Chemistry help!

Calculate the ΔfH of ethane, C2H6(g), given the enthalpy change for the following reaction and the ΔfH of ethene which is +53KJ/Mol


C2H4(g)+H2(g) =C2H6(g) ΔH=-137Kj/Mol



I got 189Kj/Mol but I'm a bit unsure, please help ^^

Hi there,

There's a useful formula when finding enthalpy changes. It states that:

Δ_rH (reaction enthalpy) = sum of Δ_fH (formation enthalpies) of the products - sum of Δ_fH of the reactants.
(Note that if you are forming two moles of a particular product then you must do 2xΔ_fH in the formula, for example)

This can be derived using Hess's law by constructing a cycle where the reactants are first broken down into elements and then form the products.

In this question, you are given the reaction enthalpy and the enthalpy of formation of ethene. The enthalpy of formation of any element (H₂ in this case) is 0. So, the equation can be rearranged to find the formation enthalpy of ethane:

-137 = Δ_fH(ethane) - 53 => Δ_fH(ethane) = -84 kJ mol^-1

Hope this was helpful,

A.T.