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
X(s) + 6O2(g) -->4CO2(g) + 3H2O(l)
Delta H for this reaction is equal to −2280
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
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
therefore the answer would be:
X = -151.4 kJ/mole
hope this helps :)