Q.The enthalpies of all elements in their standard states are:

Concept & Intuition

We are given three reactions with known enthalpy changes and asked to find $\Delta H$ for:

Notice that this reaction is not a standard combustion — it produces solid carbon and liquid water, not CO₂. The trick is to use Hess’s law: enthalpy change depends only on initial and final states, not the path. So we can algebraically manipulate the given equations (reverse them, multiply them by coefficients) so that when added, they yield the target equation. The sum of the corresponding $\Delta H$ values (with sign changes and scaling) gives the answer.

Step-by-step solution

1. Write the target equation clearly Target:

2. List the given equations with their $\Delta H^\Theta$

   1. $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)$; $\Delta H = -286$ kJ
   2. $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$; $\Delta H = -394$ kJ
   3. $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$; $\Delta H = -890$ kJ

3. Plan the combination

   The target has $\text{CH}_4$ on the left and $\text{C}(s)$ on the right. Equation (iii) has $\text{CH}_4$ on the left but produces $\text{CO}_2$, not $\text{C}$. Equation (ii) shows how to get from $\text{C}$ to $\text{CO}_2$ — if we reverse it, we go from $\text{CO}_2$ to $\text{C}$. That suggests:

   • Use (iii) as is (to get $\text{CH}_4$ on left).
   • Reverse (ii) (to turn $\text{CO}_2$ into $\text{C}$).
   • Then cancel any extra species.

4. Reverse equation (ii)

   Reversed: $\text{CO}_2(g) \rightarrow \text{C}(s) + \text{O}_2(g)$; $\Delta H = +394$ kJ (sign flips).

5. Add the reversed (ii) to equation (iii)

   (iii): $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$; $\Delta H = -890$

   • reversed (ii): $\text{CO}_2(g) \rightarrow \text{C}(s) + \text{O}_2(g)$; $\Delta H = +394$ Sum: $\text{CH}_4(g) + 2\text{O}_2(g) + \text{CO}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) + \text{C}(s) + \text{O}_2(g)$ Cancel $\text{CO}_2(g)$ on both sides. Also cancel one $\text{O}_2(g)$ from left (2 O₂) with the one on right, leaving one O₂ on left: Result: $\text{CH}_4(g) + \text{O}_2(g) \rightarrow \text{C}(s) + 2\text{H}_2\text{O}(l)$ That’s exactly the target! Enthalpy sum: $-890 + 394 = -496$ kJ.

6. Check the sign

   The sum gives $\Delta H = -496$ kJ. This is a good check: the target is not the complete combustion of methane (which gives CO₂). Here we produce solid carbon, which is less stable than CO₂, so the reaction should be less exothermic (or even endothermic) compared to full combustion. Full combustion of CH₄ gives $-890$ kJ. Our reaction gives only water, not CO₂, so it should be less negative than $-890$ kJ — and $-496$ kJ is. As an independent check, use formation enthalpies:

   • Formation of CH₄: not given directly, but we can find it. From (iii): $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$; $\Delta H = -890$ Using (ii) and (i): formation of CO₂ = $-394$, formation of H₂O(l) = $-286$ each. So $\Delta H_{\text{comb}} = [\Delta H_f(\text{CO}_2) + 2\Delta H_f(\text{H}_2\text{O})] - [\Delta H_f(\text{CH}_4) + 2\Delta H_f(\text{O}_2)]$ $-890 = [-394 + 2(-286)] - [\Delta H_f(\text{CH}_4) + 0]$ $-890 = [-394 -572] - \Delta H_f(\text{CH}_4)$ $-890 = -966 - \Delta H_f(\text{CH}_4)$ So $\Delta H_f(\text{CH}_4) = -966 + 890 = -76$ kJ/mol. Now target: $\text{CH}_4(g) + \text{O}_2(g) \rightarrow \text{C}(s) + 2\text{H}_2\text{O}(l)$ $\Delta H = [\Delta H_f(\text{C}) + 2\Delta H_f(\text{H}_2\text{O})] - [\Delta H_f(\text{CH}_4) + \Delta H_f(\text{O}_2)]$ $\Delta H_f(\text{C}) = 0$, $\Delta H_f(\text{H}_2\text{O}) = -286$, $\Delta H_f(\text{CH}_4) = -76$, $\Delta H_f(\text{O}_2)=0$ So $\Delta H = [0 + 2(-286)] - [-76 + 0] = -572 + 76 = -496$ kJ. Yes, $-496$ kJ is correct.

