Describe in words the chemical reaction N2 + 3H2 -> 2NH3.
Solution
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The chemical reaction describes the synthesis of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. This reaction is fundamental in chemistry, particularly in the industrial production of ammonia. Here's a detailed explanation of the reaction:
The chemical reaction describes the synthesis of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. This reaction is fundamental in chemistry, particularly in the industrial production of ammonia. Here's a detailed explanation of the reaction:
1. Reactants Involved:
- Nitrogen (): A diatomic molecule, consisting of two nitrogen atoms bonded together. It is a relatively inert gas under standard conditions, requiring significant energy to break its triple bond.
- Hydrogen (): Also a diatomic molecule, consisting of two hydrogen atoms. Hydrogen gas is highly reactive and forms a single bond in molecules.
2. Product Formed:
- Ammonia (): A compound consisting of one nitrogen atom and three hydrogen atoms. Ammonia is a colorless gas with a characteristic pungent smell.
3. Balancing the Equation:
The chemical equation is balanced as follows:
- The nitrogen molecule () consists of 2 nitrogen atoms. In the product side, two ammonia molecules () contain a total of 2 nitrogen atoms, thus balancing the nitrogen atoms on both sides.
- For hydrogen, there are 3 hydrogen molecules (), each containing 2 hydrogen atoms, making a total of 6 hydrogen atoms. In the product, two ammonia molecules () have a total of 6 hydrogen atoms, thus balancing the hydrogen atoms on both sides.
4. Type of Reaction:
This is a synthesis reaction where simpler substances (nitrogen and hydrogen gases) combine to form a more complex substance (ammonia).
5. Industrial Significance:
This reaction is famously known as the Haber-Bosch process in industry. It is a key industrial chemical process used for the production of ammonia, which is a critical ingredient in fertilizers and various chemicals.
6. Reaction Conditions:
The reaction typically requires high pressure and temperature, along with a catalyst (often iron-based), to proceed efficiently due to the strong triple bond in nitrogen molecules that must be broken for the reaction to occur.
In summary, the reaction represents the synthesis of ammonia from nitrogen and hydrogen. It is an exothermic reaction, releasing energy, and plays a crucial role in the industrial production of ammonia, particularly in agriculture for fertilizer manufacturing.
The Haber-Bosch process, which carries out this reaction, is a pivotal industrial process with significant implications for food production and the global economy.
Chemical Equations and Stoichiometry
Daltons Atomic Theory
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In a previous video,
we wrote the chemical equation for the combustion of glucose.
In this video, we'll see that we can use
the balanced equation to solve chemical problems.
What does this chemical equation tell us?
When we burn 1 mole of glucose,
this should be a 1 here,
but we don't write it,
and 6 moles of oxygen,
here's the 6,
we get 6 moles of carbon dioxide and 6 moles of water.
This is result of chemical bookkeeping.
It tells us how many moles we use up in order to produce a certain number of moles.
It's called stoichiometry.
Stoichiometry, difficult word to spell.
The chemical equation we had for glucose is a simple example of stoichiometry.
Stoichiometry tells us how to perform
quantitative calculations based on chemical equations.
In order to do this,
in order to solve problems,
we need to write conversion factors based,
for example, on the combustion of glucose.
Let's write conversion factors.
From 1 mole of glucose,
we get 6 moles of carbon dioxide.
We can make that into a ratio and write this is equal to 1,
1 mole of glucose is equivalent in some respect to 6 moles of carbon dioxide.
Similarly, 1 mole of glucose is equivalent to 6 moles of water.
Burning 1 mole of glucose gives us 6 moles of water.
We can make that into a ratio and write equal to 1.
Similarly, 6 moles of water is equivalent to 6 moles of carbon dioxide.
We can divide the 6,
and write 1 mole of water is equivalent to 1 mole of carbon dioxide.
All these are conversion factors.
We call them stoichiometric ratios.
We can insert them wherever it is relevant.
Let's take an example.
How many moles of carbon dioxide are produced when
1.5 moles of glucose is burnt in excess oxygen.
That means we've got so much oxygen,
we don't need to worry about it.
The number of moles of carbon dioxide is equal to the number of
moles of glucose multiplied by our conversion factor,
our stoichiometric ratio, 6 moles
of carbon dioxide being equivalent to 1 mole of glucose.
First, let's multiply the numbers.
We start off with 1.5 moles of glucose,
here it's 1.5 and here we have 6 divide by 1,
so we have 6.
Now we need the units.
We have moles of glucose divided by moles of glucose,
so we can cancel those,
and we're left with moles of carbon dioxide.
Multiply 1.5 times 6 and we get,
of course 9,
9 moles of carbon dioxide.
That's the solution to the problem.
When we burn 1.5 moles of glucose,
we get 9 moles of carbon dioxide.
That's a very simple problem.
We can also write it and we'll use this notation in the rest of the video.
Number of moles of carbon dioxide,
that's n for numbers of moles and carbon dioxide indicates of what is equal to 9 moles.
We could also write 9 moles of carbon dioxide if we want it to be very exact.
Often the calculations are more complex than just what we had before.
But the central part always involves stoichiometric conversion factors.
