Q.1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation $C(graphite) + O_2(g) \rightarrow CO_2(g)$. During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?

Statement (a) — "All enthalpies of fusion are positive." — CORRECT.

Melting means breaking down the ordered lattice of a solid into a liquid. Energy must be supplied to overcome the intermolecular/interionic attractions, so heat is absorbed:

So (a) is a true statement, not an incorrect one. That immediately rules out options (A), (C) and (D), all of which list (a).

Statement (b) — "The magnitude of enthalpy change does not depend on the strength of the intermolecular interactions." — INCORRECT.

The enthalpy of a phase transition is precisely a measure of those interactions. Water, held together by strong hydrogen bonds, has $\Delta_{vap}H^\circ = 40.79\ \mathrm{kJ\,mol^{-1}}$, while acetone, with only weak dipole–dipole forces, has about $29\ \mathrm{kJ\,mol^{-1}}$. Stronger interactions $\Rightarrow$ more energy needed $\Rightarrow$ larger $\Delta H$. Statement (b) flatly contradicts this. Incorrect.

Statement (c) — "When a chemical reaction is reversed, the value of $\Delta_r H^\circ$ is reversed in sign." — CORRECT.

This is a direct consequence of enthalpy being a state function (Lavoisier–Laplace law). If

then

Otherwise one could run the cycle and create energy from nothing. Correct statement.

Statement (d) — "The change in enthalpy is dependent of path between initial state and final state." — INCORRECT.

Enthalpy is a state function: it depends only on the state of the system, not on how that state was reached. Hence

$\Delta H = H_{final} - H_{initial}$ …

Concept and Intuition

A bomb calorimeter is a constant-volume device. The heat released by combustion is measured as a temperature rise, and from that we directly get the internal energy change, $\Delta U$, not the enthalpy change $\Delta H$.

The relationship between them is:

where $\Delta n_g$ is the change in moles of gas during the reaction.

For propane combustion:

Notice water is liquid, so only the gaseous species count.

$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 3 - (1+5) = -3$.

Thus, $\Delta H$ is more negative than $\Delta U$ by $3RT$ per mole of propane burned.

We must compute $\Delta U$ from the calorimeter data, then correct for this $PV$ work term.

Step-by-step solution

1. Find moles of propane burned

Molar mass of propane ($\mathrm{C_3H_8}$) = $3\times12 + 8\times1 = 44\ \text{g/mol}$.

Mass used = $0.4\ \text{g}$.

2. Calculate $\Delta U$ for the reaction

Heat released at constant volume:

This is the heat released by $0.0090909$ mol of propane.

So the molar internal energy change:

(The negative sign because heat is released.)

3. Compute the $\Delta n_g RT$ correction

$\Delta n_g = -3$ (as derived above).

$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} = 0.008314\ \text{kJ mol}^{-1}\text{K}^{-1}$

$T = 300\ \text{K}$

Step 1 — The key relation.

On a reaction-coordinate diagram, both forward and backward reactions climb to the same transition state. Hence

Step 2 — Backward barrier without catalyst.

(Makes sense: the reaction is exothermic, so going backward — uphill — costs more.)

Step 3 — What the catalyst does.

A catalyst provides an alternative path with a lower transition state. It is a kinetic device only: $\Delta H$ is a state function and is unchanged. So if the peak drops by $10\,\mathrm{kJ\,mol^{-1}}$, both barriers drop by $10$:

$E_{a,f}^{\text{cat}} = 50 - 10 = 40\ \mathrm{kJ\,mol^{-1}}$ …

Concept & Intuition

Enthalpy ($H$) and internal energy ($U$) are related by $H = U + PV$. For a reaction at constant pressure, the heat measured is $\Delta H$, but the actual change in the system’s internal energy is $\Delta U$. The difference comes from work done by (or on) the surroundings due to volume change. If gases are involved, the volume change is significant, and we account for it using the ideal gas law: $P\Delta V = \Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gas. So $\Delta H = \Delta U + \Delta n_g RT$, and we solve for $\Delta U$.

Step-by-step solution

1. Write the balanced combustion reaction for benzoic acid Benzoic acid: $\mathrm{C_6H_5COOH(s)}$ (or $\mathrm{C_7H_6O_2}$). Complete combustion:

Note: Water is liquid at $25^\circ\mathrm{C}$ and 1 atm, so it does not contribute to gas moles.

2. Find $\Delta n_g$, the change in moles of gas Reactants: $\frac{15}{2} = 7.5$ moles of $\mathrm{O_2(g)}$ Products: $7$ moles of $\mathrm{CO_2(g)}$

3. Recall the relationship between $\Delta H$ and $\Delta U$ At constant pressure:

Rearranging:

4. Plug in the values $\Delta H = -2546\ \mathrm{kJ/mol}$ $R = 8.314\ \mathrm{J/(mol\cdot K)} = 0.008314\ \mathrm{kJ/(mol\cdot K)}$ $T = 25^\circ\mathrm{C} = 298\ \mathrm{K}$ …

Step 1 — Build the Born–Haber-type cycle for dissolution

Dissolving an ionic solid can be split into two hypothetical steps whose enthalpies are given:

Step A — lattice dissociation (pull the ions apart into the gas phase):

This is endothermic — energy must be supplied to overcome the strong electrostatic attractions in the crystal.

Step B — hydration (surround the gaseous ions with water molecules):

This is exothermic — ion–dipole attractions form between the ions and the polar water molecules.

Step 2 — Apply Hess's law

Enthalpy is a state function, so the enthalpy change for the overall route (solid → aqueous ions) equals the sum of the steps, whatever the actual path:

Step 3 — Substitute

Step 4 — Interpret the result …

1. The reaction and its mole balance

Count gaseous moles:

This single fact drives the whole question.

2. Effect of pressure — option (C)

Le Chatelier: on increasing the pressure (compressing the system), the equilibrium shifts toward the side with fewer moles of gas. Here both sides have exactly two moles of gas, so there is no side with fewer moles — there is nothing for the system to shift towards.

Quantitatively, $K_p$ and $K_c$ are related by

and the reaction quotient is

If the total pressure is raised by a factor $\lambda$ (compressing at constant $T$), every partial pressure is multiplied by $\lambda$:

$Q$ is unchanged, so the system is still exactly at equilibrium: no shift, no disturbance whatsoever. ✓

3. Why the other three DO disturb the system

1. (A) Addition of A — raises $p_A$, so $Q = \dfrac{p_Cp_D}{p_Ap_B}$ falls below $K$. The system responds by shifting forward (left → right) to restore $Q = K$. The equilibrium is disturbed (its position moves). ✗ …

The key idea is that under identical temperature and pressure, equal volumes of gases contain equal numbers of moles (Avogadro’s law). So the volume percentage given in the problem is directly the mole percentage. Once we know the mole ratio, we convert to mass using the molar masses of He (4 g/mol) and CH₄ (16 g/mol).

1. Interpret the volume composition as mole composition.

   At the same T and P, volume % = mole %.

   So in 100 moles of mixture:

   • Moles of He = 25
   • Moles of CH₄ = 75

2. Find the mass of each component.

   • Mass of He = $25 \times 4 = 100$ g
   • Mass of CH₄ = $75 \times 16 = 1200$ g

3. Total mass of the mixture.

   $100 + 1200 = 1300$ g

4. Percentage by mass of CH₄.

   $$\frac{1200}{1300} \times 100 = \frac{12}{13} \times 100 \approx 92.3\% …$$