The concept here is bond enthalpy — the energy required to break one mole of a specific bond in the gas phase. When a molecule like hydrogen peroxide ($\mathrm{H_2O_2}$) is completely atomised into its constituent atoms, every bond in the molecule is broken. The total enthalpy change for that atomisation is simply the sum of the bond enthalpies of all bonds present.

In $\mathrm{H_2O_2(g)}$, the structure is H–O–O–H. That means there are:

• two O–H bonds
• one O–O bond

So the atomisation enthalpy $\Delta_a H^\Theta$ must equal $2 \times \Delta_{\mathrm{O-H}} H^\Theta + \Delta_{\mathrm{O-O}} H^\Theta$.

We are given $\Delta_a H^\Theta = 1066 \ \text{kJ mol}^{-1}$ and $\Delta_{\mathrm{O-H}} H^\Theta = 464 \ \text{kJ mol}^{-1}$. The only unknown is the O–O bond enthalpy.

Let’s work it through.

1. Write the bond-breaking equation for atomisation:

The enthalpy change for this is the sum of all bond enthalpies broken.

2. Express that sum:

3. Substitute the known values:

4. Solve for the O–O bond enthalpy:

In the lanthanide series the melting point depends on the number of electrons an atom contributes to the metallic (delocalised) bonding. Metals that release more bonding electrons pack more strongly and melt higher; Eu and Yb behave as divalent metals (stable half- and fully-filled 4f giving only 2 bonding electrons) and show anomalously low melting points.

Approximate melting points of the given metals:

• Ce ~ 795 C
• Sm ~ 1074 C
• Yb ~ 824 C (divalent, low)
• Dy ~ 1412 C

Dysprosium clearly has the highest melting point of the set.

ANSWER: D

The atomic radius of an element is a measure of the size of its atoms, typically defined as half the distance between the nuclei of two identical atoms bonded together. Understanding how atomic radius changes across the periodic table is crucial for solving this problem.

Concept: Periodic Trends in Atomic Radius

1. Down a Group (Vertical Column): As we move down a group, the atomic radius generally increases. This is because each successive element adds a new electron shell, which is further away from the nucleus. Even though the nuclear charge increases, the shielding effect of the inner electrons effectively reduces the attraction between the nucleus and the outermost electrons, causing the atomic size to expand.

2. Across a Period (Horizontal Row): As we move from left to right across a period, the atomic radius generally decreases. This is because electrons are added to the same principal energy shell, but the nuclear charge (number of protons) increases. The increased nuclear charge pulls the electrons closer to the nucleus, leading to a stronger attraction and a smaller atomic size. The shielding effect from electrons in the same shell is not very effective.

Let's apply these concepts to the given elements: Na, Li, H, and F.

1. Identify the positions of the elements in the periodic table:

   • H (Hydrogen): Group 1, Period 1. It is the smallest element.
   • Li (Lithium): Group 1, Period 2.
   • Na (Sodium): Group 1, Period 3.
   • F (Fluorine): Group 17, Period 2.

2. Compare elements within the same group (Li and Na):

   Both Li and Na are in Group 1. Na is below Li.

   Following the trend that atomic radius increases down a group, we can conclude:

   $\text{Atomic radius of Na} > \text{Atomic radius of Li}$

3. Compare elements within the same period (Li and F):

   Both Li and F are in Period 2. Li is on the far left (Group 1), and F is on the far right (Group 17).

