Understanding the Enthalpy Change in the Formation of Methane (CH4)
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Introduction
In the realm of chemistry, understanding the change in enthalpy for chemical reactions is crucial for predicting the energy transfer during reactions. One common reaction involves the formation of methane (CH4) from solid carbon in its graphite form and hydrogen gas. This article will guide you through the process of calculating this enthalpy change using Hess's Law, which is particularly useful when direct measurement is impractical due to slow reaction rates.
What is Enthalpy Change?
Enthalpy change, denoted as ( \Delta H ), is the total change in energy during a chemical reaction. It can be exothermic (where energy is released) or endothermic (where energy is absorbed). In the case of methane formation, we are interested in determining how much energy is either released or absorbed when solid carbon (graphite) reacts with hydrogen gas (H2).
The Challenge of Measuring Directly
The enthalpy change for methane formation cannot be measured directly in a lab environment. The main reason for this is the reaction's slow rate, making it difficult to observe temperature changes in the surrounding medium. However, by using data from combustion reactions of methane, carbon, and hydrogen, we can utilize Hess's Law to find this value indirectly.
Understanding Hess's Law
Hess's Law states that if a chemical reaction can be expressed as the sum of two or more reactions, the total enthalpy change for the reaction is equal to the sum of the enthalpy changes of the individual reactions. This principle allows chemists to calculate enthalpy changes for complex reactions easily.
Step-by-Step Calculation Using Hess's Law
Step 1: Identify the Desired Reaction
We want to determine the enthalpy change for the following reaction:
Equation:
[ \text{C (s, graphite)} + 2 \text{H}_2(g) \rightarrow \text{CH}_4(g) ]
Step 2: Reverse Reactions to Express Final Products
To leverage Hess's Law, we review the combustion reactions for the relevant substances. The combustion of methane can be expressed as:
Combustion of Methane:
[ \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2O(l) \quad (\Delta H = -890.3 , \text{kJ/mol}) ]
Reversing this equation gives us:
[ \text{CO}_2(g) + 2 \text{H}_2O(l) \rightarrow \text{CH}_4(g) + 2 \text{O}_2(g) \quad (\Delta H = +890.3 , \text{kJ/mol}) ]
Step 3: Incorporate Combustion Reactions of Carbon and Hydrogen
Next, we use the combustion of carbon and hydrogen to yield CO2 and H2O, respectively:
Combustion of Carbon:
[ ext{C (s, graphite)} + \text{O}_2 (g) \rightarrow \text{CO}_2(g) \quad (\Delta H = -393.5 , \text{kJ/mol}) ]
Combustion of Hydrogen:
[ 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2O(l) \quad (\Delta H = -571.6 , \text{kJ/mol}, 1 \text{ mole of water}) ]
To match the number of water molecules needed in our target reaction, we multiply the hydrogen combustion equation by 2: [ 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2O(l) \quad (\Delta H = -285.8 , \text{kJ/mol} \times 2) ]
Summary of Enthalpies
- Combustion of Carbon (reversed):
- ( \Delta H = +393.5 ) kJ/mol
- Combustion of Hydrogen (doubled):
- ( \Delta H = -571.6 ) kJ/mol
- Reversed Combustion of Methane:
- ( \Delta H = +890.3 ) kJ/mol
Step 4: Combine All Enthalpy Changes
Following Hess's Law, we sum these enthalpy changes: [ \Delta H_{total} = (-393.5) + (-571.6) + (+890.3) ]
- Calculate this:
[ \Delta H_{total} = -74.8 , \text{kJ/mol} ]
Final Conclusion
The enthalpy change for the formation of methane from its elements, solid carbon (graphite) and hydrogen gas, is approximately -74.8 kJ/mol. This indicates that forming methane is an exothermic reaction, which releases some energy to the surroundings. Thus, using Hess's Law allows us to effectively deduce the enthalpy changes of complex reactions based on simpler, measurable ones.
This process not only illustrates the utility of Hess's Law in thermodynamics but also enhances our understanding of the energetic relationship between chemical reactants and products within the field of chemistry.
