -394.0 -1560.0

The enthalpy of the formation of ethane is

1) $+54.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(2) $-68.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3) $-86.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4) $+97.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:

Given,
$
\begin{aligned}
& \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} \quad \Delta_{\mathrm{c}} \mathrm{H}=-286 \mathrm{~kJ} / \mathrm{mol}----(1) \\
& \mathrm{C}_{(\mathrm{s})}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2 \quad \Delta_{\mathrm{c}} \mathrm{H}=-394 \mathrm{~kJ} / \mathrm{mol}----(2)
\end{aligned}
$
$
\mathrm{C}_2 \mathrm{H}_6+\frac{7}{2} \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O} \quad \Delta_{\mathrm{C}} \mathrm{H}=-1560 \mathrm{~kJ} / \mathrm{mol}-----
$

The enthalpy of the formation of ethane :
$
2 \mathrm{C}_{(\mathrm{s})}+3 \mathrm{H}_2 \rightarrow \mathrm{C}_2 \mathrm{H}_6 \quad \Delta_{\mathrm{C}} \mathrm{H}=\text { ? }
$

The formation reaction of ethane can be obtained by $2 \times(2)+3 \times(1)-(3)$
Then,
$
\begin{aligned}
\Delta_f \mathrm{H} \text { of } \mathrm{C}_2 \mathrm{H}_6 & =2 \times(-394)+3 \times(-286)-(-1560) \\
& =-1646+1560 \\
& =-86 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$

Example 3: At 25^oC and 1 atm pressure, the enthalpy of combustion of benzene (l) and acetylene (g) are $-3268 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-1300 \mathrm{~kJ} \mathrm{~mol}^{-1}$.respectively. The change in enthalpy for the reaction $3 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})$, is

1) $+324 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2) $+632 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3) $-632 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4) $-732 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution
Given, $\mathrm{C}_6 \mathrm{H}_6(l)+\left(\frac{15}{2}\right) \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}, \Delta_{\mathrm{c}} \mathrm{H}=-3268 \mathrm{~kJ} / \mathrm{mol}$
$
\underset{\text { (Acetylene) }}{\mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})}+\left(\frac{5}{2}\right) \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}, \Delta_{\mathrm{C}} \mathrm{H}=-1300 \mathrm{~kJ} / \mathrm{mol}
$

We need to find the enthalpy change of the following reaction
$
3 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6 \text { (l), } \Delta_{\mathrm{r}} \mathrm{H}=\text { ? }
$

So, by doing $3 \times(2)-(1)$, we get the required reaction.
Thus, the enthalpy change of the reaction is given as
$
\Delta_{\mathrm{r}} \mathrm{H}=3 \times(-1300)-(-3268)
$

$\Delta_{\mathrm{r}} \mathrm{H}=-632 \mathrm{~kJ} / \mathrm{mol}$

Example 4: For the combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, $\mathrm{e}_{\mathrm{C}} \mathrm{H}^{\ominus}=-601.70 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the magnitude of change in internal energy for the reaction is _____________kJ. (Nearest integer)

(Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )

1) 600

2)400

3)300

4)0

Solution

Given,
$
\mathrm{T}=300 \mathrm{~K}
$
$\mathrm{P}=1$ bar
$
\begin{aligned}
& \Delta_{\mathrm{C}} \mathrm{H}^0=-601.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$

Reaction of Combustion-
$
\begin{aligned}
& \mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{MgO}(\mathrm{s}) \\
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{ngRT}
\end{aligned}
$

$
-601.7=\Delta \mathrm{U}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)-\frac{1 \times 8.3 \times 300}{2 \times 1000}
$$

On solving-
$
\Delta \mathrm{U}=-600.455 \mathrm{~kJ} \approx-600 \mathrm{~kJ}
$

Hence, the answer is (600).

Example 5: An average person needs about 10000 kJ of energy per day. The amount of glucose (molar mass = 180.0 g mol^–1) needed to meet this energy requirement is _________ g.

(Use : $\Delta_{\mathrm{C}} \mathrm{H}$ (glucose) $=-2700 \mathrm{~kJ} \mathrm{~mol}^{-1}$ )

1) 667

2)687

3)565

4)45

Solution
Total energy required $=10^4 \mathrm{~kJ}$
Energy obtained from 1 mole of Glucose $=2700 \mathrm{~kJ}$
$\therefore 2700 \mathrm{~kJ}$ is obtained from 180 g Glucose.
$\therefore 10^4 \mathrm{~kJ}$ will be obtained from $\frac{180}{2700} \times 10^4$
$
=667 \mathrm{~g}
$

Hence, the answer is the option (1).