   Following the trend that atomic radius decreases across a period, we can conclude:

   $\text{Atomic radius of Li} > \text{Atomic radius of F}$

4. Consider Hydrogen:

   Hydrogen is the smallest element. Its atomic radius is significantly smaller than that of Li and F.

5. Order the elements by their atomic radii:

   Combining the above comparisons, we get the following order from largest to smallest atomic radius:

   $\text{Na} > \text{Li} > \text{F} > \text{H}$

6. Match the ordered elements with the given atomic radii:

   The given radii are 152 pm, 99 pm, 64 pm, and 186 pm.

   • The largest radius (186 pm) must correspond to Na.
   • The second largest radius (152 pm) must correspond to Li.
   • The third largest radius (99 pm) must correspond to F.
   • The smallest radius (64 pm) must correspond to H.

   Therefore, the matching is:

   • A: Na $\rightarrow$ IV (186 pm)
   • B: Li $\rightarrow$ I (152 pm)
   • C: H $\rightarrow$ III (64 pm)
   • D: F $\rightarrow$ II (99 pm)

This corresponds to option (A).

The concept first

Electron gain enthalpy $\Delta_{eg}H$ is the enthalpy change when a gaseous atom accepts an electron:

It is usually negative (energy released), and the question asks for the magnitude, i.e. how strongly the atom wants that electron.

Two forces are in tension:

• Nuclear pull. A high effective nuclear charge and a small size hold the incoming electron tightly → more energy released.
• Electron–electron repulsion. The incoming electron must join an already-occupied shell. If that shell is small and compact, the crowding costs energy → less energy released.

For most of the periodic table the first effect wins, and you get the familiar rule "$\Delta_{eg}H$ becomes more negative across a period, less negative down a group." But at the top of groups 16 and 17 the second effect takes over. Fluorine's and oxygen's valence shells are $n=2$, unusually compact, so the new electron feels fierce repulsion from the electrons already there. The result is the famous anomaly:

Chlorine, not fluorine, has the most negative electron gain enthalpy of any element.

Step-by-step

1. Identify the four elements from their configurations.

• I: $[\mathrm{He}]2s^2 2p^5$ → 9 electrons → Fluorine (F), group 17, period 2.
• II: $[\mathrm{Ne}]3s^2 3p^5$ → 17 electrons → Chlorine (Cl), group 17, period 3.
• III: $[\mathrm{He}]2s^2 2p^4$ → 8 electrons → Oxygen (O), group 16, period 2.
• IV: $[\mathrm{Ne}]3s^2 3p^4$ → 16 electrons → Sulphur (S), group 16, period 3.

2. Group 17 before group 16. A halogen ($ns^2np^5$) is one electron short of a noble-gas octet, so gaining an electron is enormously favourable. A chalcogen ($ns^2np^4$) is two short, so it gains far less. Therefore

3. Within group 17: Cl beats F. Fluorine is tiny ($2p$ shell), so the seven electrons already there repel the incoming eighth strongly. Chlorine's $3p$ shell is roomier, and the added electron is accommodated far more comfortably.

So $\mathbf{II > I}$.

4. Within group 16: S beats O. Exactly the same argument, and the gap is much wider because oxygen is even more cramped relative to sulphur.

So $\mathbf{IV > III}$.

5. Combine.

6. Why the tempting options fail. Option (B) puts II > I > III > IV, i.e. O > S — that forgets the oxygen anomaly. Option (D) starts with F, forgetting the chlorine anomaly. Option (A) mixes the groups entirely.

ANSWER: C

The key concept is Hess’s law and the definition of enthalpy of formation. The enthalpy of formation ($\Delta H_f^\circ$) of a compound is the heat change when one mole of it is formed from its elements in their standard states. For a reaction, the standard enthalpy change $\Delta H^\circ$ is:

We don’t need to know the formation enthalpies of elements (like O$_2$) because they are zero by definition. So we just plug in the given numbers.

