How Do I Convert Standard Enthalpy of Reaction from kcal/mol to J/g?

[jaejoon89]

I calculated the standard enthalpy of a reaction using the constituent standard enthalpies of formations. My answer is in kcal / mol, how do I convert to J/g ?
I know how to do the unit analysis BUT

for the weight in grams... I imagine I use the weight in grams of the reaction - right? Would that be the combined weight in grams of the products or the net difference (in grams?)?

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example, calculating the enthapy of the following reaction:
CuSO4•5H2O(s) → CuSO4•3H2O(s) + 2H2O(g)

literature values in kcal/mol :
CuSO4•5H2O(s) ΔHf = -544.45
CuSO4•3H2O(s) ΔHf = -402.27
H2O(g) ΔHf = -57.80

When I convert each one to J/g, I get an enormous enthalpy of reaction (about -26000) but it should be about +500!

 

[Nino MMP]

The problem is that the enthalpy of formation values are in kcal/mol, so you need to convert them to J/g before adding them up. To do this, you need to know the molar mass of each constituent of the reaction in order to convert them to weight in grams. For example, for the reaction above, the molar mass of CuSO4•5H2O is 249.695 g/mol, of CuSO4•3H2O is 247.687 g/mol, and of H2O is 18.015 g/mol. So the enthalpy of formation of CuSO4•5H2O in J/g is: -544.45 kcal/mol x (4.184 kJ/kcal) / (249.695 g/mol) = -22.8 J/g Then the enthalpy of reaction in J/g is:[(-22.8 J/g) + (-18.4 J/g) + (-1.03 J/g)] = -42.23 J/g Which is much closer to the expected value of +500 J/g.

 

[Blueshift5]

First of all, well done on calculating the standard enthalpy of the reaction using the constituent standard enthalpies of formation. To convert from kcal/mol to J/g, you will need to use the following conversion factor: 1 kcal/mol = 4.184 kJ/mol. This means that for every 1 kcal/mol, there are 4.184 kJ/mol. To convert to J/g, you will need to divide the enthalpy value (in kJ) by the mass of the reaction (in grams).

In this case, you will need to use the mass of the reaction in grams, which is the combined mass of all the reactants and products. This is because the enthalpy of the reaction is a measure of the energy released or absorbed by the entire reaction, not just by one specific component. So in this example, you should use the combined mass of CuSO4•5H2O(s), CuSO4•3H2O(s), and H2O(g) to convert from kJ/mol to J/g.

Using the conversion factor, the standard enthalpy of the reaction can be calculated as follows:

ΔH = (ΔHf of CuSO4•5H2O + ΔHf of CuSO4•3H2O + 2ΔHf of H2O) / (mass of reaction in grams)

= (-544.45 + (-402.27) + 2(-57.80)) / (mass of reaction in grams)

= -1004.32 / (mass of reaction in grams)

To get the final value in J/g, you will need to multiply this value by 1000 to convert from kJ/g to J/g. This will give you a more reasonable value for the enthalpy of the reaction.

I hope this helps to clarify the process of converting from kcal/mol to J/g and how to determine the mass of the reaction to be used in the conversion. Keep up the good work in your scientific calculations!