The combustion reaction for Compound X isgiven.X(s) + 6O2(g) -->4CO2(g) + 3H2O(l)Delta H for this reaction is equal to −2280kJ/mol delta H of ffor CO2(g) is −393.5 kJ/moland delta H of ffor...

The combustion reaction for Compound X is
given.
X(s) + 6O2(g) -->4CO2(g) + 3H2O(l)
Delta H for this reaction is equal to −2280
kJ/mol

delta H of f
for CO2(g) is −393.5 kJ/mol
and delta H of f
for H2O(l) is −285.8 kJ/mol.
Calculate delta H of o
f for Compound X.
Answer in units of kJ/mol

Asked on by candyyyyy

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jerichorayel's profile pic

jerichorayel | College Teacher | (Level 2) Senior Educator

Posted on

Hello!

 

In order to answer the problem, we need to draw the chemical reaction.

The complete combustion of X will yield water and carbon dioxide. It can be written as:

X  + 6O2 ---->  4CO2  + 3H2O

In solving the delta H of compound X we will use the enthalpy change formula.

 

enthalpy change = (sum of nH of products) - (sum of nH of the reactants)

 

products: CO2 and H2O

reactants: O2 and X

note: O2 does not participate in the enthalpy change computation.

 

**in this case the delta H of compound x is missing

-2280kJ = [(-393.5 kJ/mol x 4 mol) + ( -285.8 kj/mol x 3mol)] - (X x 1mole)

 

-2280.0kJ     = -2431.4kJ - (X x 1mole)

+2431.4kJ    = +2431.4 kJ                     **add 2431.4 on both sides

_____________________________

 

151.4 kJ    =  -(X  x  1mole)            **divide both sides by 1mole

--------------------------

1mole             1 mole

 

151.5kJ/mole = -X                          **divide both sides by -1

----------------------

-1                       -1

 

therefore the answer would be:

X = -151.4 kJ/mole

 

hope this helps :)

 

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