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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 ^^
C2H4(g)+H2(g) =C2H6(g) ΔH=-137Kj/Mol
I got 189Kj/Mol but I'm a bit unsure, please help ^^
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(Original post by MagicalKiwi123)
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 ^^
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 ^^
There's a useful formula when finding enthalpy changes. It states that:
ΔrH (reaction enthalpy) = sum of ΔfH (formation enthalpies) of the products - sum of ΔfH of the reactants.
(Note that if you are forming two moles of a particular product then you must do 2xΔfH 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 = ΔfH(ethane) - 53 => ΔfH(ethane) = -84 kJ mol-1
Hope this was helpful,
A.T.
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(Original post by AThow)
Hi there,
There's a useful formula when finding enthalpy changes. It states that:
ΔrH (reaction enthalpy) = sum of ΔfH (formation enthalpies) of the products - sum of ΔfH of the reactants.
(Note that if you are forming two moles of a particular product then you must do 2xΔfH 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 = ΔfH(ethane) - 53 => ΔfH(ethane) = -84 kJ mol-1
Hope this was helpful,
A.T.
Hi there,
There's a useful formula when finding enthalpy changes. It states that:
ΔrH (reaction enthalpy) = sum of ΔfH (formation enthalpies) of the products - sum of ΔfH of the reactants.
(Note that if you are forming two moles of a particular product then you must do 2xΔfH 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 = ΔfH(ethane) - 53 => ΔfH(ethane) = -84 kJ mol-1
Hope this was helpful,
A.T.
This was a really helpful explanation ^^
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