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

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.