Q.71. If 3.365 g of ethanol (l) burnt completely in a bomb calorimeter at 298.15K, the heat produced

Q: What is the correct answer for JEE Main Chemistry Q71?

The correct answer is 282. First, let's calculate the moles of ethanol: Molar mass of C₂H₅OH = 2*12 + 6*1 + 1*16 = 24 + 6 + 16 = 46 g/mol. Moles of ethanol = 3.365 g / 46 g/mol = 0.073152 mol. Heat produced (q) = 99.472 kJ. This is the internal energy change (ΔU) for the combustion reaction in a bomb calorimeter. ΔU = -q = -99.472 kJ for 0.073152 mol of ethanol. ΔU for 1 mole of ethanol = -99.472 kJ / 0.073152 mol = -1359

Q: Q.71. If 3.365 g of ethanol (l) burnt completely in a bomb calorimeter at 298.15K, the heat produced is 99.472 kJ. The |ΔH| of ethanol at 298.15 K is ______ × 10² kJ mol⁻¹. (Nearest integer) Given…

The correct answer is 282. First, let's calculate the moles of ethanol: Molar mass of C₂H₅OH = 2*12 + 6*1 + 1*16 = 24 + 6 + 16 = 46 g/mol. Moles of ethanol = 3.365 g / 46 g/mol = 0.073152 mol. Heat produced (q) = 99.472 kJ. This is the internal energy change (ΔU) for the combustion reaction in a bomb calorimeter. ΔU = -q = -99.472 kJ for 0.073152 mol of ethanol. ΔU for 1 mole of ethanol = -99.472 kJ / 0.073152 mol = -1359

Question

Q.71. If 3.365 g of ethanol (l) burnt completely in a bomb calorimeter at 298.15K, the heat produced is 99.472 kJ. The |ΔH| of ethanol at 298.15 K is ______ × 10² kJ mol⁻¹. (Nearest integer)
Given :
Standard enthalpy of combustion of graphite = -393.5 kJ mol⁻¹
Standard enthalpy of formation of water = -285.8 kJ mol⁻¹
Molar mass in gmol⁻¹ of C, H, O are 12, 1 and 16 respectively.

FirstInTest · Accepted Answer

First, let's calculate the moles of ethanol:
Molar mass of C₂H₅OH = 2*12 + 6*1 + 1*16 = 24 + 6 + 16 = 46 g/mol.
Moles of ethanol = 3.365 g / 46 g/mol = 0.073152 mol.

Heat produced (q) = 99.472 kJ.
This is the internal energy change (ΔU) for the combustion reaction in a bomb calorimeter.
ΔU = -q = -99.472 kJ for 0.073152 mol of ethanol.

ΔU for 1 mole of ethanol = -99.472 kJ / 0.073152 mol = -1359.8 kJ/mol.

The combustion reaction of ethanol is:
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

Now, we need to find ΔH for this reaction. The relationship between ΔH and ΔU is:
ΔH = ΔU + Δn_g RT
Where Δn_g = (moles of gaseous products) - (moles of gaseous reactants)
For the combustion of ethanol: Δn_g = (2 moles CO₂) - (3 moles O₂) = 2 - 3 = -1.

R = 8.314 J mol⁻¹ K⁻¹ = 0.008314 kJ mol⁻¹ K⁻¹.
T = 298.15 K.

Δn_g RT = (-1) * 0.008314 kJ mol⁻¹ K⁻¹ * 298.15 K = -2.478 kJ/mol.

ΔH_combustion = ΔU_combustion + Δn_g RT
ΔH_combustion = -1359.8 kJ/mol + (-2.478 kJ/mol) = -1362.278 kJ/mol.

Now, we need to calculate the standard enthalpy of formation of ethanol (ΔH_f°(C₂H₅OH)).
Using Hess's Law for the combustion reaction:
ΔH_combustion = [2 * ΔH_f°(CO₂) + 3 * ΔH_f°(H₂O)] - [ΔH_f°(C₂H₅OH) + 3 * ΔH_f°(O₂)]

We know:
ΔH_f°(CO₂) = ΔH_combustion(graphite) = -393.5 kJ mol⁻¹
ΔH_f°(H₂O) = -285.8 kJ mol⁻¹
ΔH_f°(O₂) = 0 kJ mol⁻¹ (for an element in its standard state)

-1362.278 = [2 * (-393.5) + 3 * (-285.8)] - [ΔH_f°(C₂H₅OH) + 0]
-1362.278 = [-787.0 - 857.4] - ΔH_f°(C₂H₅OH)
-1362.278 = -1644.4 - ΔH_f°(C₂H₅OH)
ΔH_f°(C₂H₅OH) = -1644.4 + 1362.278
ΔH_f°(C₂H₅OH) = -282.122 kJ mol⁻¹.

The question asks for |ΔH| of ethanol at 298.15 K, which refers to the enthalpy of formation. The value is -282.122 kJ mol⁻¹.

Rounding to the nearest integer and expressing as × 10² kJ mol⁻¹:
|-282.122| ≈ 282.
So, 282 × 10² kJ mol⁻¹ (This seems to be a typo in the question, it should be just kJ mol⁻¹ or the value should be 2.82 × 10² kJ mol⁻¹). Assuming it means 282 kJ mol⁻¹.

The provided solution calculates ΔU_combustion = -1359.8 KJ/mole. Then ΔH_combustion = -1362.27 KJ/mole. Then it calculates ΔH_f°(C₂H₅OH) = -282.122 KJ/mole. The final answer is 282.

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