L'équation stœchiométrique de la combustion du méthanol s'écrit : CH3 OH(1)+3 / 2 O2=CO2+2 H2 O

L'équation stœchiométrique de la combustion du méthanol...

L'équation stœchiométrique de la combustion du méthanol s'écrit : $$ \mathrm{CH}_3 \mathrm{OH}(1)+3 / 2 \mathrm{O}_2=\mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} \text { (1) } $$ L'enthalpie molaire de référence de cette réaction est - $725 \mathrm{~kJ}^{-1} \mathrm{~mol}^{-1}$ à $298 \mathrm{~K}$; quelle est sa valeur à $350 \mathrm{~K}$ ?

L'équation stœchiométrique de la combustion du méthanol s'écrit :
$$
\mathrm{CH}_3 \mathrm{OH}(1)+3 / 2 \mathrm{O}_2=\mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} \text { (1) }
$$

L'enthalpie molaire de référence de cette réaction est - $725 \mathrm{~kJ}^{-1} \mathrm{~mol}^{-1}$ à $298 \mathrm{~K}$; quelle est sa valeur à $350 \mathrm{~K}$ ?

Key Concepts

Enthalpy Change of a Reaction

This concept refers to the heat absorbed or released during a chemical reaction occurring at constant pressure. It is typically measured under standard conditions and can be adjusted for different temperatures by considering other thermodynamic properties.

Kirchhoff's Law

Kirchhoff’s Law allows the calculation of the reaction enthalpy at a temperature different from the reference temperature by accounting for the temperature dependence of the heat capacities of the reactants and products. It involves integrating the difference in heat capacities over the temperature change.

Heat Capacity

Heat capacity is the amount of heat required to change the temperature of a substance by a given amount. When determining the enthalpy change of a reaction at a temperature different from the standard value, the difference in heat capacities between products and reactants is used to adjust the reaction enthalpy via integration over the temperature range.

Transcript

00:01 So what's the overall delta h of reaction if we sum up these three equations by manipulating them? and by manipulating here, i mean we're going to by flipping the first reaction and multiplying it, i should say, by two, and by multiplying each of the other two by one half.

00:30 What is the overall net reaction and what is the overall net delta h so remember what happens with delta h values here based on h's law if we flip a reaction we're going to flip the sign and if we divide a reaction we're going to divide the delta h if we multiply we're going to multiply the delta h and so let's go ahead and do these things so if i flip this reaction the products become the reactants the reactants become the products we're also dividing by two so one mole divided by two we have one half of our hcho2 we have one half water becomes one half of methanol and one half we flip the reaction so we flip the sign negative is now positive we've divided everything by two, so we're also going to divide the delta h by two.

02:00 Now if we multiply each of these by a half, we'll take one half, two times a half is one, and we have one half, and we'll have to multiply our delta h by a half.

02:22 Same thing down for the last one, which makes everything one half of a mole.

02:29 And then once we have each of these three, how does it wind up getting added together? well, you can notice that several substances wind up adding together, several substances wind up up canceling out.

02:47 So you can see that we have half of an hc -h -o -2 and a half h -c -h -o -2.

02:56 It doesn't get cancelled, out by anything on either side.

02:59 So we have a total of one mole of hcho2.

03:08 You can see that we have one mole of hydrogen.

03:13 It doesn't get canceled out by anything.

03:21 On the right hand side, you can see we have one half mole of methanol, one half mole of methanol...

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