I have a question in my textbook and I do not really understand their explanation of the answer, it is as follows:

$$\ce{2CH3OH + 3O2 -> 2CO2 + 4H2O}$$ When 2 moles of $\ce{CH3OH}$ react with 3 moles of oxygen, exactly $\pu{1354 kJ}$ of energy is released. How much heat is released when $\pu{5.00 g}$ of $\ce{CH3OH}$ is combusted in excess oxygen?

Here is the procedure to solve it:

  1. Convert the mass of $\ce{CH3OH}$ to amount of substance, which is $\pu{0.156 mol}$.
  2. $\displaystyle \pu{0.156 mol} \times \left(\frac{\pu{1 mol}}{\pu{2 mol}} ~\ce{CH3OH}\right) \times \pu{-1354 kJ} = \pu{-106 kJ}$

My first question is, why is $\pu{0.156 mol}$ multiplied by $\pu{1 mol}/\pu{2 mol}~\ce{CH3OH}$? Where is the $\pu{1 mol}$ in the numerator coming from, if for every two moles of $\ce{CH3OH}$, 3 moles of oxygen react?

My second question is, why is this expression multiplied by $\pu{-1354 kJ}$? It says in the problem that $\pu{1354 kJ}$ of energy are released. Does this mean I need to infer based on the equation whether the reaction is exothermic or endothermic? I just started learning thermochemistry so I am not quite familiar yet with these types of problems and what I need to do.

  • $\begingroup$ Regarding your second question - yes, you have to infer from the question's phrasing the direction of energy "movement". Since the definitions of endothermic and exothermic reaction are pretty straightforward as far as energy movement is concerned, it should be rather simple to deduce that when "energy is released" we are dealing with an exothermic reaction. Remember that the definitions are based on a convention, and the convention is system-centric, i.e., we are interested in what is happening in the system, and not the surroundings. $\endgroup$ – Don_S Feb 2 '17 at 5:47
  • $\begingroup$ Regarding your first question - this seems to be a poor choice of conversion factor. In this type of questions you are expected to use the "enthalpy of reaction" (here specifically it is the enthalpy of combustion) as a conversion factor. Enthalpy of reaction has the units of kJ/mol of substance. This way, when you are multiplying a certain amount of moles by the enthalpy of reaction, you should be left with a value in kJ, i.e. an amount of energy only. If you calculate 0.156 mol x -1354 kJ/2 mol CH3OH (given enthalpy), this may simplify things and make more sense. What do you think? $\endgroup$ – Don_S Feb 2 '17 at 5:57

It says that 2 moles of of $\ce{CH3OH}$ release $\text{1354 kJ}$. This is the same as saying that 1 mole of of $\ce{CH3OH}$ releases $\text{677 kJ}$. Here I just divided the 1354 by 2 to obtain the number of the energy released when one mole is burned. This way it is easier to do dimensional analysis. If we were to express this in a more mathematical way, $$\frac{677\,\text{kJ}}{\text{mol}\,\,\ce{CH3OH}}$$

Note that this is equal to $$\frac{1\ \text{mol}\,\,\ce{CH3OH}}{2\ \text{mol}\,\,\ce{CH3OH}}\cdot 1354\,\,\text{kJ}$$

Basically, this is done to find the energy that ONE mole of $\ce{CH3OH}$ releases. Your confusion stems from a poor track on the units that are involved. You do not need to worry about the oxygen since it is in excess.

As for the sign of the energy (the number in front of kJ) is due to convention. You are right that it is to denote that the process is exothermic or endothermic. But I suggest that you read more of it. This is important, and will come back again in physical chemistry.

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  • $\begingroup$ This is a great answer. Just note that when you are writing (1mol CH3OH/2mol CH3OH) x 1354 kJ you are left with a value in kJ only (the mol CH3OH may be canceled out and we are left with a value of 1/2), so this cannot be used as a conversion factor and is not equal to 677 kJ/mol CH3OH (which is indeed a conversion factor) as you wrote. Remove the nominator and you're good to go. $\endgroup$ – Don_S Feb 2 '17 at 6:02

I've never done thermochemsitry. But this is how I arrived to the same answer.

In two moles of methanol there is: $$\pu{2 mol} \times \pu{32.04 g//mol} = \pu{64 g}~\ce{MeOH}$$

Which releases $\pu{1354 kJ}$ of energy. Therefore: $$\frac{\pu{1354 kJ}}{\pu{64.08 g}}~\ce{MeOH} = \pu{21.1298 kJ}~\text{released per gram of}~\ce{MeOH}$$

This implies: $$\pu{5 g}~\ce{MeOH} \times \pu{21.1298 kJ//g} = \pu{105.649 kJ}~\text{released }$$

And yes, you are correct about the negative denotation of energy. If a reaction has a positive value, energy must be provided for the reaction to proceed (endothermic). If the reaction has a negative value, the system is losing energy as heat (exothermic).

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  • $\begingroup$ In my opinion, when doing mathematics in chemistry, it is best to write things out in a way that is similar to this. $\endgroup$ – Bob Feb 1 '17 at 23:16

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