Total enthalpy change for freezing of 1 ~mol$1 \mathrm{~mol}$ of water at 10° C$10^{\circ} \mathrm{C}$ to ice at -10° C$-10^{\circ} \mathrm{C}$ is (Given: ΔfusH = x$\Delta_{\mathrm{fus}}\mathrm{H} = \mathrm{x}$ kJ/mol, Cₚ[H₂O( )] = y J mol⁻¹ K⁻¹$\mathrm{C}_{\mathrm{p}}[\mathrm{H}_2\mathrm{O}(\ell)] = \mathrm{y}\text{ J mol}^{-1}\text{ K}^{-1}$, Cₚ[H₂O(s)] = z J mol⁻¹ K⁻¹$\mathrm{C}_{\mathrm{p}}[\mathrm{H}_2\mathrm{O}(\text{s})] = \mathrm{z}\text{ J mol}^{-1}\text{ K}^{-1}$)

Related Formula

According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided.

Core Logic

Let the given reactions be: (1) 2Fe(s) + (3)/(2)O2(g) arrow Fe₂O3(s), Δ H₁ = -822 kJ/mol$2\mathrm{Fe}_{(s)} + \frac{3}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_{3(s)}, \quad \Delta H_1 = -822 \, \mathrm{kJ/mol}$ (2) C(s) + (1)/(2)O2(g) arrow CO(g), Δ H₂ = -110 kJ/mol$\mathrm{C}_{(s)} + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}, \quad \Delta H_2 = -110 \, \mathrm{kJ/mol}$

Target Reaction (3):

To construct the target reaction:

• We need 3 CO(g)$3 \mathrm{CO}_{(g)}$ on the product side, so we multiply reaction (2) by 3.
• We need Fe₂O3(s)$\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and 2 Fe(s)$2 \mathrm{Fe}_{(s)}$ on the product side, so we reverse reaction (1).

Step 1: Calculate Net Enthalpy

Target Reaction (3) = 3 × (2) - (1)$3 \times (2) - (1)$

Chapter Mix

Class 11 Chemistry: Thermodynamics

Related Formula

Core Logic

The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph. Note that the provided graph is Volume (V$V$) on the y-axis versus Pressure (P$P$) on the x-axis. The path A arrow B arrow C arrow A$A \rightarrow B \rightarrow C \rightarrow A$ is traced in a clockwise direction on the V$V$-P$P$ graph. Clockwise on a V$V$-P$P$ graph corresponds to anti-clockwise on a standard P$P$-V$V$ graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for).

Step 1: Calculating Area

The enclosed region is a right-angled triangle. Base of triangle on P-axis = 30 - 10 = 20 kPa$30 - 10 = 20 \text{ kPa}$ Height of triangle on V-axis = 30 - 10 = 20 dm³$30 - 10 = 20 \text{ dm}^3$

Step 2: Unit conversion

1 kPa = 10³ Pa$1 \text{ kPa} = 10^3 \text{ Pa}$ 1 dm³ = 1 Litre = 10⁻³ m³$1 \text{ dm}^3 = 1 \text{ Litre} = 10^{-3} \text{ m}^3$

Pattern Recognition

1 kPa · 1 L = 1 Joule$1 \text{ kPa} \cdot 1 \text{ L} = 1 \text{ Joule}$. This direct conversion saves time without converting explicitly to standard SI units (Pa$Pa$ and m³$m^3$).

Chapter Mix

Class 11 Chemistry: Thermodynamics

Related Formula

Core Logic

For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention. Given: n = 5 moles$n = 5\text{ moles}$ R = 8.314 J K⁻¹mol⁻¹$R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$ T = 300 K$T = 300\text{ K}$ V₁ = 10 L$V_1 = 10\text{ L}$ V₂ = 100 L$V_2 = 100\text{ L}$

Step 1: Calculating Work Done

Step 2: Final Formatting

The question asks for work w = -x J$w = -x\text{ J}$. So x = 28720.713$x = 28720.713$, which rounds to 28721$28721$.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Related Formula

Step 1: Calculation

Step 2: Nearest Integer

Rounding 163.19$163.19$ to the nearest integer gives 163$163$.

Chapter Mix

Class 11 Chemistry: Thermodynamics