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 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

The heat of a chemical reaction (change of enthalý) may be calcualted as the heat of formation of the products less the heat of formation of the reactants.

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,

C_5H_{12}(g)+8O_2(g)rightarrow 5CO_2(g)+6H_2O(g)

The expression for enthalpy change is,

Delta H=sum [ntimes Delta H_f(product)]-sum [ntimes Delta H_f(reactant)]

Delta H=[(n_{H_2O}times Delta H_{H_2O})+(n_{CO_2}times Delta H_{CO_2})]-[(n_{O_2}times Delta H_{O_2})+(n_{C_5H_{12}}times Delta H_{C_5H_{12}})]

where,

n = number of moles

Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

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

Delta H=[(6times -241.8)+(5times -393.5)]-[(8times 0)+(1times -119.9)]

Delta H=-3298kJ/mol

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 Delta H^o

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

Delta H^o_{rxn}=sum [ntimes Delta H^o_f(product)]-sum [ntimes Delta H^o_f(reactant)]

The equilibrium reaction follows:

C_5H_{12}(g)+8O_2(g)rightarrow 5CO_2(g)+6H_2O(g)

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

Delta H^o_{rxn}=[(n_{(CO_2)}times Delta H^o_f_{(CO_2)})+(n_{(H_2O)}times Delta H^o_f_{(H_2O)})]-[(n_{(C_5H_{12})}times Delta H^o_f_{(C_5H_{12})})+(n_{(O_2)}times Delta H^o_f_{(O_2)})]

We are given:

Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\Delta H^o_f_{(C_5H_{12}(g))}=-119.9kJ/mol\Delta H^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

Delta H^o_{rxn}=[(5times -393.5)+(6times -241.8)]-[(1times -393.5)+(8times 0)=-3024.8kJ

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 

If I were you, I'd try to go with this one :[(6)(-241.8)+(5)(-393.5)]-[-35.1] =-481.6. I think that is a solution you have to find. Just use the sum of standard enthalpies of products and  of reactants. Hope you will find it helpful!

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts