Question

± Enthalpy of Reaction: State and Stoichiometry

Use the data below to answer the questions.

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

Keep in mind that the enthalpy of formation of an element in its standard state is zero

Part D

Suppose that 0.290 mol of methane, CH4(g), is reacted with 0.440 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?

Express your answer to three significant figures and include the appropriate units

Answer 1

CH₄ + 4F₂ ---------> CF₄ + 4HF

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.

1. Convert 0.290 mol CH4 to mols CF4. Use the coefficients in the balanced equation to do it.
0.290 mol CH4 x (1 mol CF4/1 mol CH4) = 0.290 mols CF4 produced.

2. Convert 0.4400 mol F2 to mols CF4.
0.440 mol F2 x (1 mol CF4/4 mol F2) = 0.110 mols

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 0.110 mol.

Next we must determine how much of the CH4 reacted and how much HF is produced. That is done with coefficients, too.
0.440 mol F2 x (1 mol CH4/4 mol F2) = 0.440 x 1/4 = 0.110 mol CH4 used.
We know 0.110 mol CF4 is produced; therefore, we must have produced 4 x 0.110 = 0.440 mol HF.

Now we plug data into the delta H rxn using delta Ho formation values:
dHrxn = (mols*dH products) - (mols *dH reactants). dHrxn is the heat produced by that reaction.

dHrxn = [(-0.110*679.9) + (-0.44*268.61) ] - [(-74.8*0.110)]

Qrxn = -184.75 kJ