The standard enthalpy of formation of NH3 is -46.0 kJ mol^-1. If the enthalpy of formation of H2 fro

Q: Why did we take the bond enthalpies of $H_2$ and $N_2$ as positive when the question gave negative values?

The question provided the enthalpy of *formation* of $H_2$ and $N_2$ from their atoms (e.g., $2H \rightarrow H_2$, $\Delta H = -436$). Bond enthalpy is defined as the energy required to *break* a bond ($H_2 \rightarrow 2H$), which is the exact reverse process. Therefore, we flip the sign to make it positive $+436 \text{ kJ mol}^{-1}$.

Q: Why do we multiply the bond enthalpy of $N_2$ by $1/2$ and $H_2$ by $3/2$?

The standard enthalpy of formation ($\Delta_f H^\circ$) is defined for the formation of exactly *one mole* of the product ($NH_3$). To balance the equation for 1 mole of $NH_3$, we must use $1/2$ mole of $N_2$ and $3/2$ moles of $H_2$. The bond enthalpies are per mole of bonds, so we must scale them by these stoichiometric coefficients.

Q: Why is the formula $\text{Reactants} - \text{Products}$ instead of $\text{Products} - \text{Reactants}$?

When using standard enthalpies of formation ($\Delta_f H^\circ$), the formula is Products - Reactants. However, when using Bond Enthalpies (BE), we are calculating the energy put *in* to break reactant bonds (endothermic, positive) plus the energy given *off* when product bonds form (exothermic, negative). Hence, $\Delta_r H = \sum \text{BE}(\text{Reactants}) - \sum \text{BE}(\text{Products})$.

Q: What does 'average' bond enthalpy mean?

In a polyatomic molecule like $NH_3$, breaking the first N-H bond requires a different amount of energy than breaking the second or third N-H bond because the electron cloud rearranges after each bond is broken. The 'average bond enthalpy' is simply the total energy required to atomize the entire molecule divided by the number of bonds (in this case, 3).

Q: Can bond enthalpy ever be negative?

No, bond dissociation enthalpy is always a positive value. Breaking a chemical bond always requires an input of energy to overcome the electrostatic attraction between the shared electrons and the nuclei. If you see a negative value, it refers to bond *formation*, not bond breaking.

Question

The standard enthalpy of formation of $	ext{NH}_3$ is $-46.0 	ext{ kJ mol}^{-1}$. If the enthalpy of formation of $	ext{H}_2$ from its atoms is $-436 	ext{ kJ mol}^{-1}$ and that of $	ext{N}_2$ is $-712 	ext{ kJ mol}^{-1}$, the average bond enthalpy of N—H bond in $	ext{NH}_3$ is

Options:

(A) $- 964 	ext{ kJ mol}^{-1}$

(B) $+352 	ext{ kJ mol}^{-1}$

(C) $+1056 	ext{ kJ mol}^{-1}$

(D) $-1102 	ext{ kJ mol}^{-1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Let's embark on a thermodynamic journey to uncover the strength of the bonds holding an ammonia molecule together. We start by writing the standard formation reaction for ammonia. To form exactly one mole of $	ext{NH}_3(g)$, we must assemble it from its constituent elements in their standard states. This requires half a mole of nitrogen gas and three-halves of a mole of hydrogen gas:

$\frac{1}{2} 	ext{N}_2(g) + \frac{3}{2} 	ext{H}_2(g) \longrightarrow 	ext{NH}_3(g)$

We are given that the standard enthalpy of this reaction, $\Delta_r H^\circ$, is $-46.0 	ext{ kJ mol}^{-1}$.

