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
The answer is -168kJ
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.
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.
Well, this sounds like a gas-tly situation!
To calculate the heat released in this reaction, we need to apply the concept of enthalpy change (ΔH). The enthalpy change of a reaction is the difference in the enthalpy of the products and the enthalpy of the reactants.
First, let's write down the balanced equation for the reaction:
CH4(g) + 2F2(g) -> CF4(g) + 2HF(g)
Now, let's calculate the enthalpy change (∆H) for the reaction:
ΔH = [∑H(products)] - [∑H(reactants)]
ΔH = [H(CF4)] + 2[H(HF)] - [H(CH4)] - 2[H(F2)]
ΔH = [-679.9] + 2[-268.61] - [-74.8] - 2[0]
ΔH = -679.9 + 2(-268.61) + 74.8
ΔH = -679.9 - 537.22 + 74.8
ΔH = -1142.32 kJ/mol
So, approximately -1142.32 kJ of heat is released in this reaction.
Now that we have the answer, don't go getting all heated up about it!
To calculate the amount of heat released in the reaction, we can use the concept of enthalpy change (ΔH). The enthalpy change for a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.
First, we need to determine the balanced chemical equation for the reaction:
CH4(g) + 2F2(g) -> CF4(g) + 2HF(g)
Using the given information, we can calculate the enthalpy change for the reaction as follows:
ΔH = [ΔHf(CF4) + 2ΔHf(HF)] - [ΔHf(CH4) + 2ΔHf(F2)]
Now, let's plug in the values for the enthalpies of formation:
ΔH = [(-679.9 kJ/mol) + 2(-268.61 kJ/mol)] - [(-74.8 kJ/mol) + 2(0 kJ/mol)]
ΔH = [-679.9 kJ/mol - 537.22 kJ/mol] - [-74.8 kJ/mol]
ΔH = -1217.12 kJ/mol + 74.8 kJ/mol
ΔH = -1142.32 kJ/mol
Therefore, the reaction releases 1142.32 kJ of heat.