Question

A student wishes to determine the enthalpy of combustion of liquid methanol (CH3OH(l), MW = 32.04 g/mol)

a. Write out the balanced combustion reaction: liquid water is a product.

b. A student wishes to determine the enthalpy of combustion of liquid methanol (CH3OH(l), MW = 32.04 g/mol) a. Write out the balanced combustion reaction: liquid water is a product. b. Use bond enthalpies (see chapter 8 of your book for appropriate values) to estimate the enthalpy of combustion of one mole of liquid methanol. The structure of methanol is given below c. A student uses a calorimeter setup similar to what you used in the studio. The calorimeter contains 250. grams of water and the initial temperature of the calorimeter and the water is 25.00 oC. After burning 1.00 gram of methanol the temperature of the calorimeter and its contents has increased to 37.45 oC. What is the enthalpy of combustion for one mole of liquid methanol? Ccal = 775 J/ oC and the specific heat of liquid water is 4.184 J/g oC.

Answer 1

The balanced equation is

2 CH₃OH (l) + 3 O₂ (g) 2 CO₂ (g) + 4 H₂O (l)

Mass of water = 250 g

specific heat of liquid water = C_w = 4.184 J/g^oC

calorimeter constant = 775 J / ^oC

Change in temperature = 37.45 ^oC - 25.00 ^oC = 12.45 ^oC

Heat released = [(mass of water) * C_w + C_cal] * (Change in temperature)

Heat released = [(250 g) * (4.184 J/g^oC) + 775 J / ^oC] * (12.45 ^oC)

Heat released = [1046 J/^oC + 775 J/^oC] * (12.45 ^oC)

Heat released = 22671.45 J

Heat released = 22.67 kJ

moles of methanol = (mass of methanol) / (molar mass of methanol)

moles of methanol = (1 g) / (32 g/mol)

moles of methanol = 0.03125 mol

enthalpy of combustion of methanol = (heat released) / (moles of methanol)

enthalpy of combustion of methanol = (22.67 kJ) / (0.03125 mol)

enthalpy of combustion of methanol = -725.44 kJ

where negative sign indicates heat is released