Enter your answer in the provided box. From the following data, C (graphite) + O2 (g) → CO2 (g) ΔH o rxn = - 393.5 kJ/mol H2 (g) + 1 2 O2 (g) → H2O (l) ΔH o rxn = - 285.8 kJ/mol 2C2H6 (g) + 7O2 (g) → 4CO2 (g) + 6H2O (l) ΔH o rxn = - 3119.6 kJ/mol calculate the enthalpy change for the reaction below: 2 C (graphite) + 3H2 (g) → C2H6 (g)

final reaction to get is 2C (graphite) + H2 (g) → C2H6 (g)

First step: 2C (graphite) + 2O2 (g) → 2CO2 (g) ΔH = (-393.5 x 2)

3H2 (g) + 3/2O2 (g) → 3H2O (l) ΔH = (-285.8 x 3)

2CO2 (g) + 3H2O (l) → C2H6 (g) + 7/2O2 (g) ΔH = (-3119.6 X - 1/2)

Second step: 2C (graphite) + 2O2 (g) + 3H2 (g) + 3/2O2 (g) + 2CO2 (g) + 3H2O (l) → 2CO2 (g) + 3H2O (l) + C2H6 (g) + 7/2O2 (g)

Third step: 2C (graphite) + O2 (g) → C2H6 (g)

ΔH final = (-393.5 x 2) + (-285.8 x 3) + (-3119.6 x - 1/2) = - 84.6 KJ/mol

Explanation: Using Hess Law

In the first step, rearrange the 3 given reactions in a way that they match the final reaction by multiplying or dividing (also the ΔH) by a given number and by reverse (if so the multiply by - 1).

In the second step, add all 3 reactions with reactant on their side and product on theirs.

In 3rd step, cancel out similarity in the product and reactant (such as there are (2+3/2 = 7/2) O2 which cancels with 7/2O2 in the product).

finally, add all 3 ΔH to find the final one which is - 84.6 KJ/mol