Suppose that 0.650mol of methane, CH4 (g), is reacted with 0.800mol 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?
This is a combination limiting reagent problem as well as a heat production problem. Do it in steps. First the limiting reagent and the amount of product formed.
Write and balance the equation.
CH4 + 4F2 ==> CF4 + 4HF
1. Convert 0.650 mol CH4 to mols CF4. Use the coefficients in the balanced equation to do it.
0.650 mol CH4 x (1 mol CF4/1 mol CH4) = 0.650 mols CF4 produced.
Convert 0.800 mol F2 to mols CF4.
0.800 mol F2 x (1 mol CF4/4 mol F2) = 0.800 x 1/4 = 0.200
In limiting reagent problem you are likely to obtain values for the product that are different (as is this case) so one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Thus F2 is the limiting reagent at 0200 mol.
Next we must determine how much of the CH4 reacted and how much HF is produced. That is done with coefficients, too.
0.800 mol F2 x (1 mol CH4/4 mol F2) = 0.800 x 1/4 = 0.200 mol CH4 used.
We know 0.200 mol CF4 is produced; therefore, we must have produced 4 x 0.200 = 0.800 mol HF.
Now we plug data into the delta H rxn using delta Ho formation values that you probably can find in your text.
dHrxn = (mols*dH products) - (mols *dH reactants). dHrxn is the heat produced by that reaction.
To calculate the amount of heat released in the reaction, we need to find the change in enthalpy (∆H) of the reaction.
First, let's write the balanced chemical equation for the reaction:
CH4 (g) + 2F2 (g) -> CF4 (g) + 2HF (g)
According to the stoichiometry of the reaction, 1 mole of methane reacts with 2 moles of fluorine to produce 1 mole of CF4 and 2 moles of HF.
Next, we need to find the enthalpy change for each of the products and reactants. The enthalpy change for each substance is usually given in a standard enthalpy of formation table.
Using the given data:
∆Hf(CH4) = -74.8 kJ/mol
∆Hf(F2) = 0 kJ/mol
∆Hf(CF4) = -95.3 kJ/mol
∆Hf(HF) = -271.0 kJ/mol
The enthalpy change for the reaction can be calculated using the formula:
∆H = (∑∆Hf(products)) - (∑∆Hf(reactants))
Substituting the values:
∆H = [∆Hf(CF4) + 2∆Hf(HF)] - [∆Hf(CH4) + 2∆Hf(F2)]
∆H = [(-95.3 kJ/mol) + 2(-271.0 kJ/mol)] - [(-74.8 kJ/mol) + 2(0 kJ/mol)]
∆H = -637.3 kJ/mol - (-74.8 kJ/mol)
∆H = -562.5 kJ/mol
Since the coefficients in the balanced equation represent mole ratios, we can directly correlate the heat change to the number of moles used in the reaction.
In this case, we have 0.650 mol of methane reacting, so the total heat released can be calculated using the formula:
Heat released = ∆H x number of moles of CH4
Substituting the values:
Heat released = -562.5 kJ/mol x (0.650 mol)
Heat released = -365.6 kJ
Therefore, the heat released in the reaction is approximately -365.6 kJ. Note that the negative sign indicates that heat is released (exothermic reaction).
To determine how much heat is released in this reaction, we need to use the concept of enthalpy change (ΔH), specifically the concept of heat of reaction or heat of formation.
The balanced chemical equation for the reaction is:
CH4 (g) + 2F2 (g) → CF4 (g) + 2HF (g)
Now, we need to find the ΔH values for the formation of CF4 and HF. These values can be found in a table or reference book. Let's assume the values are as follows:
ΔHf for CF4 = -74.6 kJ/mol
ΔHf for HF = -272.3 kJ/mol
Next, we need to calculate the energy change for the reaction using the ΔH values. Since the reaction produces 1 mol of CF4 and 2 mol of HF for every 1 mol of CH4, we can calculate the energy change as follows:
Energy change = (ΔHf of products) - (ΔHf of reactants)
Energy change = [(-74.6 kJ/mol) + 2(-272.3 kJ/mol)] - [0 kJ/mol + 0 kJ/mol]
Energy change = -74.6 kJ/mol - 2(272.3 kJ/mol)
Energy change = -74.6 kJ/mol - 544.6 kJ/mol
Energy change = -619.2 kJ/mol
Now, we can calculate the heat released for the given amounts of reactants by using the stoichiometry of the reaction:
Molar ratio of CH4 to ΔH = 1 mol CH4 reacts with -619.2 kJ
Therefore, 0.650 mol of CH4 would release:
Heat released = (0.650 mol CH4) * (-619.2 kJ/mol)
Heat released = -401.88 kJ
So, the reaction releases approximately 401.88 kJ of heat.