Heat Stoichiometry

What do we do when a stoichiometry problem involves heat? Here's an example:
{Note - for this example, I'm using very limited sig figs on purpose to make the math easier to follow.}

28g of nitrogen gas reacts with 7g of hydrogen gas to produce ammonia gas. How much heat is liberated by this reaction?

Approach it just like any other stoichiometry problem following the 4 steps (review here):
1. Write a balanced chemical equation. The only twist here is that since we're dealing with heat, we'll also need the {delta}H for the reaction as well.

N₂(g)  +  3 H₂(g)   -->   2 NH₃(g)

{delta}H_rxn = 1(0 kJ/mol) + 3(0 kJ/mol) + 2(-46 kJ/mol) = -92 kJ/mol

2. Find moles. Again, there's a little twist here... this is a limiting reactant problem. So let's find moles of ammonia from each reactant:

(28g / 28g/mol N₂) (2mols NH₃(g) / 1mol N₂(g)) (17g/mol NH₃) = 34g NH₃(g)

(7g / 2g/mol N₂) (2mols NH₃(g) / 3mol N₂(g)) (17g/mol NH₃) = 40.g NH₃(g)

So we know that N₂(g) is the limiting reactant.

3. Use the mole ratio from the balanced chemical equation to find "moles of interest". We already kind of blew past this in the last step... let's back up a bit now that we know the limiting reactant.

(28g / 28g/mol N₂) (1mol of "reaction" / 1mol N₂(g)) = 1 "mole of reaction"

{What's a "mole of reaction"?}

4. Convert moles of interest to whatever you're looking for:

(28g / 28g/mol N₂) (1mol of "reaction" / 1mol N₂(g)) (92kJ/mole of reaction) = 92kJ liberated

Same process, different details. Learn the process and the details will seem less detailed.

Balanced chemical formulas

An email question:
------------------------------

I just had some confusion regarding the homework for Chapter 2. I am confused about how to get the ions for the balanced chemical formulas. An example that I got incorrect was: 

The balanced chemical formula for magnesium chlorate contains___magnesium ion(s) and___chlorate ion(s)

I'm not sure how to go about getting the answer for this example. 

------------------------------

Balancing chemical formulas is a critical skill, but it's also one that many students struggle with until that magic moment when it clicks. Let's see if we can break this question apart a bit...

Many monoatomic ions have charges that can be reasonably predicted based upon where they are on the Periodic Table. Basically, we know from observation that noble gases have very stable numbers of electrons (we'll get deeper into that later in the semester...), so other elements in the Periodic Table tend to lose or gain electrons to get to the same number of electrons as a noble gas. This is especially true when the neutral element has a number of electrons that's pretty close to the number in the noble gas.

Metals:

The alkali metals (lithium, sodium, potassium, etc) almost always lose 1 electron to become +1 in ionic formulas, the alkaline earth metals (beryllium, magnesium, calcium, etc) are almost always +2, aluminum and gallium are almost always +3.

Transition metals and the metals toward the bottom of the B/C/N/O/F columns can have more than one common charge, so those usually have to be specified with a roman numeral; nickel(III) is +3, ruthenium(II) is +2, manganese(V) is +5, etc. The one common exception is zinc, which is (almost) always +2 in ionic compounds.

Non-Metals:

These have similar trends, but now as anions. Halogens (fluorine, chlorine, etc) gain an electron to become -1 charged halides in ionic formulas; chalcogens (oxygen, sulfur, etc) gain 2 electrons to become -2; pnictogens (N, P, As, etc) gain 3 electrons to become -3.

In the context of this specific problem, it's reasonable to assume that the magnesium is Mg⁺².

The other half of this specific problem is a polyatomic ion. There are a number of trends that we could explore in polyatomic ions, but at this point in this course, this is something you should just memorize. Find the list of polyatomic ions in your book, make up some flashcards, and just drill these into your brain. Chlorate is ClO₃^-1.

When writing balanced ionic formulas, the "balance" refers to the balance of charge... nature does not allow random positive or negative charges to be running loose, so for every positive charge in an ionic formula, we must have a corresponding negative charge. Since we have a +2 cation and a -1 anion, we need two anions so the +2 charge of the cation is "balanced" by two negative charges from the anions.

Balance in all things. A yin for every yang.

More Email Questions...