This problem is from chapter five of the Kotz, Treichel, Townsend Chemistry and Chemical Reactivity textbook. So they tell us, suppose you want to know the enthalpy
change-- so the change in total energy-- for the formation of methane, CH4, from solid carbon as a graphite-- that's right there-- and hydrogen gas.
So we want to figure out the enthalpy change of this reaction. How do we get methane-- how much energy is absorbed or
released when methane is formed from the reaction of-- solid carbon as graphite and hydrogen gas? So they tell us the enthalpy change for this reaction
cannot to be measured in the laboratory because the reaction is very slow. So normally, if you could measure it you would have this
reaction happening and you'd kind of see how much heat, or what's the temperature change, of the surrounding solution. Maybe this is happening so slow that it's very hard to
measure that temperature change, or you can't do it in any meaningful way. We can, however, measure enthalpy changes for the
combustion of carbon, hydrogen, and methane. So they're giving us the enthalpy changes for these combustion reactions-- combustion of carbon,
combustion of hydrogen, combustion of methane. And they say, use this information to calculate the change in enthalpy for the formation of
methane from its elements. So any time you see this kind of situation where they're giving you the enthalpies for a bunch of reactions and they
say, hey, we don't know the enthalpy for some other reaction, and that other reaction seems to be made up of similar things, your brain should immediately say, hey,
maybe this is a Hess's Law problem. Hess's Law. And all Hess's Law says is that if a reaction is the sum
of two or more other reactions, then the change in enthalpy of this reaction is going to be the sum of the change in enthalpies of those reactions.
Now, when we look at this, and this tends to be the confusing part, how can you construct this reaction out of these reactions over here?
And what I like to do is just start with the end product. So I like to start with the end product, which is methane in a gaseous form.
And when we look at all these equations over here we have the combustion of methane. So this actually involves methane, so
But what we can do is just flip this arrow and write it as methane as a product. So if we just write this reaction, we flip it.
So now we have carbon dioxide gas-- let me write it down here-- carbon dioxide gas plus-- I'll do this in another color-- plus two waters-- if we're thinking of these as
moles, or two molecules of water, you could even say-- two molecules of water in its liquid state. That can, I guess you can say, this would not happen
spontaneously because it would require energy. But if we just put this in the reverse direction, if you go in this direction you're going to get two waters-- or two
oxygens, I should say-- I'll do that in this pink color. So two oxygens-- and that's in its gaseous state-- plus a gaseous methane.
in reverse order. I'm going from the reactants to the products. When you go from the products to the reactants it will
release 890.3 kilojoules per moles of the reaction going on. But if you go the other way it will need 890 kilojoules.
So the delta H here-- I'll do this in the neutral color-- so the delta H of this reaction right here is going to be the reverse of this.
So it's positive 890.3 kilojoules per mole of the reaction. All I did is I reversed the order of this
reaction right there. The good thing about this is I now have something that at least ends up with what we eventually
want to end up with. This is where we want to get. This is where we want to get eventually.
Now, if we want to get there eventually, we need to at some point have some carbon dioxide, and we have to have at some point some water to deal with.
So how can we get carbon dioxide, and how can we get water? Well, these two reactions right here-- this combustion
reaction gives us carbon dioxide, this combustion reaction gives us water. So we can just rewrite those.
Let me just rewrite them over here, and I will-- let me use some colors. So if I start with graphite-- carbon in graphite form--
carbon in its graphite form plus-- I already have a color for oxygen-- plus oxygen in its gaseous state, it will produce carbon dioxide in its gaseous form.
It will produce carbon-- that's a different shade of green-- it will produce carbon dioxide in its gaseous form. And this reaction, so when you take the enthalpy of the
carbon dioxide and from that you subtract the enthalpy of these reactants you get a negative number. Which means this had a lower enthalpy, which means energy
So this is essentially how much is released. But our change in enthalpy here, our change in enthalpy of this reaction right here, that's reaction one.
So the reaction occurs a mole times. This would be the amount of energy that's essentially released.
