Question

1. for the reaction \\(\ce{n_{2(g)} + 3h_{2(g)} -> 2nh_{3(g)}}\\), the standard enthalpy change is \\(\dots\\) following statements regarding their bond energies would be true?

a. the bond energy of h-h must be greater than that of n-h
b. the sum of bond energies in the products is greater than the bond energies in the reactants
c. the n-h triple bond is weaker than the n-h single bond
d. the enthalpy of the reactants is lower than the enthalpy of the products
e. the reaction is endothermic because bonding always requires energy
which is incorrect?

Answer:

To solve this, we analyze each statement using bond energy and enthalpy concepts:

Step 1: Recall reaction enthalpy formula

Reaction enthalpy ($\Delta H$) = Sum of bond energies of reactants - Sum of bond energies of products. For $\ce{N2 + 3H2 -> 2NH3}$, reactants have $\ce{N#N}$ (triple bond) and $\ce{H-H}$ bonds; products have $\ce{N-H}$ bonds.

Analyze each option:

• Option a: To form $\ce{N-H}$ bonds (exothermic step), if $\ce{H-H}$ bond energy > $\ce{N-H}$ bond energy, breaking $\ce{H-H}$ (endothermic) would require more energy than forming $\ce{N-H}$ (exothermic) releases. But the reaction is exothermic (standard enthalpy change is negative, as $\ce{NH3}$ formation is exothermic). So this is false.

• Option b: $\Delta H$ = (Bond energy of $\ce{N#N}$ + 3×Bond energy of $\ce{H-H}$) - (6×Bond energy of $\ce{N-H}$). For the reaction to be exothermic ($\Delta H < 0$), (Bond energy of $\ce{N#N}$ + 3×Bond energy of $\ce{H-H}$) < (6×Bond energy of $\ce{N-H}$). Thus, sum of reactant bond energies < sum of product bond energies. So this is false.

• Option c: $\ce{N#N}$ (triple bond) is stronger than $\ce{N-H}$ (single bond) because triple bonds have higher bond energy. So this is false.

• Option d: Enthalpy of reactants = Enthalpy of products + $\Delta H$. Since the reaction is exothermic ($\Delta H < 0$), Enthalpy of reactants = Enthalpy of products + (negative value) → Enthalpy of reactants < Enthalpy of products? No—wait, exothermic means reactants have higher enthalpy than products (energy released). Wait, no: $\Delta H = H_{products} - H_{reactants}$. For exothermic, $H_{products} - H_{reactants} < 0$ → $H_{reactants} > H_{products}$. Wait, the option says "enthalpy of the reactants is lower than the enthalpy of the products"—but if $H_{reactants} > H_{products}$, this is false? Wait, no, maybe I messed up. Wait, let's re-express: $\Delta H = \sum E_{reactant bonds broken} - \sum E_{product bonds formed}$. For exothermic, $\sum E_{reactant bonds broken} < \sum E_{product bonds formed}$, so enthalpy of reactants (which includes bond energies) is lower? Wait, no—bond energy is the energy required to break bonds (endothermic), so higher bond energy means more energy is needed to break, so higher enthalpy for reactants? Wait, maybe the key is: in exothermic reactions, reactants have higher enthalpy than products (since energy is released). But the option says "enthalpy of the reactants is lower than the enthalpy of the products"—which would mean $H_{reactants} < H_{products}$, so $\Delta H = H_{products} - H_{reactants} > 0$ (endothermic), but the reaction is exothermic. Wait, no—wait, the standard enthalpy change for $\ce{NH3}$ formation is negative (exothermic). Wait, maybe the original problem has a typo, but assuming the reaction is exothermic (as $\ce{NH3}$ formation is exothermic), then $H_{reactants} > H_{products}$, so this option is false? Wait, no—wait, maybe I misread. Wait, the standard enthalpy change for $\ce{N2 + 3H2 -> 2NH3}$ is negative (exothermic), so $\Delta H = H_{products} - H_{reactants} < 0$ → $H_{products} < H_{reactants}$. So "enthalpy of the reactants is lower than the enthalpy of the products" would mean $H_{reactants} < H_{products}$, which is false. Wait, maybe the option is correct? Wait, no—wait, maybe the reaction is endothermic? No, $\ce{NH3}$ formation is exothermic. Wait, maybe the original problem’s standard enthalpy change is positive? Wait, the user’s text says "the standard enthalpy change is an…" (cut off), but typically $\ce{N2 + 3H2 -> 2NH3}$ has $\Delta H = -92\ \text{kJ/mol}$ (exothermic). But if we assume the reaction is endothermic (maybe a typo), but no—bond breaking always requires energy (endothermic), bond forming releases energy (exothermic). So "bonding always requires energy" is false, but "bond breaking always requires energy" is true. Wait, the option says "bonding always requires energy"—bonding releases energy, so this is false. Wait, maybe I made a mistake. Let's re-express:

Wait, the correct statement is Option d? Wait, no—wait, enthalpy of reactants: for exothermic reaction, $H_{reactants} > H_{products}$, so "enthalpy of the reactants is lower than the enthalpy of the products" is false. Wait, maybe the original problem’s standard enthalpy change is positive (endothermic), but that contradicts $\ce{NH3}$ formation. Alternatively, maybe the question has a typo, but among the options, let's re-express:

