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# Chem ocr f322 bond enthalpy calculations Watch

1. From the following equations and enthalpy changes, calculate the standard enthalpy of formation of but-1-yne, C4H4(g).

C(s)+O2(g) =>CO2(g) - ΔH = -394kJmol-1

H2(g)+1/2 O2(g) => H2O - ΔH=-286kJmol-1

C4H4(g)+5O2(g) => 4CO2(g) + 2H2O(l) ΔH=-2597kJmol-1

Can anyone help?
2. (Original post by bloomingblossoms)
From the following equations and enthalpy changes, calculate the standard enthalpy of formation of but-1-yne, C4H4(g).

C(s)+O2(g) =>CO2(g) - ΔH = -394kJmol-1

H2(g)+1/2 O2(g) => H2O - ΔH=-286kJmol-1

C4H4(g)+5O2(g) => 4CO2(g) + 2H2O(l) ΔH=-2597kJmol-1

Can anyone help?
Alright, so you're given the other two equations to aid you to answer the question. It's very rare they put information in the question for no reason.

For most people, I find that drawing the enthalpy cycle is easier. The arrows will point upwards towards your products/reactants.

Essentially these problems are rearranging equations, much like you would in Maths. If we use the information given, we can equate the enthalpy changes to your third equation:

If ΔH = -2597, then, according to Hess' law, the equations would be, in the case, (enthalpy change of but-1-yne + enthalpy change of O2, + ΔH) = (enthalpy change of CO2 + enthalpy change of H2O)

So, simplified:

Δf(but-1-yne) + 0 + -2597 => 4(-394) + 2(-286)
NOTE: 0 for O2 because all elements have a 0 formation value in their natural states.

We can thus rearrange to: Δf (but-1-yne) = -2148 + 2597 (adding 2597 to both sides).
Hence I think the Δf of but-1-yne should be 449kJmol.
3. Thanks so much!

Updated: May 3, 2012
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