Answer:
∆Hrxn = ∑Hproducts - ∑Hreactants
= 392 kj
Explanation:
∆H (heat of reaction) of a reaction is the heat that accompanies the entire reaction. This is, it represents the heat released or absorbed during the reaction
. This value can be positive or negative, and will depend on whether the process is exothermic or endothermic. A process is considered exothermic when this happens, heat is released to the surroundings, this means that the system loses heat, so ∆H <0. The opposite is true for endothermic processes, which are characterized by absorbing heat from the surroundings, which implies that ∆H> 0.
For the calculation of the heat of reaction you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (quantity of molecules of each compound that participates in the reaction) and finally subtract them:
<em>Enthalpy of reaction = ΔH = ∑Hproducts - ∑Hreactants
</em>
In this case you have the reaction
<em>CH₃OH (l) + 2 O₂(g) ⇒ CO₂ (g) + 2 H₂O (g)</em>
<u><em>PRODUCTS:
</em></u>
CO₂: It is O=C=O. You have two bonds C=O, so the bond energy = 2*799 kJ/mol = 1598 kJ/mole*1 mol (
Upon observing the reaction, 1 mol of CO₂ is stoichiometrically produced) = 1598 kJ
H₂O: It is H-O-H. You have two bons O - H, so the bond energy = 2*464 kJ/mol = 928 kJ/mole*2 moles (
Upon observing the reaction, 2 mol of H₂O is stoichiometrically produced) = 1856 kJ
<em>
∑Hproducts= 1598 kJ + 1856 kJ = 3454 kJ
</em>
<u><em>REACTANTS:
</em></u>
CH₃OH: has 3 C-H bonds, 1 C-O bond and 1 O-H bond, so the bond energy = 3*414 kJ/mol + 360 kJ/mol + 464 kJ/mol = 2066 kJ
/mol*1 mol (Upon observing the reaction, stoichiometrically reacts 1 mol of CH₃OH)= 2066 kJ/mol
O₂: It is O=O, so the bond energy = 498 kJ/mol*2 moles (Upon observing the reaction, stoichiometrically reacts 2 mol of O₂)= 996 kJ
<em>∑Hreactants = 2066 kJ + 996 kJ = 3062 kJ
</em>
<em>
∆Hrxn = ∑Hproducts - ∑Hreactants
=3454 kJ - 3062 kJ= 392 kj</em>
The sign is negative indicating an endothermic reaction.
