Im really stuck here and i will apreciate any help.
How do i find the energy given out in the exothermic reaction of methane with oxygen?
Then how would i find the energy given out in the exothermic reation of hydrochloric acid and pottasium?
If you help me with those two i get the idea by comparing them...
Lastly,
To find the endothermic energy reqired to intiate a reaction do i just reverse what i did for the above two?
Thanks
Hi. for a reaction, its heat is given by:
Q=sum( ni*hi,f) of products-sum( ni*hi,f) of reactants
where "ni" are the moles of specimen "i" in the equation, and hi,f is the formation enthalpy of specimen "i" in the equation.
Some importan things!
I`m not sure about translation of words
, so I don`t know if the word "reactants" actually exists, but it means the elements that form the products (usually at left side of reaction equation). In spanish it would be "reactantes" if you know some spanish, or know some good dictionary.
In the same way, I don`t know if "formation enthalpy" is a correct translation. In spanish is "entalpía de formación". It is the enthalpy of the reaction that forms the specimen "i" from its most common 'reactants'.
Note that the heat of reaction is the same of enthalpy of reaction (but not 'formation enthalpy', because in general, you`re not forming only one product).
The sign of Q tells you if the reaction is an endothermic reaction or an exothermic reaction.
The 'formation enthalpies' that you need to get the answer are tabulated.
Pure molecules (like O2, H, He2) have 'formation enthalpy' zero.
That`s what I know about this subject. I hope it helps.
Alejandro
Q=sum( ni*hi,f) of products-sum( ni*hi,f) of reactants
where "ni" are the moles of specimen "i" in the equation, and hi,f is the formation enthalpy of specimen "i" in the equation.
Some importan things!
I`m not sure about translation of words
, so I don`t know if the word "reactants" actually exists, but it means the elements that form the products (usually at left side of reaction equation). In spanish it would be "reactantes" if you know some spanish, or know some good dictionary.In the same way, I don`t know if "formation enthalpy" is a correct translation. In spanish is "entalpía de formación". It is the enthalpy of the reaction that forms the specimen "i" from its most common 'reactants'.
Note that the heat of reaction is the same of enthalpy of reaction (but not 'formation enthalpy', because in general, you`re not forming only one product).
The sign of Q tells you if the reaction is an endothermic reaction or an exothermic reaction.
The 'formation enthalpies' that you need to get the answer are tabulated.
Pure molecules (like O2, H, He2) have 'formation enthalpy' zero.
That`s what I know about this subject. I hope it helps.
Alejandro
can you give me a worked out example please. Im not good at interpreting formulas i have not used before.
Sure. I`m going to write in [] the chemical compounds or elements, to make cleared the difference with moles.
The reaction is:
0.3[C3H8]+0.7[C4H10]+28.81{0.21[O2]+0.79[N2]} --> 3.7[CO2]+4.7[H20]+22.76[N2]
Some notes about this:
-This represent a perfect and stequiometric combustion of 1 mole of fuel (propane and butane) with 28.81 moles of air.
-Maybe you know that moles should be integers. Actually the relationship between species is the important thing. Of couse this doesn`t have necessarily physical meaning.
-Following the previous note, the heat of the reaction you`ll get, is not the same if you use other mole quantities. That`s because heat depends of mass, and more moles gives more mass for a given thing. However, the heat you`ll get here is proportional to the heat with other moles quantities, in the same proportion that moles change (that`s, heat is linear with mole quantity, since you`re multiplying both sides of reaction equation for the same quantity to change moles quantity).
Now the heat of reaction/enthalpy of reaction/energy of reaction:
Q=sum( ni*hi,f) of products-sum( ni*hi,f) of reactants
Here I will use the following notation: using as example the properties for N2
n[N2]=moles (number of moles) of N2
h,f[N2]=formation enthalpy of N2
This equation for the reaction`s heat is at 298 K of temperature , and 1 atm. of pressure. That`s because moles and enthalpies are given in these conditions (the combustion is supposed to be in these conditions)
The formation enthalpy are 0 for N2, O2, because they are diatomic compunds (the natural way the pure element is found, requires no energy to form it).
I got the other values of enthalpies of formation from a table.
So:
Q={n[CO2]h,f[CO2]+n[H20]h,f[H20]+n[N2]h,f[N2]}-{n[C3H8]h,f[C3H8]+n[C4H10]h,f[C4H10]+n[O2]h,f[O2]+n[N2]h,f[N2]}
Q={3.7*(-393546)+4.7*(-241845)+22.76*0}-{0.3*(-103847)+0.7*(-124733)+6.05*0+22.76*0}
Where enthalpies are given in J/mol.
Q=-2474324.5 J/mol
You can consider it as J/(mol of fuel), since the energy comes from the fuel.
Because the heat is negative, it`s leaving the system, thus the reaction is exothermical.
If you put this heat per unit of mass of fuel, you get the HHV, or high heat value of the fuel.
Alejandro
The reaction is:
0.3[C3H8]+0.7[C4H10]+28.81{0.21[O2]+0.79[N2]} --> 3.7[CO2]+4.7[H20]+22.76[N2]
Some notes about this:
-This represent a perfect and stequiometric combustion of 1 mole of fuel (propane and butane) with 28.81 moles of air.
-Maybe you know that moles should be integers. Actually the relationship between species is the important thing. Of couse this doesn`t have necessarily physical meaning.
-Following the previous note, the heat of the reaction you`ll get, is not the same if you use other mole quantities. That`s because heat depends of mass, and more moles gives more mass for a given thing. However, the heat you`ll get here is proportional to the heat with other moles quantities, in the same proportion that moles change (that`s, heat is linear with mole quantity, since you`re multiplying both sides of reaction equation for the same quantity to change moles quantity).
Now the heat of reaction/enthalpy of reaction/energy of reaction:
Q=sum( ni*hi,f) of products-sum( ni*hi,f) of reactants
Here I will use the following notation: using as example the properties for N2
n[N2]=moles (number of moles) of N2
h,f[N2]=formation enthalpy of N2
This equation for the reaction`s heat is at 298 K of temperature , and 1 atm. of pressure. That`s because moles and enthalpies are given in these conditions (the combustion is supposed to be in these conditions)
The formation enthalpy are 0 for N2, O2, because they are diatomic compunds (the natural way the pure element is found, requires no energy to form it).
I got the other values of enthalpies of formation from a table.
So:
Q={n[CO2]h,f[CO2]+n[H20]h,f[H20]+n[N2]h,f[N2]}-{n[C3H8]h,f[C3H8]+n[C4H10]h,f[C4H10]+n[O2]h,f[O2]+n[N2]h,f[N2]}
Q={3.7*(-393546)+4.7*(-241845)+22.76*0}-{0.3*(-103847)+0.7*(-124733)+6.05*0+22.76*0}
Where enthalpies are given in J/mol.
Q=-2474324.5 J/mol
You can consider it as J/(mol of fuel), since the energy comes from the fuel.
Because the heat is negative, it`s leaving the system, thus the reaction is exothermical.
If you put this heat per unit of mass of fuel, you get the HHV, or high heat value of the fuel.
Alejandro