1. Identify the reaction:

2. Write the formation enthalpies given:

   • $\Delta H_f^\circ(\text{N}_2\text{O}) = +82.0\ \text{kJ mol}^{-1}$
   • $\Delta H_f^\circ(\text{NO}) = +90.0\ \text{kJ mol}^{-1}$
   • $\Delta H_f^\circ(\text{O}_2) = 0$ (element in standard state)

3. Apply the formula:

4. Substitute the numbers:

The concept here is the definition of standard enthalpy of formation. $\Delta_f H^\circ$ is the enthalpy change when one mole of a compound is formed from its elements in their standard states. The key phrase is "standard states" — for any element, the standard state is its most stable physical form at 1 bar pressure and 298 K. By convention, the $\Delta_f H^\circ$ of an element in its standard state is exactly zero. This is a reference point, like setting sea level as zero altitude.

So the question reduces to: which form of carbon is the most stable under standard conditions? Let's examine each option.

1. Graphite is the thermodynamically most stable allotrope of carbon at 298 K and 1 bar. Its layered structure has the lowest Gibbs free energy among all carbon allotropes. Therefore, by definition, its $\Delta_f H^\circ$ is zero.

2. Diamond is a metastable allotrope of carbon. It has a higher enthalpy than graphite (about 1.9 kJ/mol higher at 298 K). Since it is not the most stable form, its $\Delta_f H^\circ$ is positive, not zero.

3. Fullerene (like C$_{60}$) is another allotrope with a higher enthalpy than graphite. Its formation from graphite is endothermic, so $\Delta_f H^\circ$ is positive.

4. Bituminous coal is not a pure element — it is a complex mixture of carbon, hydrogen, oxygen, nitrogen, and sulfur compounds. It is not an elemental form of carbon at all, so the concept of $\Delta_f H^\circ$ being zero does not apply.

The core idea is that enthalpy is a state function — it doesn’t matter how you get from reactants to products, only where you start and end. So if we can add, subtract, or reverse the given reactions to exactly match the target equation, we can simply add and subtract their $\Delta H$ values in the same way. This is Hess’s law, and it’s the only reliable tool for this kind of problem.

Let’s label the three given reactions for clarity:

1. $\mathrm{H}_2(g) + \frac12 \mathrm{O}_2(g) \rightarrow \mathrm{H}_2\mathrm{O}(l)$; $\Delta H_1 = -285\ \mathrm{kJ}$
2. $\mathrm{N}_2\mathrm{O}_5(g) + \mathrm{H}_2\mathrm{O}(l) \rightarrow 2\mathrm{HNO}_3(l)$; $\Delta H_2 = -76.6\ \mathrm{kJ}$
3. $\mathrm{N}_2(g) + 3\mathrm{O}_2(g) + \mathrm{H}_2(g) \rightarrow 2\mathrm{HNO}_3(l)$; $\Delta H_3 = -348.2\ \mathrm{kJ}$

We want: $2\mathrm{N}_2(g) + 5\mathrm{O}_2(g) \rightarrow 2\mathrm{N}_2\mathrm{O}_5(g)$; $\Delta H = ?$

1. Start with the target’s reactants. We need $2\mathrm{N}_2$ on the left. Reaction 3 has one $\mathrm{N}_2$, so multiply it by 2:

   $2\mathrm{N}_2(g) + 6\mathrm{O}_2(g) + 2\mathrm{H}_2(g) \rightarrow 4\mathrm{HNO}_3(l)$; $\Delta H = 2 \times (-348.2) = -696.4\ \mathrm{kJ}$

   But we have too much $\mathrm{O}_2$ and unwanted $\mathrm{H}_2$ and $\mathrm{HNO}_3$ — we’ll cancel them later.

2. We need $2\mathrm{N}_2\mathrm{O}_5$ on the right. Reaction 2 has one $\mathrm{N}_2\mathrm{O}_5$ on the left, so reverse it and multiply by 2:

   $4\mathrm{HNO}_3(l) \rightarrow 2\mathrm{N}_2\mathrm{O}_5(g) + 2\mathrm{H}_2\mathrm{O}(l)$; $\Delta H = -2 \times (-76.6) = +153.2\ \mathrm{kJ}$

   (Reversing flips the sign; multiplying by 2 doubles the magnitude.)

3. Now cancel the $\mathrm{H}_2\mathrm{O}$ that appears. Reaction 1 produces one $\mathrm{H}_2\mathrm{O}$ from $\mathrm{H}_2$ and $\frac12\mathrm{O}_2$. We have $2\mathrm{H}_2\mathrm{O}$ on the right from step 2, so we need to consume them. Reverse reaction 1 and multiply by 2:

   $2\mathrm{H}_2\mathrm{O}(l) \rightarrow 2\mathrm{H}_2(g) + \mathrm{O}_2(g)$; $\Delta H = -2 \times (-285) = +570\ \mathrm{kJ}$

4. Add all three modified reactions and their $\Delta H$ values.

   From step 1: $2\mathrm{N}_2 + 6\mathrm{O}_2 + 2\mathrm{H}_2 \rightarrow 4\mathrm{HNO}_3$; $\Delta H = -696.4$

   From step 2: $4\mathrm{HNO}_3 \rightarrow 2\mathrm{N}_2\mathrm{O}_5 + 2\mathrm{H}_2\mathrm{O}$; $\Delta H = +153.2$

   From step 3: $2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{H}_2 + \mathrm{O}_2$; $\Delta H = +570$

   Sum the left sides: $2\mathrm{N}_2 + 6\mathrm{O}_2 + 2\mathrm{H}_2 + 4\mathrm{HNO}_3 + 2\mathrm{H}_2\mathrm{O}$

   Sum the right sides: $4\mathrm{HNO}_3 + 2\mathrm{N}_2\mathrm{O}_5 + 2\mathrm{H}_2\mathrm{O} + 2\mathrm{H}_2 + \mathrm{O}_2$

   Cancel identical species on both sides: $4\mathrm{HNO}_3$, $2\mathrm{H}_2\mathrm{O}$, and $2\mathrm{H}_2$ cancel.

   Left remains: $2\mathrm{N}_2 + 6\mathrm{O}_2$

   Right remains: $2\mathrm{N}_2\mathrm{O}_5 + \mathrm{O}_2$

   Subtract the extra $\mathrm{O}_2$ from both sides: $2\mathrm{N}_2 + 5\mathrm{O}_2 \rightarrow 2\mathrm{N}_2\mathrm{O}_5$ — exactly the target.

5. Add the enthalpy changes:

   $\Delta H = -696.4 + 153.2 + 570 = 26.8\ \mathrm{kJ}$

The enthalpy change of a reaction ($\Delta_r H^\circ$) can be estimated using bond enthalpies. Bond enthalpy is the energy required to break one mole of a particular type of bond in the gaseous state. Conversely, the same amount of energy is released when one mole of that bond is formed.

The core idea is that a chemical reaction involves breaking existing bonds in reactants and forming new bonds in products.

• Breaking bonds is an endothermic process: It requires energy input, so the enthalpy change associated with bond breaking is positive.
• Forming bonds is an exothermic process: It releases energy, so the enthalpy change associated with bond formation is negative.

Therefore, the net enthalpy change for a reaction can be calculated as:

Let's apply this to the given problem.

1. Write the balanced chemical equation for the formation of D$_2$O: The reaction involves heavy hydrogen (D$_2$) and oxygen (O$_2$) forming heavy water (D$_2$O).

We assume the product D$_2$O is in the gaseous state because bond enthalpies are defined for gaseous molecules. If the question implied liquid D$_2$O, an additional enthalpy of condensation would be needed, which is not provided.

2. Identify the bonds broken in reactants:

* In D$_2(g)$, one D–D bond is broken.

* In $\frac{1}{2} \text{O}_2(g)$, half of an O=O bond is broken.