A couple more email questions:

-----Question-----
On exam b spring 2011 #5 why should this be changed to Ka instead of Kb? Also on exam b spring 2010 #2 the 3rd line its says that HCl  works as an effective buffer. I thought an effective buffer had to be a weak conjugate acid/base?
-----Answer-----
Here's question #5

This is K_a because ammonium ion is a conjugate acid. You could also do this as a K_b equilibrium, but you'd be starting with products and shifting to the left to form reactants. Either way should give the same answer.

Here's #2:

Correct, an effective buffer is an approximately equimolar combination of a weak conjugate acid and its weak conjugate base. In the 3rd line here, HCl(aq) is protonating the carbonate in solution 1.5 times so the resulting mixture is 0.64mols HCO₃^-1(aq) and 0.64mols H₂CO₃(aq). This is an equimolar mixture of a weak acid and its conjugate base, so it should be a good buffer. {NOTE: because carbonic acid decomposes and the resulting CO₂ can escape from solution, this might not be the best buffer in the real world, but it works fine as a sample problem.} This is similar to how you are making the carbonate buffer that you will measure in lab this week.

Problem Set 2

There's a key posted for the problem set we did in class that was titled "Problem Set #2" on my Gen Chem webpage:
http://www.drbodwin.com/teaching/genchem.php
Direct link to the key:  http://www.drbodwin.com/teaching/problemsets/c150gps02k.pdf

Don't forget, there are also old exams posted at:
http://www.drbodwin.com/teaching/examarchive.php

Other questions, let me know…

Redox Lab question...

A few people have asked about the chemical reactions for the Redox lab hand-in {Redox Hand-in}, and this is often a source of confusion for students, so let me answer it here. First of all, I think a number of people get confused because you're trying to over-think the question and make it more complex than it really is.

So you need to draw a voltaic cell... you can draw this by hand, or you can draw it electronically. DO NOT just find one online and copy-paste it into your hand-in, if that's all I wanted I would have pasted this {voltaic cell} into the hand-in myself before I posted it.

The part that causes some confusion is the "...write a correctly balanced net ionic equation for the spontaneous process..." Don't over-read that! The metal cation solutions you were using in lab were probably nitrate salts, but nitrate (or whatever anion might have been present) was a spectator in all of your reactions. Net ionic equations are actually easier than full-formula equations because they're not cluttered up with with a bunch of extra stuff, and net ionic equations actually just describe the CHEMISTRY that's happening rather than distracting you with a bunch of spectator ions and species.

Now, I'm not going to write out a net ionic equation that's the exact answer to one that you have to write, but here's an example. Let's say I made up the voltaic cell Fe|Fe⁺³||Cd⁺²|Cd with the black/negative lead of my meter hooked up to the Fe(s) electrode and the red/positive lead connected to the Cd(s). The potential I measure is -0.32V. Because the measured potential is negative, the cell is running backwards, so the spontaneous cell reaction is Cd|Cd⁺²||Fe⁺³|Fe. Translating that into a reaction, we can write the two half-reactions as:

Cd(s) <=> Cd⁺²(aq) + 2 e^- 

3e^- + Fe⁺³(aq) <=> Fe(s)

Adding those up gives the overall (or "net") reaction:

3Cd(s) + 2 Fe⁺³(aq) <=> 2 Fe(s) + 3 Cd⁺²(aq)

{Remember to multiply each half-reaction by an appropriate integer to make all the electrons cancel...}

That's a "correctly balanced net ionic equation for the spontaneous process" in this case. Now go do that for all the cells you measured in the experiment.

Qualitative analysis of metal cations

This week in lab, you'll be using the chemical tests you observed last week to separate (and identify) the metal cations in an unknown mixture of cations. It's probably better to think about this as a separation rather than just an identification because you will be given 1 sample and through a series of chemical (or physical) steps you will end up with up to 5 different metals in 5 different container. Some tips:
1. Flow chart - You have to organize your procedure to do well. The logical way to do this is with a flow chart for this type of a problem... a sequence of steps with decision points and branches along the way. You might not be a "flow chart person", but it makes it MUCH easier to follow through on a logical set of tests if you become a bit of a flow chart person for this experiment. Here's an example of a flow chart for separating and identifying anions if you're looking for an example: http://chemlab.truman.edu/chemlab_backup/CHEM131Labs/QualFiles/Figure2.gif
2. Assume you have all 5 metals - OK, there's not much chance that you'll get the sample that has all 5 metals, but you might. Even if you don't, design your flow chart and approach so that it will work for any possible unknown in this experiment.
3. The first step is the key! - Any step in which you are potentially making 2 or more precipitates represents a potential problem because you need to be able to separate those solids from one another. If you can't separate the solids from one another, it's not a useful test. For example, chromate makes a BUNCH of precipitates with the metals we are using, but there's no way (in our list of chemical tests) to separate those solids from one another. Adding chromate would be a horrible first step in your flow chart, but it might be handy later on when you only have to test for the presence of a single metal cation.
4. Positive tests - You need to have a positive confirmation test for all the metals, do not try to infer the presence or absence of a specific metal cation based upon negative results.
5. Chemical equations - If you come to lab prepared with a good plan, the "wet" part of this lab will not take three hours. This is a lab report experiment and as part of your lab report, you will have to write out balanced chemical equations for ALL of the tests you performed in the first week of the experiment. {quit rolling your eyes, it's not that hard and it's good practice!} Take advantage of your time in lab to make sure you understand the chemical equations and ask your instructor and lab assistant for feedback.

