Answer:
This is an exothermic reaction.
Explanation:
You can calculate the reaction using the calculation below:
ΔH reac= 3(413Kj/mol) +358Kj/mol +467 Kj/mol +1070 kj/mol = 3134 kj/mol
ΔH prod = 3(413 kj/mol) + 347 kj/mol + 358kj/mol +467kj/mol +745 kj/mol = 3156kj/mol
ΔH = 3134 kj/mol - 3156kj/mol = -22 Kj/mol
Negative results indicate an exothermic reaction.
An image of the reaction is attached below.
The heat of the reaction has been -22 kJ/mol. The negative sign has been the indication of the exothermic reaction.
The heat of reaction has been given as the amount of energy absorbed or released by the system to form the products.
The heat of reaction has been given by:
[tex]\Delta H=\Delta H_{reactant}-\Delta H_{product}[/tex]
The heat of reaction has been calculated based on the bond energies. The energy constituent by each bond in the reactant has been the heat of the reactant, and the energy of product has been the heat of product.
The reactant molecules have been [tex]\rm CH_3OH[/tex] and [tex]\rm C\equiv O[/tex].
The bond energy of the reactant has been:
[tex]\Delta H_{reactant}=3\;\times \;\text {C-H}\;+\;\text{C-O + O-H}+\;\text C\equiv \text O\\ \Delta H_{reactant}=3\;\times \;413\;+\;358+467+1070\;\text{kJ/mol}\\ \Delta H_{reactant}=3134\;\text{kJ/mol}[/tex]
The heat of reactant has been 3134 kJ/mol.
The product molecule has been [tex]\rm CH_3COOH[/tex]
The bond energy of the product has been:
[tex]\Delta H_{product}=3\;\times \;\text {C-H}\;+\;\text{C-O + O-H +\;C-C}+\;\text C=\text O\\\Delta H_{reactant}=3\;\times \;413\;+\;358+467+\;347\;+\;745\;\text{kJ/mol}\\\Delta H_{reactant}=3156\;\text{kJ/mol}[/tex]
The heat of product has been 3156 kJ/mol.
The heat of reaction has been given as:
[tex]\Delta H=\Delta H_{reactant}-\Delta H_{product}\\ \Delta H=3134-3156\;\text{kJ/mol}\\ \Delta H=-22\;\rm kJ/mol[/tex]
The heat of the reaction has been -22 kJ/mol. The negative sign has been the indication of the exothermic reaction.
Learn more about bond energy, here:
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