# Enthelpy of Reactions - can someone help me real quick?

• sugarandspice

#### sugarandspice

I hate stoichiometry.

Here's the problem:

Let's look at an example of how delta H is found for a reaction. Find the enthalpy for the following reaction: 2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)

Step 1:

*

You can look up the enthalpy, H, for each of the products in a reaction
*

multiply by the number of moles of that substance,
*

add them up to get Hproducts.

A table of enthalpies for substances can be found in different chemical handbooks. This is a table of the heat needed to create these substances.

*

Watch the state (s,l,g) carefully. It makes a difference.
*

It would take more heat to create H2O(g) than H2O(l).

HCO2(g) = -393.5 kJ/mol x 4 mol = -1574 kJ

HH2O(g) = -241.8 kJ/mol x 6 mol = -1450.8 kJ
_____________________________________

Hproducts = -3024.8 kJ

Step 2:

*

Now look up the enthalpy, H, for each of the reactants
*

multiply by the number of moles of each
*

HC2H6(g) = -84.7 kJ/mol x 2 mol = -169.4 kJ

HO2(g) = 0 kJ/mol (All elements are zero as they take no heat to form.)

_____________________

Hreactants = -169.4 kJ

How do I figure out how many mols to multiply for each product and reactant? I've already been given the other part of the equation.

Thanks

The question should be worded "Complete combustion of" or its usually implied that the combustion goes to completion and that all the reactants are turned into products.

Stoichiometrically you must find what ratios (how many moles of each reactant is needed) to achieve this complete combustion, that is, no reactants are left unburned/unreacted.

Thats already provided in the starting reaction equation, but it is usually simple to work it out.

Usually a question like this sais, Ethane is burned in "excess" oxygen, this implies that you have as much oxygen as you need. The total amount of Carbon and Hydrogen after the reaction must be balenced with the amount before the reaction, you can't get matter from nowhere.

C2H6(g)
2x Carbon which when burned becomes Carbon Dioxide (CO2) requiring 2 O atoms ( or 1 O2 molecule, oxygen is naturally diatomic and found as paired Oxygen )
6x hydrogen which when burned becomes Water (H2O) requiring another Oxygen atom for every 2 Hydrogen.

this is written:
C2H6 + ?O2 -> 2CO2 + 3H2O

you can see that for it to balance you require 3.5 oxygen molecules to every 1 Ethane molecule, or 7 mol oxygen to 2 mol of Ethane.

When considering the energy involved in the reaction, you must take into account these quantities of reactants. The more you burn, the more heat you get out.

How do I figure out how many mols to multiply for each product and reactant? I've already been given the other part of the equation.

..
HCO2(g) = -393.5 kJ/mol x 4 mol = -1574 kJ
..etc

You multiplied it by the 4 moles you used in the reaction.

The overall energy transfer in the reaction is given by the dH of Products - dH of Reactants.

-3024.8 - (-169.4) = -3024.8 + 169.4 = -12855.4 kJ

But remember that you used 2 moles of C2H6, and therefor need to divide by 2.

-2855.4 kJ / 2 mol = -1427.9 kJ/mol

You can then say that the Hreaction for the combustion of Ethane in Oxygen is -1427.9 kJ/mol.

I hope i didnt confuse :s

3trQN said:

You multiplied it by the 4 moles you used in the reaction.

I hope i didnt confuse :s

Whoops! This is an example question! Sorry sorry - here's the real problem! This is what I'm trying to figure out how many mols to times by. :D

Balance the equation (Already did this) and Calculate the enthalpy change for the following reaction:

NH3(g) + O2(g) --> N2(g) + H2O(l)

NH3(g)= -46.11 kj/mol

O2(g) = 0 kJ/mol

N2 = 0 kJ/mol

H20= - 285.830 kJ/mol

ETA- Thankyou soooo much for taking the time to respond. It means so much to me.

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Balance the equation (Already did this) and Calculate the enthalpy change for the following reaction:

NH3(g) + O2(g) --> N2(g) + H2O(l)

Not quite, check again.

[you wrote]
Elements in Reactants:
N x 1 (mono atomic in NH3)
H x 3
O x 2
Elements in Products:
N x 2 ( youve gained Nitrogen from nowhere ) (Diatomic in N2)
H x 2 ( Youve lost Hydrogen somwhere )
O x 1 ( youve lost oxygen also )

This is like factoring numbers, you need to find a common factor between reactants/products that will allow you to make a complete reaction. To do this you mix the reactants in the correct ratios and *poof* you get the product.

3trQN said:
Not quite, check again.

[you wrote]
Elements in Reactants:
N x 1 (mono atomic in NH3)
H x 3
O x 2
Elements in Products:
N x 2 ( youve gained Nitrogen from nowhere ) (Diatomic in N2)
H x 2 ( Youve lost Hydrogen somwhere )
O x 1 ( youve lost oxygen also )

This is like factoring numbers, you need to find a common factor between reactants/products that will allow you to make a complete reaction. To do this you mix the reactants in the correct ratios and *poof* you get the product.

Ok! So here's the correct balanced equation:

4NH3 + 3O2 ----> 2N2 + 6H2O

Keeping in mind the problem:
Balance the equation and Calculate the enthalpy change for the reaction

Now, I already know this (I didn't figure this out, it was provided to me):

NH3(g)= -46.11 kj/mol

O2(g) = 0 kJ/mol

N2 = 0 kJ/mol

H20= - 285.830 kJ/mol

So now I need to figure out how many mols to times each amount by. This doesn't seem too tough so maybe I'm just missing something? How do I go about doing that?

------------

Also, the example problem I was given, from the original post, this is the answer: http://osd.flvs.net/webdav/educator_chemistry_v5/module5/imagmod5/5_16_e.gif
Just thought it might help.

Thanks SO much!

Last edited by a moderator:
i have to do this problem... (its in green)
Assignment Problem
Balance the equation and calculate the enthalpy change for the following reaction:

NH3(g) + O2(g) --> N2(g) + H2O(l)

NH3(g)= -46.11 kj/mol

O2(g) = 0 kJ/mol

N2 = 0 kJ/mol

H20= - 285.830 kJ/mol

and i don't get it at all. i think the balanced equation is 4NH3 + 3O2 --> 2N2 +6H2O, but I am not sure how to do the enthalpy change.

can anyone help me please?! :uhh: