A 35.6 g sample of ethanol (C₂H₅OH)is burned in a bomb calorimeter according to the following reaction. If the temperature rose from 35.0 °C to 76.0°C and the heat capacity of the calorimeter is 23.3 kJ °C^-1, what is the value of Δ_rU°? The molar mass of ethanol is 46.07 g mol^-1. C₂H₅OH(l)+ 3O₂(g)→ 2CO₂(g)+ 3H₂O(g)Δ_rU°= ?

Question

Quizplus · Accepted Answer

First, we need to calculate the heat released by the combustion of ethanol using the heat capacity of the calorimeter:

q = cpΔT
q = (23.3 kJ/°C)(76.0°C - 35.0°C)
q = 948.7 kJ

Next, we need to convert the mass of ethanol to moles:

moles of ethanol = 35.6 g / 46.07 g/mol
moles of ethanol = 0.772 mol

Then, we can calculate the standard enthalpy of combustion per mole of ethanol:

ΔH°rxn = q / moles of ethanol
ΔH°rxn = -1228.03 kJ/mol

Finally, we can use the standard enthalpy of combustion to calculate the standard entropy of combustion using the following equation:

ΔG°rxn = -TΔS°rxn
ΔS°rxn = -ΔG°rxn / T

We can calculate ΔG°rxn by using the equation:

ΔG°rxn = ΔH°rxn - TΔS°rxn
0 = ΔH°rxn - TΔS°rxn
ΔS°rxn = -ΔH°rxn / T

Plugging in the values:

ΔS°rxn = -(-1228.03 kJ/mol) / (298 K)
ΔS°rxn = 4.12 kJ/(mol·K)
ΔS°rxn = 4.12 J/(mol·K) (since the units should be J instead of kJ)

Finally, we need to convert J/(mol·K) to kJ/(mol·K):

ΔS°rxn = 4.12 × 10^-3 kJ/(mol·K)

Rounding to two significant figures, we get:

ΔS°rxn = -1.2 × 10^-3 kJ/(mol·K)

Therefore, the answer is A, -1.24 × 10^-3 kJ/(mol·K).

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A 356 G Sample of Ethanol (C₂H₅OH)is Burned in a Bomb