The combustion of pentane, c5h12, occurs via the reaction c5h12(g)+8o2(g)→5co2(g)+6h2o(g) with heat of formation values given by

Question

The combustion of pentane (C5H12) is represented by the following reaction: 
C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(g). 
To analyze this reaction further, we can consider the heat of formation values provided in the following table: 
Substance         ΔH°f (kJ/mol)
C5H12(g)        -35.1
CO2(g)         -393.5
H2O(g)         -241.8  
With this information, how can we calculate the total enthalpy change for the combustion of pentane?

Prof. Walker Berge · Accepted Answer

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) - 0ΔH° = -3,383.2 kJ/molThen, the heat of the combustion of pentane is -3,383.2 kj/mol

Anonymous · Answer

-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 zeroNow put all the given values in this expression, we getTherefore, the enthalpy change for this reaction is, -3298 kJ/mol

Anonymous · Answer

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

Anonymous · Answer

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

Jocelyn Kessler · Answer

If I were you, I'd try to go with this one :. 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!

Anonymous

The combustion of pentane, c5h12, occurs via the reaction c5h12(g)+8o2(g)→5co2(g)+6h2o(g) with heat of formation values given by

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