Question

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?

Answer 1

The balanced chemical eqaution for the given reaction:

2CH₄ + 5F₂ --> 2CF₄ + 2HF

The heats of formation of the compounds are as follows:  

CH₄= -74.8kJ/mole  

CF₄ = -925.0 kJ/mole  

HF = -271.1 kJ/mole  

0.29 moles CH₄ x (5 moles of F₂ / 2 moles of CH₄) = 0.725 moles F₂ required.

Given, moles of F₂ = 0.44

Therefore, F₂ is the limiting reactant  

Now, F₂ will determine the moles of CF₄ and HF produced in the reaction.

0.44 moles of  F₂  x (2 moles of CF₄/5 moles of  F₂)  = 0.176 moles CF₄

0.44 moles of  F₂  x (2 moles of HF/5 moles of  F₂)  = 0.176 moles HF

Since the molar ratio between F₂: CF₄ and F₂: HF is 1:1

So 0.176 moles CF₄  and 0.176 moles HF will require 0.176  moles CH₄  

ΔHrxn = ∑(moles x H products) - ∑ (moles x H reactants)

ΔHrxn = (0.176 x -271.1 kJ/mole + 0.176 x -925 kJ/mole) - (0.176 x -74.8 kJ/mole)  

ΔHrxn = - 197.35 kJ

- 197.35 kJ of heat is released.