Question

Part A How much energy (in klojoules) is released when 22.0 g of capacity is 113 J/(K mol) for the liquid and 65.7 J/(K-mol) for the vapor ethanol vapor at 98.5 C is cooled to-12.0 C Ehanol has mp k.J Submit de Feedback Next

Answer 1

You are asked about the heat released by vapor ethanol so we have to do 3 heat calculations

1. Ethanol vapor going from 98.5 C to the boiling point (78.4)

2. Ethanol going through a phase change (gas to liquid also known as latent heat)

3. Ethanol (liquid) going from 78.4 to -12C

The ethanol turns to solid at -114 C so we do not have solid ethanol here

1. vapor ethanol

∆H = m * Cp * ∆T

∆T = Final temperature – Initial temperature

m is mass

Cp is heat capacity

H is enthalpy so

calculate moles of ethanol, moles = mass / molar mass (molar mass of ethanol is 46)

22 / 46 = 0.478 moles

H = 0.478 moles * 65.7 J / mole * (78.4 - 98.5) = -631.57 J , the negative sign is because heat is being released

2. calculate the heat because of phase change

H = moles * Enthalpy of vaporization

H = 0.478 * 38.56 KJ / mole = 18.441 KJ or 18 441 Joules

3. Calculate the heat for liquid ethanol

∆H = m * Cp * ∆T

H = 0.478 * 113 * ( -12 - 78.4) = -4885.53 Joules

Total heat released = 4885.53 + 18431 + 631.57 = 23 958.84 Joules

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