Unit 7 2. Consider the reaction Calculate the AlF for the above reaction. The st

Question

Unit 7 2. Consider the reaction Calculate the AlF for the above reaction. The standard enthalpies of formation for Cll,OH(g, COAg), and H2O(g) are-201.0 ki/mol,-3935 kJ/rnol, and-241.8 kJ/mol, respectively a. b. Would Aft for the reaction be different if liquid (instead of gaseous) water were formed? c. Calculate the heat evolved when 50.0 g of methanol reacts with excess oxygen. d. If the heat evolved in a is used to heat up a sample of an unknown liquid frorm 25.0 "C to 91.2 °C, calculate the mass of this sample, assuming the molar specific heat of the liquid is 61.2 J/mol-K with molar mass 31.8 g/mol (and no heat lost to surroundings). e. If the heat evolved in a were used to warm an Fe sample (0.449 J/g.K) of equal mass (to the liquid in d, explain whether the temperature change of the iron is higher or lower than that of the liquid? No calculation is necessary

Explanation / Answer

Ans 2

Part a

Enthalpy change

H = sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants

= Hf(CO2 g ) + 2*Hf(H2O g ) - 0.5Hf(O2 g ) - Hf(CH3OH g )

= (-393.5) + 2*(-241.8) - 0 - (-201)

H = - 676.1 kJ/mol

Part b

If liquid water formed instead of gaseous water

Enthalpy of formation of H2O(l) = - 285.8 kJ/mol

Then enthalpy change

= Hf(CO2 g ) + 2*Hf(H2O l ) - 0.5Hf(O2 g ) - Hf(CH3OH g )

= (-393.5) + 2*(-285.8) - 0 - (-201)

H = - 764.1 kJ/mol

Part C

Heat evolved when 50 g methanol reacts

Moles of methanol reacts = 50g / 32.04g/mol

= 1.56 mol

Heat evolved = - 676.1 kJ/mol x 1.56 mol

= - 1054.716 kJ

Part d

H = m x Cp x (T2-T1)

676.1 kJ/mol x 1000J/kJ = m x 61.2 J/mol·K x (91.2 - 25)K

676100 = 4051.44 m

m = 166.88 mol

Mass of unknown = 166.88 mol x 31.8 g/mol

= 5306.75 g

= 5.306 kg

Part e

Temperature change would be higher because it has lower Cp than liquid.

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