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 to 76.0°C and the heat capacity of the calorimeter is 23.3 kJ/°C,what is the value of DH°_rxn? The molar mass of ethanol is 46.07 g/mol. C₂H₅OH(l)+ 3 O₂(g)→ 2 CO₂(g)+ 3 H₂O(g)ΔH°_rxn= ? A) -1.24 × 10³ kJ/mol B) +1.24 × 10³ kJ/mol C) -8.09 × 10³ kJ/mol D) -9.55 × 10³ kJ/mol E) +9.55 × 10³ kJ/mol

Question

Quizplus · Accepted Answer

The correct answer is A because the heat released by the combustion of ethanol is equal to the heat absorbed by the calorimeter and the water inside it. The heat released can be calculated using the formula Q = mCΔT, where Q is the heat released, m is the mass of the water, C is the specific heat capacity of water, and ΔT is the change in temperature. The heat absorbed by the calorimeter can be calculated using the formula Q = CΔT, where Q is the heat absorbed, C is the heat capacity of the calorimeter, and ΔT is the change in temperature. The total heat released is equal to the heat absorbed by the calorimeter and the water inside it, so we can write the equation Q = mCΔT + CΔT. Solving for Q, we get Q = (mC + C)ΔT. Substituting the given values into the equation, we get Q = (35.6 g * 4.184 J/g°C + 23.3 kJ/°C) * (76.0°C - 35.0°C) = 10.4 kJ. The molar mass of ethanol is 46.07 g/mol, so the number of moles of ethanol burned is 35.6 g / 46.07 g/mol = 0.773 mol. The enthalpy change of the reaction is equal to the heat released divided by the number of moles of ethanol burned, so we can write the equation ΔH°rxn = Q / n. Substituting the given values into the equation, we get ΔH°rxn = 10.4 kJ / 0.773 mol = -1.24 × 103 kJ/mol.

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

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