Question

Calculate the enthalpy change for the following combustion reaction:C2H6(g) + 7/2 O2(g) --> 2CO2(g) + 3H2O(l)Using the following reactions and their corresponding enthalpies of formation:C2H6(g) + 3 1/2 O2(g) --> 2CO2(g) + 3H2O(l) ΔH°f = -1560 kJ/mol2H2(g) + O2(g) --> 2H2O(l) ΔH°f = -484 kJ/mol.

Answer 1

To calculate the enthalpy change for the combustion reaction, we can use Hess's Law, which states that the total enthalpy change for a reaction is the same regardless of the path taken. We can manipulate the given reactions to match the target reaction and then add their enthalpies.First, let's look at the target reaction:C2H6 g  + 7/2 O2 g  --> 2CO2 g  + 3H2O l Now, let's manipulate the given reactions:1  C2H6 g  + 3 1/2 O2 g  --> 2CO2 g  + 3H2O l  Hf = -1560 kJ/molThis reaction is already part of the target reaction, so we don't need to change it.2  2H2 g  + O2 g  --> 2H2O l  Hf = -484 kJ/molThis reaction is not needed for the target reaction, so we can ignore it.Now, let's add the manipulated reactions:C2H6 g  + 3 1/2 O2 g  --> 2CO2 g  + 3H2O l  Hf = -1560 kJ/molThe manipulated reactions match the target reaction, so we can add their enthalpies:H = -1560 kJ/molTherefore, the enthalpy change for the combustion reaction is -1560 kJ/mol.