The central part is always converting moles to moles,
moles of products to moles of reactants,
moles of reactant to moles of products,
depending on the question.
Let's write it out like this.
The information we're given about the reactants,
we have to convert to the moles of reactants.
Then the moles of reactants have to be converted to the moles of products.
That's a central part of the problem.
The moles of products have to be converted to
the information required about the products.
Let's take an example.
What volume of water is produced when 270 grams of glucose is burnt in excess oxygen.
Let's convert the scheme that we had above into something a little bit more precise.
We're given the mass of glucose.
We want to convert it into the moles of glucose.
This is always the central part of the problem.
Then we have to convert it into the moles of water.
From the moles of water,
we need to convert it into the mass of water,
and from the mass of water to the volume of water,
which is what we're asked about.
We're asked what volume of water.
We can label this 1,
2, 3, 4.
These are the 4 steps.
Before we can start to solve this problem.
We need to recall some things that we learned before.
The first thing we need to remember is how to convert the mass to number of moles.
We need the following equation.
We need the number of moles is equal to the mass divided by the molar mass.
Mw, we're going to write as molar mass.
I know it doesn't sound right. It's Mw.
It's because once we used to call it molecular weight and the name has stuck.
But it's n = m,
the mass divided by the molar mass,
so Mw is molar mass.
At the end, we need to convert the number of moles to mass.
That's the opposite of what we've just done.
We take this equation,
n = m divided by Mw.
We want m,
so we can multiply both sides by Mw and we get the mass is equal to n times Mw.
These are just 2 versions of the same equation.
Now before we can proceed,
we need to know what the molar masses are and we can easily calculate
the molar mass of glucose is equal to 180 grams.
The molar mass of water is 18 grams, 18.0 grams.
We saw how to do that before.
We need the molar mass of each atom.
We've 6 carbon atoms, 12 hydrogen atoms,
6 oxygen atoms and we add them up and we get to a 180 grams.
In addition, at the end,
we need to convert the mass to the volume,
and the mass of water to the volume of water.
How do we do that?
We learnt this right at the beginning of the course.
We use the equation for density,
that's d = m/v.
The density is equal to the mass divided by the volume.
But we need the volume,
so we multiply both sides by v,
we get vd = m,
and from that we get v = m/d by dividing by d. Divide both sides by d,
we get v = m/d.
This is the equation we require.
d of course is the density and v is the volume.
We need to know what the density of water is
and this is something we should perhaps remember,
that the density of water is 1 gram per milliliter.
The density of water is 1 gram per milliliter or 1 kilogram per liter.
In other words, if you carry a 1 liter bottle of water,
you are carrying 1 kilogram of water.
Let's start to solve the problem.
The first step, the step 1 is the mass of glucose to the moles of glucose.
We have to convert the mass of glucose to the moles of glucose.
The number of moles of glucose is 270 grams.
That's the mass of glucose divided by its molar mass,
180 grams per mole.
We divide the 2,
we get 1.5 moles.
We have 1.5 moles of glucose.
Now we have Step 2,
which is the central step.
We have to convert the moles of glucose to moles of water.
The number of moles of water,
that's what we need to find out.
After writing moles of water,
you don't really need to write it,
but I want to be very clear,
at least at the beginning.
Number of moles of water and the units are, of course,
moles of water is equal the number of moles of glucose,
and I've written moles of glucose,
times our conversion factor,
the one that relates water to glucose,
so 6 moles of water is equivalent to 1 mole of glucose.
Here it is. Let's solve the numbers first.
We know we have 1.5 moles of glucose,
we can write 1.5,
and here we have 6 divided by 1, so 6.
What about the units?
We have moles of glucose on the top,
moles of glucose in the bottom.
We're left with moles of water,
1.5 times 6 is 9.
Here's our answer, 9 moles of water.
Now here's Step 3.
We need to convert the moles of water to the mass of water.
Our equation is, the mass is equal to the number of moles times the molar mass.
We have 9 moles of water.
The molar mass of water is 18.0 grams per mole.
We multiply 9 times 18, we get 162.
Moles cancels with mole minus 1.
Moles times moles minus 1 is just 1.
We're left just with a gram.
We have a 162 grams.
Now we have the mass of the water produced.
Here we have the final step, Step 4.
The mass of water to the volume of water.
We know from above that the volume's equal to the mass divide by density.
The mass is 162 grams,
as we saw up here,
divided by the density,
which is 1 gram per milliliter,
so 162 divided by 1 is just a 162.
Grams cancels with grams and we're leftover 1 over milliliter to minus 1,
which is just milliliter.
Here we have our final answer, 162 milliliters.
That's the answer to the problem.
In this video, we learned about the connection
between chemical equations and stoichiometry.
We learned how to use stoichiometry to solve numerical problems.
There will be many more examples in the exercises.
This video explains how to use stoichiometry to solve chemical problems. It begins by introducing the concept of stoichiometry, which is the process of using a balanced chemical equation to determine the number of moles of reactants and products. It then explains how to use conversion factors to convert moles of reactants to moles of products. An example is given to illustrate how to solve a problem involving the combustion of glucose. Finally, the video explains how to use the density of water to convert mass to volume.
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