El calor de formación molar (también llamado entalpía de formación estándar) de un compuesto (Δh f ) es igual a su cambio de entalpía (Δh) cuando se forma un mol de compuesto a 25 ° C y 1 atm de los elementos en su forma estable. necesita conocer los valores de calor de formación para calcular la entalpía y para otros problemas de termoquímica.
Esta es una tabla de los calores de formación para una variedad de compuestos comunes.
Como puede ver, la mayoría de los calores de formación son cantidades negativas, lo que implica que la formación de un compuesto a partir de sus elementos generalmente es un proceso exotérmico .
tabla de calores de formación
| compuesto | Δh f (kj / mol) | compuesto | Δh f (kj / mol) |
| agbr (s) | -99,5 | c 2 h 2 (g) | +226,7 |
| agcl (s) | -127,0 | c 2 h 4 (g) | +52.3 |
| agi (s) | -62,4 | c 2 h 6 (g) | -84,7 |
| ag 2 o (s) | -30,6 | c 3 h 8 (g) | -103,8 |
| ag 2 s (s) | -31,8 | nc 4 h 10 (g) | -124,7 |
| al 2 o 3 (s) | -1669,8 | nc 5 h 12 (l) | -173,1 |
| bacl 2 (s) | -860,1 | c 2 h 5 oh (l) | -277,6 |
| baco 3 (s) | -1218,8 | coo (s) | -239,3 |
| bao (s) | -558,1 | cr 2 o 3 (s) | -1128.4 |
| baso 4 (s) | -1465,2 | cuo (s) | -155,2 |
| cacl 2 (s) | -795,0 | cu 2 o (s) | -166,7 |
| caco 3 | -1207.0 | ser tío) | -48,5 |
| cao (s) | -635,5 | cuso 4 (s) | -769,9 |
| ca (oh) 2 (s) | -986,6 | fe 2 o 3 (s) | -822,2 |
| caso 4 (s) | -1432,7 | fe 3 o 4 (s) | -1120,9 |
| ccl 4 (l) | -139,5 | hbr (g) | -36,2 |
| ch 4 (g) | -74,8 | hcl (g) | -92,3 |
| chcl 3 (l) | -131,8 | hf (g) | -268,6 |
| ch 3 oh (l) | -238,6 | hola (g) | +25,9 |
| diente) | -110.5 | hno 3 (l) | -173,2 |
| co 2 (g) | -393,5 | h 2 o (g) | -241,8 |
| h 2 o (l) | -285,8 | nh 4 cl (s) | -315,4 |
| h 2 o 2 (l) | -187,6 | nh 4 no 3 (s) | -365,1 |
| h 2 s (g) | -20,1 | no (g) | +90,4 |
| h 2 so 4 (l) | -811,3 | no 2 (g) | +33,9 |
| hgo (s) | -90,7 | nio (s) | -244,3 |
| hgs (s) | -58,2 | pbbr 2 (s) | -277,0 |
| kbr (s) | -392,2 | pbcl 2 (s) | -359,2 |
| kcl (s) | -435,9 | pbo (s) | -217,9 |
| kclo 3 (s) | -391,4 | pbo 2 (s) | -276,6 |
| kf (s) | -562,6 | pb 3 o 4 (s) | -734,7 |
| mgcl 2 (s) | -641,8 | pcl 3 (g) | -306,4 |
| mgco 3 (s) | -1113 | pcl 5 (g) | -398,9 |
| mgo (s) | -601,8 | sio 2 (s) | -859,4 |
| mg (oh) 2 (s) | -924,7 | sncl 2 (s) | -349,8 |
| mgso 4 (s) | -1278,2 | sncl 4 (l) | -545,2 |
| mno (s) | -384,9 | sno (s) | -286,2 |
| mno 2 (s) | -519,7 | sno2(s) | -580.7 |
| nacl(s) | -411.0 | so2(g) | -296.1 |
| naf(s) | -569.0 | so3(g) | -395.2 |
| naoh(s) | -426.7 | zno(s) | -348.0 |
| nh3(g) | -46.2 | zns(s) | -202.9 |
reference: masterton, slowinski, stanitski, chemical principles, cbs college publishing, 1983.
points to remember for enthalpy calculations
when using this heat of formation table for enthalpy calculations, remember the following:
- calculate the change in enthalpy for a reaction using the heat of formation values of the reactants and products.
- the enthalpy of an element in its standard state is zero. however, allotropes of an element not in the standard state typically do have enthalpy values. for example, the enthalpy values of o2 is zero, but there are values for singlet oxygen and ozone. the enthalpy of solid aluminum, beryllium, gold, and copper are zero. the vapor phases of these metal have enthalpy values.
- when you reverse the direction of a chemical reaction, the magnitude of Δh is the same, but the sign changes.
- when you multiply a balanced equation for a chemical reaction by an integer value, the value of Δh for that reaction must be multiplied by the integer, too.
sample heat of formation problem
for example, heat of formation values are used to find the heat of reaction for acetylene combustion:
2c2h2(g) + 5o2(g) → 4co2(g) + 2h2o(g)
1) check to make sure the equation is balanced.
you'll be unable to calculate enthalpy change if the equation isn't balanced. if you're unable to get a correct answer to a problem, it's a good idea to check the equation. there are many free online equation balancing programs that can check your work.
2) use standard heats of formation for the products:
Δhºf co2 = -393.5 kj/mole
Δhºf h2o = -241.8 kj/mole
3) multiply these values by the stoichiometric coefficient.
in this case, the value is 4 for carbon dioxide and 2 for water, based on the numbers of moles in the balanced equation:
vpΔhºf co2 = 4 mol (-393.5 kj/mole) = -1574 kj
vpΔhºf h2o = 2 mol ( -241.8 kj/mole) = -483.6 kj
4) add the values to get the sum of the products.
sum of products (Σ vpΔhºf(products)) = (-1574 kj) + (-483.6 kj) = -2057.6 kj
5) find enthalpies of the reactants.
as with the products, use the standard heat of formation values from the table, multiply each by the stoichiometric coefficient, and add them together to get the sum of the reactants.
Δhºf c2h2 = +227 kj/mole
vpΔhºf c2h2 = 2 mol (+227 kj/mole) = +454 kj
Δhºf o2 = 0.00 kj/mole
vpΔhºf o2 = 5 mol ( 0.00 kj/mole)= 0.00 kj
sum of reactants (Δ vrΔhºf(reactants)) = (+454 kj) + (0.00 kj) = +454 kj
6) calculate the heat of reaction by plugging the values into the formula:
Δhº = Δ vpΔhºf(products) - vrΔhºf(reactants)
Δhº = -2057.6 kj - 454 kj
Δhº = -2511.6 kj
finally, check the number of significant digits in your answer.