This is our change in enthalpy. So if this happens, we'll get our carbon dioxide. Now we also have-- and so we would release this much energy
and we'd have this product to deal with-- but we also now need our water. And this reaction right here gives us our water, the
combustion of hydrogen. So we have-- and I haven't done hydrogen yet, so let me do hydrogen in a new color.
That's not a new color, so let me do blue. So right here you have hydrogen gas-- I'm just rewriting that reaction-- hydrogen gas plus 1/2 O2--
pink is my color for oxygen-- 1/2 O2 gas will yield, will it give us some water. Will give us H2O, will give us some liquid water.
Now, before I just write this number down, let's think about whether we have everything we need. To make this reaction occur, because this gets us to our
final product, this gets us to the gaseous methane, we need a mole. Or we can even say a molecule of carbon dioxide, and this
reaction gives us exactly one molecule of carbon dioxide. So that's a check. And we need two molecules of water.
Now, this reaction only gives us one molecule of water. So let's multiply both sides of the equation to get two molecules of water.
So this is a 2, we multiply this by 2, so this essentially just disappears. You multiply 1/2 by 2, you just get a 1 there.
And then you put a 2 over here. So I just multiplied this second equation by 2. So I just multiplied-- this is becomes a 1, this becomes a 2.
And if you're doing twice as much of it, because we multiplied by 2, the delta H now, the change enthalpy of the reaction, is now going to be twice this.
Let's get the calculator out. It's now going to be negative 285.8 times 2. Because we just multiplied the whole reaction times 2.
Now, let's see if the combination, if the sum of these reactions, actually is this reaction up here. And to do that-- actually, let me just copy and paste this
top one here because that's kind of the order that we're going to go in. You don't have to, but it just makes it hopefully a little
bit easier to understand. So let me just copy and paste this. Actually, I could cut and paste it.
Cut and then let me paste it down here. That first one. And let's see now what's going to happen.
To see whether the some of these reactions really does end up being this top reaction right here, let's see if we can cancel out reactants and products.
Let's see what would happen. So this produces carbon dioxide, but then this mole, or this molecule of carbon dioxide, is then used up in
Let me do it in the same color so it's in the screen. This reaction produces it, this reaction uses it. Now, this reaction right here produces the two
molecules of water. And now this reaction down here-- I want to do that same color-- these two molecules of water.
Now, this reaction down here uses those two molecules of water. Now, this reaction right here, it requires one molecule of
molecular oxygen. This one requires another molecule of molecular oxygen. So these two combined are two molecules of molecular oxygen.
So those are the reactants. And in the end, those end up as the products of this last reaction.
So those, actually, they go into the system and then they leave out the system, or out of the sum of reactions unchanged.
So they cancel out with each other. So we could say that and that we cancel out. And so what are we left with?
What are we left with in the reaction? Well, we have some solid carbon as graphite plus two moles, or two molecules of molecular hydrogen yielding--
all we have left on the product side is some methane. So it is true that the sum of these reactions is exactly what we want.
All we have left on the product side is the graphite, the solid graphite, plus the molecular hydrogen, plus the gaseous hydrogen-- do it in that color-- plus
two hydrogen gas. And all we have left on the product side is the methane. All we have left is the methane in the gaseous form.
So it is true that the sum of these reactions-- remember, we have to flip this reaction around and change its sign, and we have to multiply this reaction by 2 so that the sum
of these becomes this reaction that we really care about. So this is the sum of these reactions. Its change in enthalpy of this reaction is going to be the
So we just add up these values right here. So we have negative 393.-- no, that's not what I wanted to do.
Let me just clear it. So I have negative 393.5, so that step is exothermic. And then we have minus 571.6.
That is also exothermic. Those were both combustion reactions, which are, as we know, very exothermic.
And we have the endothermic step, the reverse of that last combustion reaction. So plus 890.3 gives us negative 74.8.
It gives us negative 74.8 kilojoules for every mole of the reaction occurring. Or if the reaction occurs, a mole time.
Nowhere near as exothermic as these combustion reactions right here, but it is going to release energy. And we're done.