Calculate the enthalpy for the following reaction (using the Standard Enthalpy of Formation Table): 2F2(g) + 2H2O(l) --- 4HF(g) + O2(g)

Answer 1

Hi Rayan,

You'll need to look up the standard enthalpies of formation (_^H_f^o)for your reactants and products. The standard enthapy of formation for a substance is the enthalpy change involved in forming that substance from its elements under standard conditions (1 atm and 25^oC). By the way, that means the standard enthalpy of formation of any element in its standard state is zero, since it is already formed.

Now, the values for _^H_f^o are in units of kJ/mol, so you'll have to multiply each substance by the number of moles (given by the coeffient in the balanced equation). I looked up the values for HF(g) and H₂O(l) and got -273.3 kJ/mol and -285.8 kJ/mol, respectively.

Finally, you can set up the calculation using the important concept that the change in enthalpy for a reaction is the sum of the enthalpies of the products minus the sum of the enthalpies for the products. Using your equation, the calculation will look like this:

_^H_f^o = [4 (HF) + 1 (O₂)] - [2(F2) + 2(H₂O)]

Substituting, we get:

_^H_f^o = [4(-273.3) + 1(0.0)] - [2(0.0) + 2(-285.8)]

_^H_f^o = [-1093.2 kJ] - [-571.6 kJ] = -521.6 kJ

If you'd like some more specific chemistry help, I am here to help. Best wishes to you, Rayan!