# 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

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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 :)