Summary

The heat of combustion is the heat evolved during the complete oxidation of a substance in oxygen, as in a bomb calorimeter, and gives its energy content as fuel. Enthalpy of dissociation provides the energy required to break specific chemical bonds in a molecule and hence allows estimation of the strength of bonds. The enthalpy of atomization expresses the energy required to break all bonds in a compound to form individual atoms. Finally, enthalpy of change refers to the associated energy of the phase change of the substance; either a melt or boil. Knowing these enthalpies gives an estimate of the changes in energy during the progress of a chemical reaction.

Frequently Asked Questions (FAQs)

1. What is bond dissociation energy?

Bond dissociation energy is the amount of energy required to break one mole of a particular type of bond in gaseous molecules under standard conditions. It's a measure of bond strength and stability.

2. How does bond dissociation energy relate to molecular stability?

Higher bond dissociation energy indicates a stronger, more stable bond. Molecules with higher overall bond dissociation energies are generally more stable and require more energy to break apart.

3. How does the standard enthalpy of formation of CO compare to CO2?

The standard enthalpy of formation of CO2 is more negative than that of CO. This is because CO2 has more bonds and is more stable, releasing more energy when formed from its elements compared to CO.

4. Why do most ionic compounds have negative standard enthalpies of formation?

Most ionic compounds have negative standard enthalpies of formation because the strong electrostatic attractions between ions release a large amount of energy when the compound forms from its elements.

5. How does the standard enthalpy of formation relate to bond strengths?

The standard enthalpy of formation is related to bond strengths as it represents the energy required to break bonds in the reactants and form new bonds in the products. Compounds with stronger bonds generally have more negative enthalpies of formation.

6. What is the standard enthalpy of combustion?

The standard enthalpy of combustion is the heat released when one mole of a substance completely burns in excess oxygen under standard conditions (298 K and 1 atm). It's always exothermic, resulting in a negative value.

7. How does the standard enthalpy of combustion relate to fuel efficiency?

The standard enthalpy of combustion is directly related to fuel efficiency. Fuels with more negative enthalpies of combustion release more energy per mole when burned, making them more efficient energy sources.

8. Why are standard enthalpies of combustion always negative?

Standard enthalpies of combustion are always negative because combustion reactions are exothermic, releasing energy in the form of heat and light. The products (usually CO2 and H2O) are more stable than the reactants.

9. How does the molecular structure affect the standard enthalpy of combustion?

Molecular structure affects the standard enthalpy of combustion through factors like bond strengths and molecular size. Larger molecules with more C-H bonds tend to have more negative enthalpies of combustion, releasing more energy when burned.

10. What's the relationship between the standard enthalpy of formation and the standard enthalpy of combustion?

The standard enthalpy of combustion can be calculated using the standard enthalpies of formation of the reactants and products. It's the difference between the sum of the enthalpies of formation of the products and the reactants.

11. What is the standard enthalpy of formation?

The standard enthalpy of formation is the heat change when one mole of a compound is formed from its constituent elements in their standard states at 298 K and 1 atm pressure. It's a measure of the stability of a compound relative to its elements.

12. Why are standard enthalpies of formation important in chemistry?

Standard enthalpies of formation are crucial because they allow us to calculate the enthalpy changes for many reactions without directly measuring them. They serve as reference points for thermochemical calculations and help predict the feasibility of chemical reactions.

13. Can the standard enthalpy of formation of an element in its standard state ever be non-zero?

No, the standard enthalpy of formation of an element in its standard state is always zero. This is because there's no energy change involved in forming an element from itself under standard conditions.

14. What's the difference between endothermic and exothermic formation reactions?

Endothermic formation reactions have positive standard enthalpies of formation, meaning they absorb heat from the surroundings. Exothermic formation reactions have negative standard enthalpies of formation, releasing heat to the surroundings.

15. How do intermolecular forces affect standard enthalpies of formation?

Intermolecular forces contribute to the overall stability of a compound and thus affect its standard enthalpy of formation. Stronger intermolecular forces generally lead to more negative enthalpies of formation as more energy is released when the compound forms.

16. Can you explain why some compounds have positive standard enthalpies of formation?

Compounds with positive standard enthalpies of formation are less stable than their constituent elements under standard conditions. Forming these compounds requires energy input, often because the bonds in the elements are stronger than those in the compound or due to unfavorable electron arrangements.

17. Can you use standard enthalpies of formation to predict the spontaneity of a reaction?

While standard enthalpies of formation contribute to predicting spontaneity, they alone are not sufficient. You need to consider the Gibbs free energy change, which includes both enthalpy and entropy changes. A reaction with products having lower enthalpies of formation than reactants is more likely to be spontaneous, but entropy must also be considered.

