Question

1. Methane (CH4) combusts as shown in the equation below. If 3 moles of methane were used in the combustion, what would be the change in enthalpy (H) for the reaction? CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) H = 802 kJ

Answer 1

Answer:

The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ

Explanation:

Step 1: The balanced equation

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)   ΔH = -802 kJ

Step 2: Given data

We notice that for 1 mole of methane (CH4), we need 2 moles of O2 to produce : 1 mole of CO2 and 2 moles of H20.

The enthalpy change of combustion, given here as  Δ H , tells us how much heat is either absorbed or released by the combustion of one mole of a substance.

In this case: we notice that the combustion of 1 mole of methane gives off (because of the negative number),  802.3 kJ  of heat.

Step 3: calculate the enthalpy change  for 3 moles

The -802 kj is the enthalpy change for 1 mole

The change in enthalpy for 3 moles = 3* -802 kJ = -2406 kJ

The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