> [!WARNING]

> The problem states "O–O" bond enthalpy as +498 kJ mol$^{-1}$. This value is characteristic of an O=O (double) bond in diatomic oxygen, not a single O–O bond (which is much weaker, around 140-200 kJ mol$^{-1}$). In such problems, when elemental oxygen is involved and a value like 498 kJ mol$^{-1}$ is given for "O–O", it refers to the O=O bond in O$_2$. We will proceed with this interpretation.

3. Identify the bonds formed in products:

   • In D$_2$O$(g)$, two D–O bonds are formed. The structure of D$_2$O is similar to H$_2$O, with a central oxygen atom bonded to two deuterium atoms.

4. List the given bond enthalpies:

   • Bond enthalpy of D–D = +400 kJ mol$^{-1}$
   • Bond enthalpy of O=O = +498 kJ mol$^{-1}$ (as interpreted above)
   • Bond enthalpy of D–O = +490 kJ mol$^{-1}$

5. Calculate the total energy required to break bonds in reactants:

   • Energy for D–D bond = $1 \times (+400 \text{ kJ mol}^{-1}) = +400 \text{ kJ mol}^{-1}$
   • Energy for O=O bond = $\frac{1}{2} \times (+498 \text{ kJ mol}^{-1}) = +249 \text{ kJ mol}^{-1}$
   • Total energy to break bonds = $400 + 249 = +649 \text{ kJ mol}^{-1}$

6. Calculate the total energy released when bonds are formed in products:

   • Energy for D–O bonds = $2 \times (+490 \text{ kJ mol}^{-1}) = +980 \text{ kJ mol}^{-1}$
   • Total energy released from forming bonds = $+980 \text{ kJ mol}^{-1}$ (Remember, this energy is released, so it contributes negatively to $\Delta_r H^\circ$).

7. Calculate the standard enthalpy of reaction ($\Delta_r H^\circ$):

   Using the formula:

Concept & Intuition:

Calorific value is the heat released when one mole of a fuel is completely burned. Different fuels have different chemical bonds, so the energy released per mole varies. The key is to recall standard enthalpies of combustion for common fuels. Hydrogen burns to water, methane to CO₂ and water, and LPG (liquefied petroleum gas, mainly propane and butane) releases much more energy per mole because it has more carbon and hydrogen atoms to oxidize.

Step-by-step reasoning:

1. Identify the fuels and their typical composition

   • (A) H₂(g): pure hydrogen.
   • (B) CH₄(g): methane, the main component of natural gas.
   • (C) CNG(g): compressed natural gas, primarily methane (~90%+).
   • (D) LPG: liquefied petroleum gas, mainly propane (C₃H₈) and butane (C₄H₁₀).

2. Recall standard molar enthalpies of combustion (kJ/mol)

   • H₂(g) + ½O₂ → H₂O(l): ΔH° = –286 kJ/mol
   • CH₄(g) + 2O₂ → CO₂ + 2H₂O(l): ΔH° = –890 kJ/mol
   • C₃H₈(g) + 5O₂ → 3CO₂ + 4H₂O(l): ΔH° = –2220 kJ/mol
   • C₄H₁₀(g) + 6.5O₂ → 4CO₂ + 5H₂O(l): ΔH° = –2877 kJ/mol

   Since CNG is essentially methane, its value is ~890 kJ/mol. LPG is a mixture, but a typical value for propane (the major component) is ~2220 kJ/mol.

3. Compare the magnitudes

   • H₂: 286 kJ/mol
   • CH₄ / CNG: 890 kJ/mol
   • LPG: ~2220 kJ/mol (or higher for butane)

   Clearly, LPG releases the most energy per mole.

4. Address a common pitfall

   Watch out

   Some might think hydrogen has the highest calorific value per gram (because it is very light), but the question asks per mole. Per mole, larger hydrocarbon molecules release more energy because they contain more bonds to break and form.

5. Conclusion

   Among the options, LPG has the highest calorific value in kJ/mole.