Suppose that 0.490 mol of methane, CH4(g), is reacted with 0.640 mol 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?
CH4 + 4F2 ==> CF4 + 4HF
Determine the limiting reagent (LR).
mols CH4 = 0.490
mols F = 0.640
F must be the LR.
Then
dHo rxn = (n*dHo products) = (n*dHo reactants) and that will be for the reaction as written. You didn't have 4 mols F2, you had only 0.640. Correct for that. Post your work if you get stuck AND post the dHo values you used.
To determine how much heat is released during the reaction, we need to use the balanced chemical equation and the molar enthalpies of formation for the reactants and products.
The balanced chemical equation for the reaction is:
CH4(g) + 2F2(g) -> CF4(g) + 2HF(g)
The molar enthalpy of formation (∆Hf) for methane (CH4), fluorine (F2), carbon tetrafluoride (CF4), and hydrogen fluoride (HF) are as follows:
∆Hf(CH4) = -74.8 kJ/mol
∆Hf(F2) = 0 kJ/mol
∆Hf(CF4) = -94.6 kJ/mol
∆Hf(HF) = -271.0 kJ/mol
Let's calculate the heat released step-by-step below:
1. Calculate the moles of reactants:
Moles of CH4 = 0.490 mol
Moles of F2 = 0.640 mol
2. Determine the limiting reactant (the reactant that is completely consumed):
From the balanced equation, we can see that the stoichiometry is 1:2 for CH4:F2. Since we have more moles of F2 than CH4, CH4 is the limiting reactant.
3. Calculate the moles of products formed:
From the stoichiometry of the balanced equation, we can determine the moles of products formed:
Moles of CF4 = 0.490 mol (CH4) x (1 mol CF4 / 1 mol CH4) = 0.490 mol
Moles of HF = 0.490 mol (CH4) x (2 mol HF / 1 mol CH4) = 0.980 mol
4. Calculate the heat released for each product:
Heat released for CF4 = Moles of CF4 x ∆Hf(CF4)
Heat released for HF = Moles of HF x ∆Hf(HF)
Heat released for CF4 = 0.490 mol x (-94.6 kJ/mol) = -46.294 kJ
Heat released for HF = 0.980 mol x (-271.0 kJ/mol) = -265.58 kJ
5. Calculate the total heat released:
Total heat released = Heat released for CF4 + Heat released for HF
Total heat released = (-46.294 kJ) + (-265.58 kJ) = -311.874 kJ
Therefore, the heat released during the reaction is approximately -311.874 kJ.
To determine the amount of heat released in this reaction, we need to use the concept of enthalpy change (∆H) and the stoichiometry of the reaction. The balanced chemical equation for the reaction is:
CH4(g) + 2F2(g) -> CF4(g) + 4HF(g)
The stoichiometric coefficients tell us that 1 mole of methane reacts with 2 moles of fluorine to produce 1 mole of carbon tetrafluoride and 4 moles of hydrogen fluoride.
Now, we need to look up the standard enthalpy change (∆H°) values for each of the reactions involved. The given standard enthalpy changes are as follows:
∆H°f(CH4) = -74.80 kJ/mol
∆H°f(F2) = 0 kJ/mol
∆H°f(CF4) = -95.44 kJ/mol
∆H°f(HF) = -273.30 kJ/mol
Since we are given the amount of methane (0.490 mol) and fluorine (0.640 mol), we can calculate the moles of product formed using the stoichiometric ratios.
According to the stoichiometry,
1 mole of methane produces 1 mole of CF4, and
1 mole of methane produces 4 moles of HF.
So, using the number of moles of methane, we have:
0.490 mol CH4 x (1 mol CF4 / 1 mol CH4) = 0.490 mol CF4
0.490 mol CH4 x (4 mol HF / 1 mol CH4) = 1.96 mol HF
Now that we have the moles of each product, we can calculate the total heat released using the equation:
∆H = ∑(∆H°f(products)) - ∑(∆H°f(reactants))
Calculating the heat released:
∆H = [∆H°f(CF4) + ∆H°f(HF)] - [∆H°f(CH4) + 2∆H°f(F2)]
= [-95.44 kJ/mol + (-273.30 kJ/mol)] - [-74.80 kJ/mol + 2(0 kJ/mol)]
= -368.74 kJ/mol + 74.80 kJ/mol
= -293.94 kJ/mol
So, for the given amount of reactants, the reaction releases -293.94 kJ of heat. Note that the negative sign indicates that the reaction is exothermic (releasing heat).