Answer to Question #312211 in General Chemistry for James

Question

Suppose that 0.250mol

m

o

l

 of methane, CH

4

(g)

C

H

4

(

g

)

, is reacted with 0.400 mol

m

o

l

 of fluorine, F

2

(g)

F

2

(

g

)

, forming CF

4

(g)

C

F

4

(

g

)

 and HF(g)

H

F

(

g

)

 as sole products. Assuming that the reaction occurs at constant pressure, how much heat is released?

Answer 1

CH₄ (g) + 4F₂ (g) = CF₄ (g) + 4HF (g)

According to the equation, methane requires 4 times more mole of fluorine gas to react completely:

0.25 x 4 = 1 mol, but only 0.400 is provided. So, fluorine is a limiting reagent. And only: 0.400 / 4 = 0.100 of CH₄ would react.

Standard enthalpies of formation (kJ/mol):

CH₄ (g) = -74.81

CF₄ (g) = -925.0

HF (g) = -271.1 x 4 (moles)

F₂ (g) = 0!

The enthalpy of the reaction:

Products: -925.0 + (-271.4 x 4) = -2010.6

Reactants: -74.81

Delta: products – reactants = -2010.6 –(-74.81) = -1935.79 kJ per mol of methane.

-1935.79 x 0.100 = -193.579 kJ per given amount of substance.