Question

Determine the enthalpy of this reaction

2 CH4(g) + 3 O2(g) → 2 CO(g) + 4 H2O(l) ∆Hrxn = ?
Given:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
∆Hrxn = −802 kJ
2 CO(g) + O2(g) → 2 CO2(g) ∆Hrxn = −566 kJ

Answer 1

To calculate the enthalpy of the reaction, the enthalpy changes of the given reactions are used by reversing the second reaction and doubling the first. The sum of the enthalpy changes equals the enthalpy for the reaction of interest, which is -1038 kJ.

Step-by-step explanation:

To determine the enthalpy of the reaction 2 CH4(g) + 3 O2(g) → 2 CO(g) + 4 H2O(l), we need to use the given reactions and their enthalpies of reaction.

First, we have:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

ΔHrxn = −802 kJ

Secondly, we have:
2 CO(g) + O2(g) → 2 CO2(g)

ΔHrxn = −566 kJ

By reversing the second reaction, we get:
2 CO2(g) → 2 CO(g) + O2(g)

ΔH = +566 kJ

To obtain CO(g) in the products and cancel out CO2, we can combine the reversed form of the second reaction with twice the first reaction, because we need 2 moles of CH4 for the desired reaction.

So, enthalpy change for 2 moles of CH4 will be 2 × (−802 kJ) = −1604 kJ.

Then the total enthalpy change for the desired reaction will be the sum of −1604 kJ and +566 kJ, which gives −1038 kJ.

Therefore, the enthalpy of the reaction 2 CH4(g) + 3 O2(g) → 2 CO(g) + 4 H2O(l) is −1038 kJ.