Using table 9.4, calculate an approximate enthalpy (in kj) for the reaction of 1.02 g gaseous methanol (ch3oh) in excess molecular oxygen to form gaseous carbon dioxide and gaseous water. (hint, remember to first write the balanced equation.)

Given:

Mass of methanol = 1.02 g

To determine:

Enthalpy for the reaction of 1.02 g of methanol with excess O2

Explanation:

Balanced equation-

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

The reaction enthalpy is given as:

ΔHrxn = ∑nH°f (products) - ∑nH°f (reactants)

where n = number of moles

H°f = standard enthalpy of formation.

ΔHrxn = [2H°f (CO2 (g)) + 4H°f (H2O (g)) ] - [2H°f (CH3OH (g)) + 3H°f (O2 (g)) ]

= [2 (-393.5) + 4 (-241.8) ]-[2 (-201.5) + 3 (0) ] = - 1351.2 kJ

Now, 1 mole of CH3OH = 32 g

The calculated ΔHrxn corresponds to 2 moles of CH3OH. i. e.

The enthalpy change for 64 g of Ch3OH = - 1351.2 kJ

Therefore, for 1.02g gaseous methanol we have:

ΔH = 1.02 * - 1351.2/64 = - 21.5 kJ

Ans: The enthalpy for the given reaction is - 21.5 kJ