Question

Calculate the standard enthalpy of formation of C2H6(g) given the following information: C(s) + O2(g) → CO2(g); ΔH° = -393.5 kJ/mol H2(g) + 1/2 O2(g) → H2O(l); ΔH° = -285.8 kJ/mol 2C(s) + 3H2(g) → C2H6(g); ΔH° = ?

Answer 1

To calculate the standard enthalpy of formation of C2H6 g , we need to use Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of its individual steps.First, we need to manipulate the given reactions to match the target reaction:Target reaction: 2C s  + 3H2 g   C2H6 g 1. 2 C s  + O2 g   CO2 g  ; H = 2 -393.5 kJ/mol  = -787 kJ/mol2. 3 H2 g  + 1/2 O2 g   H2O l  ; H = 3 -285.8 kJ/mol  = -857.4 kJ/mol3. 2CO2 g  + 3H2O l   C2H6 g  + 4O2 g ; H = ?Now, we can add reactions 1 and 2 and reverse reaction 3 to obtain the target reaction:2C s  + 2O2 g  + 3H2 g  + 3/2 O2 g   2CO2 g  + 3H2O l  + C2H6 g  + 4O2 g Simplifying, we get:2C s  + 3H2 g   C2H6 g Now, we can use Hess's Law to find the enthalpy change for the target reaction:H target  = H 1  + H 2  - H 3 H target  = -787 kJ/mol +  -857.4 kJ/mol  - H 3 H 3  = -787 kJ/mol -  -857.4 kJ/mol H 3  = 70.4 kJ/molSo, the standard enthalpy of formation of C2H6 g  is 70.4 kJ/mol.