The combustion of pentane, c5h12, occurs via the reaction c5h12(g)+8o2(g)→5co2(g)+6h2o(g) with heat of formation values given by the following table: substance δh∘f (kj/mol) c5h12(g) −35.1 co2(g) −393.5 h2o(g) −241.8

### Answers

Then, for the given equation, the total heat of the combustion of pentnae is:

ΔH°f = 5* ΔH°f CO2(g) + 6ΔH°f H2O(g) - ΔH°f C5H12(g) - 8ΔH°f O2(g)

The standard heat of formation of O2(g) is zero because that is its natural state.

Now you can replace in the equation the values given:

ΔH° = 5*(-393.5 kJ/mol) + 6*(-241.8kj/mol) - (-35.1kJ/mol) - 0

ΔH° = -3,383.2 kJ/mol

Then, the heat of the combustion of pentane is -3,383.2 kj/mol

-3298.4 kJ/mol

Explanation:

The balanced chemical reaction is,

The expression for enthalpy change is,

where,

n = number of moles

(as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

Therefore, the enthalpy change for this reaction is, -3298 kJ/mol

Answer : The enthalpy for the combustion of 1 mole of pentane is -3024.8 kJ

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as

The equation used to calculate enthalpy change is of a reaction is:

The equilibrium reaction follows:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

Therefore, the enthalpy for the combustion of 1 mole of pentane is -3024.8 kJ

The enthalpy of the reaction in an aqueous solution can be determined by taking the difference between the summation of enthalpies of the products multiplied to their respective stoichiometric coefficient and the summation of enthalpies of the reactants multiplied to their respective stoichiometric coefficient. In this case, the equation is -241(6) + -393.5 (5) -[-119.9] equal to 641.4 kJ