18. How does the concept of resonance affect standard enthalpies of formation?

Resonance typically lowers the standard enthalpy of formation of a compound. The delocalization of electrons in resonance structures increases stability, resulting in a more negative enthalpy of formation compared to similar compounds without resonance.

19. Can you explain why the standard enthalpy of formation of an allotrope (like graphite or diamond) is non-zero?

The standard enthalpy of formation of an allotrope is non-zero because it's measured relative to the most stable form of the element under standard conditions. For carbon, graphite is the standard state, so diamond has a positive enthalpy of formation, representing the energy required to convert graphite to diamond.

20. Why might the experimental value of a standard enthalpy of formation differ from the theoretical calculation?

Experimental values might differ from theoretical calculations due to factors like measurement uncertainties, non-ideal experimental conditions, or simplifications in theoretical models. Real-world complexities like solvation effects, crystal structure imperfections, or trace impurities can also contribute to discrepancies.

21. How does the standard enthalpy of formation of a compound relate to its stability?

Generally, a more negative standard enthalpy of formation indicates greater stability of a compound relative to its constituent elements. Compounds with highly negative formation enthalpies are typically more stable and less reactive, as more energy would be required to decompose them back into their elements.

22. How does electronegativity affect standard enthalpies of formation in covalent compounds?

In covalent compounds, a larger difference in electronegativity between bonded atoms often leads to a more negative standard enthalpy of formation. This is because the resulting polar bonds are generally stronger and more stable, releasing more energy when formed from the elements.

23. Why do double bonds usually have higher dissociation energies than single bonds?

Double bonds usually have higher dissociation energies than single bonds because they involve sharing two pairs of electrons instead of one. This creates a stronger attraction between the atoms, requiring more energy to break.

24. How does bond dissociation energy affect reaction rates?

Higher bond dissociation energies generally lead to slower reaction rates. This is because more energy is required to break the bonds in the reactants, which is often the rate-limiting step in chemical reactions.

25. Can bond dissociation energy be negative?

No, bond dissociation energy cannot be negative. It always requires energy to break a chemical bond, so the value is always positive. A negative value would imply that breaking a bond releases energy, which is not possible.

26. How does electronegativity difference affect bond dissociation energy?

Generally, a larger electronegativity difference between bonded atoms leads to a higher bond dissociation energy. This is because the more polar bond has a stronger electrostatic component, making it harder to break.

27. Why is the C-C bond dissociation energy in ethane different from the C-C bond in a larger alkane?

The C-C bond dissociation energy in ethane is slightly different from that in larger alkanes due to different electronic and steric effects. In larger alkanes, neighboring groups can stabilize the resulting radicals, slightly lowering the bond dissociation energy.

28. How do resonance structures affect bond dissociation energies?

Resonance structures can increase bond dissociation energies by distributing electron density over multiple bonds. This electron delocalization makes the bonds more stable and harder to break.

29. What's the relationship between bond length and bond dissociation energy?

Generally, there's an inverse relationship between bond length and bond dissociation energy. Shorter bonds are usually stronger and have higher dissociation energies, while longer bonds are weaker with lower dissociation energies.

30. How does the concept of bond dissociation energy apply to polyatomic molecules?

In polyatomic molecules, each bond has its own dissociation energy. The total energy required to completely dissociate the molecule into its constituent atoms is the sum of all individual bond dissociation energies.

31. Why are average bond enthalpies used in calculations instead of exact bond dissociation energies?

Average bond enthalpies are used because the exact bond dissociation energy can vary slightly depending on the molecular environment. Using averages simplifies calculations and provides a good approximation for many chemical processes.

32. How do standard enthalpies of formation, combustion, and bond dissociation energies relate to each other?

These concepts are interconnected. Standard enthalpies of formation and combustion depend on the bond strengths in molecules, which are quantified by bond dissociation energies. All three provide different perspectives on the energetics of chemical systems.

33. Can you use bond dissociation energies to calculate the enthalpy of a reaction?

Yes, you can estimate the enthalpy of a reaction using bond dissociation energies. The process involves calculating the energy required to break bonds in the reactants and the energy released when forming bonds in the products. The difference gives an approximation of the reaction enthalpy.

34. Why might the actual enthalpy of a reaction differ from the calculation using bond dissociation energies?

The actual enthalpy might differ because bond dissociation energies are average values and don't account for specific molecular environments or intermolecular forces. Additionally, factors like changes in molecular geometry or electron delocalization aren't considered in simple bond energy calculations.

35. How does pressure affect standard enthalpies of formation and combustion?

Standard enthalpies of formation and combustion are defined at 1 atm pressure. Changing the pressure can affect the enthalpy values, especially for reactions involving gases, due to the work done by or on the system as it expands or contracts.

