CH4 + 4F2 ==> CF4 + 4HF
dHrxn = (n*dHf products) - (n*dHf reactants)
dHrxn = [(1*-679.9) + (4*-268.61)] - [(1*-74.8) + (4*0)]
Solve for dHrxn in kJ for 1 mol
Then determine the limiting reagent (which I think is F2 but you should confirm that), determine the amount of CH4 used (Ithink that's 0.1 mol), then correct dHrxn you found above from 1 mol to mols CH4 used in the reaction.
Please help
Suppose that 0.250mol of methane, CH4(g), is reacted with 0.400mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole products. Assuming that the reaction occurs at constant pressure, how much heat is released?
Substance ΔH∘f (kJ/mol)
C(g) 718.4
CF4(g) −679.9
CH4(g) −74.8
H(g) 217.94
HF(g) −268.61
3 answers
I ended up with the wrong answer can I ask what I did wrong?
So I got the dHrxn which I found to be -1,679.54
I then determined the limited reagent was indeed F2 and that it produced .1 mol of CF4
I then divided -1,679.54 by .1 to get -16,795.4 and converted it to kilojoules to be 16.8 (sig figs) but it was not right.
So I got the dHrxn which I found to be -1,679.54
I then determined the limited reagent was indeed F2 and that it produced .1 mol of CF4
I then divided -1,679.54 by .1 to get -16,795.4 and converted it to kilojoules to be 16.8 (sig figs) but it was not right.
The answer is -168kJ
3 answers