Question

Calculate ΔHrxnΔHrxn for the following reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH4(g)+2O2(g)→CO2(g)+2H2O(l) Use the following reactions and given ΔHΔH values. CH4(g)+O2(g)→CH2O(g)+H2O(g)CH4(g)+O2(g)→CH2O(g)+H2O(g), ΔH=−ΔH=−284 kJkJ CH2O(g)+O2(g)→CO2(g)+H2O(g)CH2O(g)+O2(g)→CO2(g)+H2O(g), ΔH=−ΔH=−527 kJkJ H2O(l)→H2O(g)H2O(l)→H2O(g), ΔH=ΔH= 44.0 kJ

Answer 1

By using Hess's Law and manipulating the given reactions to yield the target reaction, we find that the ∆Hrxn for the combustion of methane is -899 kJ.

Step-by-step explanation:

To calculate the ∆Hrxn for the combustion of methane (CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)), we can use Hess's law and the given reactions. We have the following equations with their associated enthalpy changes:

CH4(g) + O2(g) → CH2O(g) + H2O(g), ∆H = -284 kJ

CH2O(g) + O2(g) → CO2(g) + H2O(g), ∆H = -527 kJ

H2O(l) → H2O(g), ∆H = 44 kJ

To obtain the target reaction, we manipulate the given reactions:

Reverse the third reaction: H2O(g) → H2O(l), ∆H = -44 kJ (since we want water in the liquid state).

The first equation remains unchanged.

Add the reverse of the third reaction to the first equation twice since there are two moles of H2O.

Add the second equation to the total to get the target reaction.

Combining the enthalpy changes, we get:

∆Hrxn = (-284 kJ) + 2(-44 kJ) + (-527 kJ) = -899 kJ