### The Master Equation
How does the overall enthalpy of a reaction relate to the individual bond energies? Think of a chemical reaction as a two-step demolition and construction project. First, we must supply energy to break all the bonds in the reactants, turning them into a cloud of isolated atoms. Then, these atoms recombine to form the products, releasing energy in the process. The net difference between the energy put in and the energy released is our reaction enthalpy:

$\Delta_r H^\circ = \sum 	ext{BE}(	ext{Reactants}) - \sum 	ext{BE}(	ext{Products})$

Let's expand this formula for our specific reaction. On the reactant side, we need to break $\frac{1}{2}$ mole of $	ext{N}\equiv	ext{N}$ triple bonds and $\frac{3}{2}$ moles of $	ext{H—H}$ single bonds. On the product side, a single ammonia molecule contains three $	ext{N—H}$ bonds, so forming one mole of ammonia means forming $3$ moles of $	ext{N—H}$ bonds.

$\Delta_r H^\circ = \left[ \frac{1}{2} 	ext{BE}(	ext{N}\equiv	ext{N}) + \frac{3}{2} 	ext{BE}(	ext{H—H}) ight] - 3 	ext{BE}(	ext{N—H})$

### Navigating the Sign Convention Trap
Here is a crucial catch where many students make a silly mistake. The problem gives us the enthalpy of *formation* of hydrogen and nitrogen from their atoms as negative values ($-436 	ext{ kJ mol}^{-1}$ and $-712 	ext{ kJ mol}^{-1}$). This negative sign simply means that forming these bonds releases energy.

However, bond dissociation enthalpy is defined as the energy required to *break* a bond, which is an endothermic process. Therefore, we must flip the signs! The bond dissociation enthalpies are strictly positive: $+436 	ext{ kJ mol}^{-1}$ for $	ext{H}_2$ and $+712 	ext{ kJ mol}^{-1}$ for $	ext{N}_2$. Let's substitute these values into our master equation:

$-46 = \left[ \frac{1}{2}(712) + \frac{3}{2}(436) ight] - 3 	ext{BE}(	ext{N—H})$

### Final Calculation
Now, we execute the atomic compute. Half of $712$ is $356$, and three-halves of $436$ is $654$.

$-46 = [356 + 654] - 3 	ext{BE}(	ext{N—H})$

Adding the reactant bond energies gives us exactly $1010 	ext{ kJ}$.

$-46 = 1010 - 3 	ext{BE}(	ext{N—H})$

Let's rearrange the terms to isolate our unknown variable. We move the negative $3	ext{BE}$ term to the left side to make it positive, and move the $-46$ to the right side:

$3 	ext{BE}(	ext{N—H}) = 1010 + 46 = 1056$

Finally, we divide by three to find the average bond enthalpy of a single $	ext{N—H}$ bond:

$	ext{BE}(	ext{N—H}) = \frac{1056}{3} = \mathbf{352 	ext{ kJ mol}^{-1}}$

**Why do we call it the 'average' bond enthalpy?** In a polyatomic molecule like ammonia, breaking the first $	ext{N—H}$ bond takes a slightly different amount of energy than breaking the second or the third, because the electronic environment of the molecule changes after each bond is severed. The value we just calculated ($352 	ext{ kJ mol}^{-1}$) is the beautiful, clean average of all three successive bond-breaking energies!

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Visualized Solution

$Formation Reaction of NH_{3}$

$Enthalpy and Bond Energies$

$Expanding the Formula$

$Substituting Values$

$Calculating Reactant Bond Energies$

$Simplifying the Equation$

$Isolating the Unknown$

$Final Calculation$

$Why Average Bond Enthalpy?$

The Sigma Insight: Enthalpy and Hess's Law

Analyzing the Setup Let's embark on a thermodynamic journey to uncover the strength of the bonds holding an ammonia molecule together

The Master Equation How does the overall enthalpy of a reaction relate to the individual bond energies? Think of a chemical reaction as a two-step demolition and construction project

Navigating the Sign Convention Trap Here is a crucial catch where many students make a silly mistake

Final Calculation Now, we execute the atomic compute

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formation Reaction of NH3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Enthalpy and Bond Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Reactant Bond Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating the Unknown

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Why Average Bond Enthalpy?