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

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8