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Enthalpy of Formation of Ethanol from Bomb Calorimetry Data

If 3.365 g3.365\text{ g} of ethanol (ll) is burnt completely in a bomb calorimeter at 298.15 K298.15\text{ K}, the heat produced is 99.472 kJ99.472\text{ kJ}. The ∣ΔHf∘∣|\Delta H_f^\circ| of ethanol at 298.15 K298.15\text{ K} is _______ ×102 kJ mol−1\times 10^2\text{ kJ mol}^{-1}. (Nearest integer)

Given: Standard enthalpy for combustion of graphite =−393.5 kJ mol−1= -393.5\text{ kJ mol}^{-1}
Standard enthalpy of formation of water (ll) =−285.8 kJ mol−1= -285.8\text{ kJ mol}^{-1}
Molar mass in g mol−1\text{g mol}^{-1} of C, H, O\text{C, H, O} are 1212, 11 and 1616 respectively

Official Numerical Answer3

Topics & Concepts

ThermodynamicsEnthalpyFirst Law of Thermodynamics

Step-by-Step Solution

To find the magnitude of the standard enthalpy of formation (∣ΔHf∘∣|\Delta H_f^\circ|) of ethanol at 298.15 K298.15\text{ K}, we follow these step-by-step thermodynamic calculations:

Step 1: Molar Mass and Moles of Ethanol

The chemical formula of ethanol is C2H5OH\text{C}_2\text{H}_5\text{OH}. Using the given molar masses: Molar mass of ethanol (M)=2(12)+6(1)+16=46 g mol−1\text{Molar mass of ethanol } (M) = 2(12) + 6(1) + 16 = 46\text{ g mol}^{-1}

The number of moles of ethanol (nn) burnt is: n=3.365 g46 g mol−1≈0.073152 moln = \frac{3.365\text{ g}}{46\text{ g mol}^{-1}} \approx 0.073152\text{ mol}

Step 2: Calculate Molar Internal Energy of Combustion (ΔUc∘\Delta U_c^\circ)

A bomb calorimeter operates at constant volume, so the heat produced corresponds to the internal energy change (ΔUc\Delta U_c). For 3.365 g3.365\text{ g} of ethanol, the heat produced is 99.472 kJ99.472\text{ kJ}. Thus, ΔUc=−99.472 kJ\Delta U_c = -99.472\text{ kJ}.

The molar internal energy change of combustion (ΔUc∘\Delta U_c^\circ) is: ΔUc∘=−99.472 kJ0.073152 mol=−1359.80 kJ mol−1\Delta U_c^\circ = \frac{-99.472\text{ kJ}}{0.073152\text{ mol}} = -1359.80\text{ kJ mol}^{-1}

Step 3: Calculate Molar Enthalpy of Combustion (ΔHc∘\Delta H_c^\circ)

The balanced equation for the complete combustion of ethanol is: C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)

The change in the number of gaseous moles (Δng\Delta n_g) is: Δng=nproducts, gas−nreactants, gas=2−3=−1 mol\Delta n_g = n_{\text{products, gas}} - n_{\text{reactants, gas}} = 2 - 3 = -1\text{ mol}

Using the relation between enthalpy change and internal energy change: ΔHc∘=ΔUc∘+ΔngRT\Delta H_c^\circ = \Delta U_c^\circ + \Delta n_g R T

Substitute the known values (R=8.314×10−3 kJ K−1mol−1R = 8.314 \times 10^{-3}\text{ kJ K}^{-1}\text{mol}^{-1} and T=298.15 KT = 298.15\text{ K}): ΔngRT=(−1)×(8.314×10−3 kJ K−1mol−1)×298.15 K=−2.48 kJ mol−1\Delta n_g R T = (-1) \times (8.314 \times 10^{-3}\text{ kJ K}^{-1}\text{mol}^{-1}) \times 298.15\text{ K} = -2.48\text{ kJ mol}^{-1}

ΔHc∘=−1359.80 kJ mol−1+(−2.48 kJ mol−1)=−1362.28 kJ mol−1\Delta H_c^\circ = -1359.80\text{ kJ mol}^{-1} + (-2.48\text{ kJ mol}^{-1}) = -1362.28\text{ kJ mol}^{-1}

Step 4: Calculate Standard Enthalpy of Formation (ΔHf∘\Delta H_f^\circ)

The standard enthalpy of combustion is given by: ΔHc∘=2ΔHf∘(CO2,g)+3ΔHf∘(H2O,l)−ΔHf∘(C2H5OH,l)\Delta H_c^\circ = 2\Delta H_f^\circ(\text{CO}_2, g) + 3\Delta H_f^\circ(\text{H}_2\text{O}, l) - \Delta H_f^\circ(\text{C}_2\text{H}_5\text{OH}, l)

Given:

  • ΔHf∘(CO2,g)=ΔHc∘(graphite)=−393.5 kJ mol−1\Delta H_f^\circ(\text{CO}_2, g) = \Delta H_c^\circ(\text{graphite}) = -393.5\text{ kJ mol}^{-1}
  • ΔHf∘(H2O,l)=−285.8 kJ mol−1\Delta H_f^\circ(\text{H}_2\text{O}, l) = -285.8\text{ kJ mol}^{-1}

Substitute these values into the equation: −1362.28=2(−393.5)+3(−285.8)−ΔHf∘(C2H5OH,l)-1362.28 = 2(-393.5) + 3(-285.8) - \Delta H_f^\circ(\text{C}_2\text{H}_5\text{OH}, l) −1362.28=−787.0−857.4−ΔHf∘(C2H5OH,l)-1362.28 = -787.0 - 857.4 - \Delta H_f^\circ(\text{C}_2\text{H}_5\text{OH}, l) −1362.28=−1644.4−ΔHf∘(C2H5OH,l)-1362.28 = -1644.4 - \Delta H_f^\circ(\text{C}_2\text{H}_5\text{OH}, l)

ΔHf∘(C2H5OH,l)=−1644.4+1362.28=−282.12 kJ mol−1\Delta H_f^\circ(\text{C}_2\text{H}_5\text{OH}, l) = -1644.4 + 1362.28 = -282.12\text{ kJ mol}^{-1}

Step 5: Magnitude and Final Value

The magnitude of the standard enthalpy of formation is: ∣ΔHf∘∣=282.12 kJ mol−1=2.8212×102 kJ mol−1|\Delta H_f^\circ| = 282.12\text{ kJ mol}^{-1} = 2.8212 \times 10^2\text{ kJ mol}^{-1}

Rounding off to the nearest integer gives: ∣ΔHf∘∣≈3×102 kJ mol−1|\Delta H_f^\circ| \approx 3 \times 10^2\text{ kJ mol}^{-1}

Final Answer: 33

Enthalpy of Formation of Ethanol from Bomb Calorimetry Data | Chemistry PYQ Solution - JEE Challenger