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Sagot :
Answer: [tex]\Delta H_{rxn}=-4652[/tex] kJ/mol
Explanation: In a reaction, a bond of a molecule can break or rearrange itself. When it happens, it absorb or release energy.
Bond Energy is the energy necessary to break or to make a particular bond, producing gaseous fragments at atmospheric temperature. This type of energy can be used to show how stable a compound is or how easy a bond can break.
Enthalpy of Reaction is the energy absorption or release at a constant temperature resulting from a chemical reaction.
To calculate the Enthalpy using bond energy, you have:
1) Write the balanced equation with all the all the reactants and products in gaseous form:
[tex]C_{5}H_{12}_{(g)}+8O_{2}_{(g)}[/tex] ⇒ [tex]5CO_{2}_{(g)}+6H_{2}O_{(g)}[/tex]
2) Count each bond of each molecule from both sides, including double or triple bonding, if they exists:
Reactants:
[tex]C_{5}H_{12}[/tex] : 12 C-H
4 C-C
O₂ : 1 O=O
Products:
CO₂ : 2 C=O
H₂O : 2 H-O
3) From the balanced reaction, we there are 8 moles of oxygen gas, 5 moles of carbon dioxide and 6 moles of water, which has to be included when adding each side.
[tex]\Sigma H_{reactants}=[/tex] 12.257 + 4.370 + 8.490 = 8484
[tex]\Sigma H_{products}=[/tex] 5.2.740 + 6.2.478 = 13136
4) Calculate Enthalpy of Reaction by:
[tex]\Delta H_{rxn}= \Sigma H_{reactants}-\Sigma H_{products}[/tex]
[tex]\Delta H_{rxn}=8484-13136[/tex]
[tex]\Delta H_{rxn}=-4652[/tex]
The enthalpy of reaction for the combustion of 1 mole of pentane on a just discovered imaginary planet is an exothermic process of magnitude [tex]\Delta H_{rxn}=[/tex] - 4652kJ/mol
Using the planet's given bond energies to calculate the enthalpy of reaction, the combustion of 1 mole of pentane is -4652 kJ.
The balanced chemical reaction for the combustion of pentane is well represented as:
[tex]\mathbf{C_5H_{12}+8O_2 \to 5CO_2+6H_2O}[/tex]
From the above reaction, the bonds broken are 12 Carbon-Hydrogen bonds, 4 Carbon-Carbon bonds, and 8 oxygen-oxygen bonds.
Similarly, the bonds formed are 10 Carbon-Oxygen bonds and 12 Oxygen-hydrogen bonds.
Hence, the energy absorbed when bonds are broken are:
= 12(C-H) + 4(C-C) + 8(O-O)
= 12(257 kJ) + 4(370 kJ) + 8(490 kJ)
= 3084 kJ + 1480 kJ+ 3920 kJ
= 8484 kJ
Also, the energy released as a result of the formation of bond is:
= 10 (C=O) + 12 (O-H)
= 10 (740 kJ) + 12 (478 kJ)
= 7400 kJ + 5736 kJ
= 13136 kJ
From above, we can notice that the energy released during the formation of bonds is greater than the energy absorbed while breaking bonds.
Thus, the enthalpy of the combustion reaction is the difference in energy.
i.e.
The enthalpy of the combustion reaction = 8484 kJ - 13136 kJ
= -4652 kJ
Therefore, we can conclude that the enthalpy of reaction from the combustion of 1 mole of pentane is -4652 kJ. The negative sign indicates that more energy is released compared to the amount of energy absorbed.
Learn more about the enthalpy of reaction here:
https://brainly.com/question/1657608?referrer=searchResults
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