Good luck on your unknowns. One final note on your flow chart: this is one of those experiments where there is not a single correct answer. There are probably half a dozen or more variations on the "correct" flow chart that all work well.

Titration question...

Email question:
------------
I am having trouble with this question from a previous exam: You find that 25.00mL of phosphorous acid requires 41.39 mL of Sodium hydroxide to reach the second equivalence point. What is the concentration of the phosphorous acid solution?

I understand the process of what to do in this question, but am really struggling understanding why we are supposed to use the equation that you used in the explanation: H3PO3(aq) +2 OH-1(aq)†(一)2O(l) +HPO3-2(aq)
Earlier in this problem, we broke down this into three different equations, and I don't understand why we don't use 
H2PO3-1 + OH-1 --> H2O + HPO3-2?
------------
The equation you list gets us from the first equivalence point to the second equivalence point. In the titration described in the equation, we're going from phosphoric acid all the way to the second equivalence point, so we're doing 2 steps of the potential 3 step process. To get to the second equivalence point, have to go through the following steps:
H₃PO₃(aq) + OH^-1(aq)(aq) <=> H₂O(l) + H₂PO₃^-1(aq)

H₂PO₃^-1(aq) + OH^-1(aq) <=> H₂O(l) + HPO₃^-2(aq)

So the whole 2-step process to get from the beginning of the titration to the second equivalence point has the net equation:

H₃PO₃(aq) + 2 OH^-1(aq) <=> 2 H₂O(l) + HPO₃^-2(aq)

That's where the 2:1 stoichiometry comes from.

Balancing Redox Reactions

Balancing redox reactions can be pretty simple for some system, but some redox reactions can be exceptionally challenging. To balance any redox reaction, we can follow a systematic set of steps, and every Gen Chem book happily provides a set of steps. In my experience, most books use rules that require a little bit of faith and function like a black-box. There are a LOT of places that mistakes can be made when balancing redox reactions, so I prefer to use rules that have built-in places to check my answer before I go through the whole process. Here they are:

Balancing redox rules (in acidic or neutral aqueous solutions):

1. Assign oxidation numbers to all atoms in the equation

2. Identify the oxidation and reduction half reaction

3. In each half reaction, balance all atoms except hydrogen and oxygen

4. In each half reaction, add electrons to the reactant or product side to balance the change in oxidation state. Note: you are not adding electrons to balance charge

5. In each half reaction, add water molecules to balance any oxygen atoms

6. In each half reaction, add H⁺(aq) to balance any hydrogen atoms

7. At this point, the half reactions should be balanced, check the charge balance to confirm

8. Multiply each half reaction by an appropriate integer to balance the electrons involved in the oxidation and reduction processes

9. Add the half reactions together

10. Again, the resulting reaction should be balanced, check the charge balance to confirm

11. Cancel out any spectator species, sit back and pat yourself on the back for writing such a lovely balanced redox equation.

These rules work well, BUT rely upon some assumptions. First, since we're using water and H⁺(aq), the reaction must be taking place in aqueous solution that is neutral or acidic. That's OK for most redox reactions in Gen Chem, but once in a while we'll run into a rxn that takes place in basicaqueous solution. How do we handle that? Think about it... basic solutions have excess (relatively speaking) hydroxide ions. Hydroxide ions react with H⁺(aq) ions to form water. If a reaction is taking place in basic aqueous solution, balance it according to the above rules and then add:

12. Add enough OH^-1(aq) to each side to react with all the H⁺(aq) that is present

13. Check charge balance

14. Cancel any excess water

As with any process, practice is the key, so practice balancing redox rxns, then practice a little more, and when you think you have it all figured out, practice a couple more times. We'll do some of that in class...