Wait, the key is: Bond breaking requires energy (endothermic), bond forming releases energy (exothermic). So "bonding (forming) always requires energy" is false. But let's re-analyze:

Wait, the correct answer is Option d? No, wait—let's check each option again:

• Option a: $\ce{H-H}$ bond energy > $\ce{N-H}$ bond energy? If true, breaking $\ce{H-H}$ (endothermic) needs more energy than forming $\ce{N-H}$ (exothermic) releases. But the reaction is exothermic, so total energy released (forming $\ce{N-H}$) > total energy absorbed (breaking $\ce{N#N}$ and $\ce{H-H}$). So $\ce{H-H}$ bond energy must be less than $\ce{N-H}$ bond energy? No—wait, bond energy is the energy required to break the bond. So higher bond energy = stronger bond. So if $\ce{H-H}$ bond energy > $\ce{N-H}$ bond energy, breaking $\ce{H-H}$ is harder (needs more energy) than breaking $\ce{N-H}$. But in the reaction, we break $\ce{H-H}$ and $\ce{N#N}$, form $\ce{N-H}$. For the reaction to be exothermic, (energy to break reactant bonds) < (energy released to form product bonds). So (Bond energy of $\ce{N#N}$ + 3×Bond energy of $\ce{H-H}$) < (6×Bond energy of $\ce{N-H}$). So if $\ce{H-H}$ bond energy > $\ce{N-H}$ bond energy, 3×Bond energy of $\ce{H-H}$ > 3×Bond energy of $\ce{N-H}$, so (Bond energy of $\ce{N#N}$ + 3×Bond energy of $\ce{H-H}$) > (Bond energy of $\ce{N#N}$ + 3×Bond energy of $\ce{N-H}$). But we need this to be < 6×Bond energy of $\ce{N-H}$. So unless $\ce{N#N}$ bond energy < 3×Bond energy of $\ce{N-H}$, it might not hold. But $\ce{N#N}$ bond energy is ~945 kJ/mol, $\ce{H-H}$ is ~436 kJ/mol, $\ce{N-H}$ is ~391 kJ/mol. So 945 + 3×436 = 945 + 1308 = 2253; 6×391 = 2346. So 2253 < 2346, so $\Delta H = 2253 - 2346 = -93$ kJ/mol (exothermic). So $\ce{H-H}$ (436) > $\ce{N-H}$ (391), so Option a is true? Wait, the question says "the standard enthalpy change is an…" (cut off), but if the reaction is exothermic (as $\ce{NH3}$ formation is), then Option a: $\ce{H-H}$ bond energy > $\ce{N-H}$ bond energy is true (436 > 391). Wait, maybe I was wrong earlier. Let's re-express:

If $\ce{H-H}$ bond energy (436) > $\ce{N-H}$ bond energy (391), then breaking $\ce{H-H}$ (endothermic, +436) requires more energy than forming $\ce{N-H}$ (exothermic, -391) releases per bond. But in the reaction, we break 3 $\ce{H-H}$ bonds (total +3×436) and 1 $\ce{N#N}$ bond (+945), and form 6 $\ce{N-H}$ bonds (-6×391). So total energy: (945 + 1308) - 2346 = 2253 - 2346 = -93 (exothermic). So even though $\ce{H-H}$ bond energy > $\ce{N-H}$ bond energy, the total energy released from forming $\ce{N-H}$ bonds (6×391) is more than the total energy absorbed from breaking $\ce{N#N}$ and $\ce{H-H}$ bonds (945 + 3×436). So Option a is true? Wait, the question’s options:

Wait, let's re-express all options with actual bond energies:

• a: $\ce{H-H}$ bond energy (436) > $\ce{N-H}$ (391) → True.
• b: Sum of reactant bond energies (945 + 3×436 = 2253) > sum of product bond energies (6×391 = 2346)? 2253 < 2346 → False.
• c: $\ce{N#N}$ (triple bond) is weaker than $\ce{N-H}$ (single bond)? $\ce{N#N}$ bond energy (945) > $\ce{N-H}$ (391) → False (triple bond is stronger).
• d: Enthalpy of reactants < enthalpy of products? For exothermic reaction, $H_{reactants} > H_{products}$ (since energy is released, reactants have more enthalpy) → False.
• e: Reaction is endothermic because bonding always requires energy? Bonding releases energy, so this is False (and the reaction is exothermic).

Wait, but the standard enthalpy change for $\ce{NH3}$ formation is negative (exothermic). So the correct statement is Option a? Wait, but earlier I thought it was false, but with actual numbers, $\ce{H-H}$ bond energy (436) is greater than $\ce{N-H}$ (391). So Option a is true.

But let's confirm:

• Option a: $\ce{H-H}$ bond energy (436 kJ/mol) > $\ce{N-H}$ bond energy (391 kJ/mol) → True.
• Option b: Sum of reactant bond energies (2253) > sum of product bond energies (2346)? No, 2253 < 2346 → False.
• Option c: $\ce{N#N}$ (triple) is weaker than $\ce{N-H}$ (single)? No, triple bond is stronger → False.
• Option d: Enthalpy of reactants < enthalpy of products? For exothermic, $H_{reactants} > H_{products}$ → False.
• Option e: Reaction is endothermic (it’s exothermic) and bonding requires energy (bonding releases energy) → False.

So the correct answer is a. The bond energy of H-H must be greater than that of N-H (since 436 > 391).