How to solve enthalpy of formation when energy produced is given

This is the question:

When ethanol burns in oxygen under standard conditions, carbon dioxide, water and 1368 kJmol-1 of energy are produced. Calculate the enthalpy of formation of ethanol, given that the enthalpies of formation of carbon dioxide and water are -393.7 and -285.9 kJmol-1 respectively.

Im confused as first I worked out the sum of the products and i got an answer of -1645.1kJmol-1. Im not able to figure out what the enthalpy of formation of the reactants would be as from the equation it needs to total to 1368 kJmol-1

Reply 1

Bookworm_88

Have you tried a Hess's cycle?

Reply 2

Fa123123

Original post by Bookworm_88

Have you tried a Hess's cycle?

I have tried to but im kinda confused on how to go about it

Reply 3

Hellllpppp

Original post by Fa123123

I have tried to but im kinda confused on how to go about it

Have you written the balanced equation for combustion of ethanol?

Reply 4

Fa123123

Original post by Hellllpppp

Have you written the balanced equation for combustion of ethanol?

Yes I did it as C2H5OH + 6 1/2 O2 --> 2CO2 + 3H2O

Reply 5

Hellllpppp

Original post by Fa123123

Yes I did it as C2H5OH + 6 1/2 O2 --> 2CO2 + 3H2O

Is that 6 x 1/2 for CO2

It won’t make a difference with the calculations of O2 is wrong but yeah it’s C2H5OH + 3 O2 --> 2CO2 + 3H2O

Do you remember the definition for enthalpy of formation?

Reply 6

Fa123123

Original post by Hellllpppp

Is that 6 x 1/2 for CO2

It won’t make a difference with the calculations of O2 is wrong but yeah it’s C2H5OH + 3 O2 --> 2CO2 + 3H2O

Do you remember the definition for enthalpy of formation?

yes and yes Standard enthalpy change of formation (ΔHfo) is the energy change when 1 mole of substance made from its elements in their standard state.

Reply 7

Hellllpppp

Original post by Fa123123

yes and yes Standard enthalpy change of formation (ΔHfo) is the energy change when 1 mole of substance made from its elements in their standard state.

Yeah so at the bottom of the Hess’ cycle put the elements in their standard state then draw arrows from the elements to the equation.

Reply 8

Fa123123

Original post by Hellllpppp

Yeah so at the bottom of the Hess’ cycle put the elements in their standard state then draw arrows from the elements to the equation.

This is what ive done

Reply 9

Hellllpppp

As it’s combustion the enthalpy change is always negative

I haven’t got a calculator on me so I’ll assume the calculation is right.

Follow the arrows and use Hess’ law

(edited 2 years ago)

Reply 10

gogrizz123

Original post by Fa123123

This is what ive done

Provided the calculation for formation of co2 and h2o is correct and assuming you mean -1368, formation of ethanol from its constituent elements is -286.1 kj/mol. You work this out by (-1654.1 + 1368) = -286.1

(edited 2 years ago)

Reply 11

Fa123123

Original post by Hellllpppp

As it’s combustion the enthalpy change is always negative

I haven’t got a calculator on me so I’ll assume the calculation is right.

Follow the arrows and use Hess’ law

yeah but i still need to figure out the sum of reactants as thats what the question is asking so how would i do that?

Reply 12

Hellllpppp

Original post by Fa123123

yeah but i still need to figure out the sum of reactants as thats what the question is asking so how would i do that?

What’s the enthalpy of formation of oxygen?

Reply 13

Fa123123

Original post by gogrizz123

Provided the calculation for formation of co2 and h2o is correct and assuming you mean -1368, formation of ethanol from its constituent elements is -286.1 kj/mol

thank you so much, how did you get to this?

Reply 14

gogrizz123

Original post by Fa123123

thank you so much, how did you get to this?

▲hF for ethanol = ▲Hf(co2+h20) - ▲H combustion of ethanol (but remember we are going against the arrow so you have to reverse the sign e.g turn a negative into a positive)

Reply 15

Hellllpppp

Original post by Fa123123

thank you so much, how did you get to this?

The enthalpy change is independent of the route taken the up arrow is -1645.1 by your maths and the arrow to the left is 1368

-1645.1 + 1368 = 1368 - 1645.1