36. Why is the standard state important in enthalpy measurements?

The standard state (usually 298 K and 1 atm) is important because it provides a consistent reference point for comparing enthalpy values across different substances and reactions. This standardization allows for meaningful comparisons and calculations in thermochemistry.

37. How does aromaticity affect bond dissociation energies?

Aromaticity generally increases bond dissociation energies within the aromatic system. The delocalized electrons in aromatic compounds create a stable resonance structure, making it more difficult to break individual bonds compared to similar non-aromatic compounds.

38. Why is the standard enthalpy of combustion of carbon monoxide (CO) different from that of methane (CH4)?

The difference in standard enthalpies of combustion between CO and CH4 is due to their different molecular structures and bond types. CH4 has more C-H bonds to break and forms more H2O upon combustion, resulting in a more exothermic reaction compared to CO, which only forms CO2.

39. How does hydrogen bonding affect the standard enthalpy of formation of compounds like water or ammonia?

Hydrogen bonding makes the standard enthalpies of formation of compounds like water or ammonia more negative (exothermic). The additional stability provided by hydrogen bonds results in more energy being released when these compounds form from their elements.

40. Why are standard enthalpies of combustion useful in comparing different fuels?

Standard enthalpies of combustion are useful for comparing fuels because they indicate the amount of heat energy released per mole of fuel when completely burned. Fuels with more negative enthalpies of combustion release more energy and are generally more efficient.

41. How does the presence of a catalyst affect standard enthalpies of formation or combustion?

A catalyst does not affect the standard enthalpies of formation or combustion. Catalysts change the reaction pathway and lower the activation energy, but they do not alter the overall energy change (ΔH) of the reaction, which is determined by the initial and final states.

42. How do standard enthalpies of formation relate to lattice energies in ionic compounds?

The standard enthalpy of formation of an ionic compound is closely related to its lattice energy. A more negative lattice energy (indicating stronger ionic bonds) generally corresponds to a more negative enthalpy of formation, as more energy is released when the ions come together to form the compound.

43. How does electron affinity relate to standard enthalpies of formation for ionic compounds?

Electron affinity affects the standard enthalpy of formation of ionic compounds because it represents the energy change when an atom gains an electron to form an anion. A more negative electron affinity of the non-metal contributes to a more negative enthalpy of formation for the ionic compound.

44. Can you use bond dissociation energies to explain why some reactions are endothermic while others are exothermic?

Yes, bond dissociation energies can explain the endo- or exothermic nature of reactions. If more energy is required to break bonds in the reactants than is released by forming bonds in the products, the reaction is endothermic. Conversely, if bond formation releases more energy than bond breaking absorbs, the reaction is exothermic.

45. Why are standard enthalpies of combustion always measured with excess oxygen?

Standard enthalpies of combustion are measured with excess oxygen to ensure complete combustion of the substance. This standardization allows for consistent and comparable results across different compounds, as it guarantees that all of the substance reacts to form the most oxidized products.

46. How do standard enthalpies of formation and combustion change with temperature?

Standard enthalpies of formation and combustion can change with temperature due to the heat capacity of the substances involved. As temperature increases, the enthalpy values generally become more positive (or less negative) because some of the energy goes into increasing the thermal energy of the molecules.

47. Can you explain why the standard enthalpy of formation of a diatomic molecule (like H2 or N2) is zero, but its bond dissociation energy is not?

The standard enthalpy of formation of a diatomic molecule in its standard state is zero by definition, as it's already in its elemental form. However, its bond dissociation energy is non-zero because energy is required to break the bond between the atoms. The bond dissociation energy represents the strength of this bond.

48. Why might the standard enthalpy of combustion per mole of carbon atoms decrease as the length of a hydrocarbon chain increases?

The standard enthalpy of combustion per mole of carbon atoms often decreases (becomes less negative) as hydrocarbon chain length increases because the proportion of C-H bonds to C-C bonds increases. C-H bonds release more energy when broken and reformed into H2O than C-C bonds do when forming CO2.

49. How do standard enthalpies of formation relate to the energy changes in chemical reactions?

Standard enthalpies of formation can be used to calculate the enthalpy change of a reaction using Hess's Law. The enthalpy change is the difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants, multiplied by their respective stoichiometric coefficients.

50. Can you explain why some bond dissociation energies (like O-H in water) can vary depending on whether it's the first or second bond being broken?

The bond dissociation energy can vary for subsequent bonds in the same molecule because breaking the first bond changes the electronic environment of the remaining atoms. For example, in water, breaking the first O-H bond requires more energy than breaking the second because the resulting OH radical has a different electronic structure than the original H2O molecule.