Question

The enthalpies of formation of the compounds in the combustion of methane,

CH4 (9)+20,(g) → C02(g) +2H2O(g),

are CH4(g): AH = -74.6 kJ/mol; CO2(g): AH+ = -393.5 kJ/mol; and H2O(g): AHp= -241.82 kJ/mol.

How much heat is released by the combustion of 2 mol of methane?

Use AH xn=(AHi products) - (AHereactants).

-80.3 kJ

-802.5 kJ

O-1,605.1 kJ

O -6,420.3 kJ

Answer 1

-1605.1 kJ

Step-by-step explanation:

The enthalpy of formation is the enthalpy of the reaction that forms the substance only by its constituents, so, substances formed by one element, such as O2, in its ambient temperature phase, have an enthalpy of formation 0.

The enthalpy is a measure of how much heat the system contains, and so, the variation of it measures the heat lost (ΔH <0) or gained (ΔH >0), and for a reaction:

ΔHrxn = ∑ni*ΔHi products - ∑ni*Hi reactants

Where ni represents the coefficient of the substance, so by the data given:

ΔHrxn = [2*(-241.82) + (-393.5)] - [-74.6]

ΔHrxn = -802.54 kJ/ mol of CH4

Thus, the heat released is the enthalpy multiplied by the number of moles of CH4:

Q = -802.54*2

Q = -1605.1 kJ