Suppose that 0.650 mol of methane, CH4(g), is reacted with 0.800 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?
Substance & Enthalpy of Form. (kJ/mol)
C(g) 718.4
CF4(g) -679.9
CH4(g) -74.8
H(g) 217.94
HF(g) -268.61
I have figured out the equation:
CH4(g)+4F2(g) -> CF4(g)+4HF(g)
I also figured out the enthalpy of reaction:
Delta H = -1679.54kJ
I realize that I must figure out which of the reactants are limiting. How do I do that? What do I do after I find the limiting reactant?
because CH4 and 4F_2 ratio is 1:4
Well, finding the limiting reactant is like finding the star of a clown show - the one who steals the spotlight. It's all about the moles, my friend.
To determine the limiting reactant, you need to compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation. In this case, you have 0.650 mol of methane and 0.800 mol of fluorine.
For the methane, the stoichiometric coefficient is 1, while for the fluorine, it's 4. So, you need to multiply the number of moles of methane by 4 to compare them equally.
0.650 mol CH4 * 4 = 2.6 mol CH4
Now you can see that you have 2.6 mol of CH4 and 0.800 mol of F2. Since you need 1 mol of F2 for every 1 mol of CH4, it's clear that 0.800 mol of F2 is actually the limiting reactant.
After identifying the limiting reactant, it's time to do the math and calculate the amount of heat released. The enthalpy change is given as -1679.54 kJ.
Since the stoichiometric coefficient of F2 is 4, you need to multiply the enthalpy change by 4 to account for the molar ratio.
-1679.54 kJ * 4 = -6718.16 kJ
So, approximately -6718.16 kJ of heat is released. And that, my friend, is how the chemistry circus performs its fiery tricks!
To determine the limiting reactant, you need to compare the number of moles of the reactants involved in the reaction. In this case, the reactants are methane (CH4) and fluorine (F2).
First, calculate the number of moles of methane and fluorine. Given that you have 0.650 mol of methane and 0.800 mol of fluorine, you can use the stoichiometry of the balanced equation to find out how many moles of CF4 and HF are formed.
From the balanced equation:
1 mole of CH4 reacts with 4 moles of F2 to produce 1 mole of CF4 and 4 moles of HF.
So,
Number of moles of CF4 = 0.650 mol of CH4 × (1 mol of CF4 / 1 mol of CH4) = 0.650 mol
Number of moles of HF = 0.650 mol of CH4 × (4 mol of HF / 1 mol of CH4) = 2.600 mol
Now, you can determine the limiting reactant by comparing the moles of each reactant and considering the stoichiometry of the reaction. The reactant that is completely consumed is the limiting reactant.
From the stoichiometry:
1 mole of CH4 is required to produce 1 mole of CF4
4 moles of F2 is required to produce 1 mole of CF4
So, if you have 0.650 mol of CH4, you would need 4 × 0.650 = 2.600 mol of F2 to fully react.
Given that you have 0.800 mol of F2, which is more than the required 2.600 mol, CH4 is the limiting reactant.
After identifying the limiting reactant, you can use the balanced equation and the enthalpies of formation to calculate the amount of heat released in the reaction.
From the balanced equation:
1 mole of CH4 reacts to release -1679.54 kJ of heat.
Since the balanced equation shows a 1:1 mole ratio between CH4 and heat released, the amount of heat released can be calculated as follows:
Heat released = -1679.54 kJ/mol × 0.650 mol of CH4 = -1091.71 kJ
Therefore, in this reaction, approximately -1091.71 kJ of heat is released.
Use the coefficients in the balanced equation to convert moles CH4 to a product (say CF4) and to convert moles F2 to moles CH4. The smaller number of moles produced is the correct value and the reagent producing that value is the limiting reagent. For example, for 0.650 moles CH4, converted to moles CF4 is
0.650 moles CH4 x (1 mole CF4/1 mole CH4) = 0.650 x (1/1) = 0.650 moles CF4 produced if we started with 0.650 moles CH4 and had all of the F2 we needed.
After identifying the limiting reagent, then heat released is
q = delta Hrxn x (moles of limiting reagent/4 = ??