If one did not notice the mistake one could not find the solution by means of the classical approach. What would happen if we mistakenly mixed it up a bit? To solve this system is to find the value of variables a, b, c which are the desired coefficients. With mathematics in mind you can write such a simple equation with variables a, b, c denoting the coefficients to be found:Īll you have to do is balance the quantity for each element on the left and the right side of the equation. Use the calculator with the explanation box checked to see the output. To see how the program finds the solution let`s start from a simple example.Īll the examples can be cut and pasted into the calculator That students will find it very useful in determining the To prove that this algebraic technique works. It was a real shock when I realized the power of It provides better solutions than chemistry itself! SuchĪn idea might seem a bit strange for a chemist. Mathematics provides a general way to find reaction coefficients. Luckily it turns out that oxidation numbers and half-reaction are unnecessary! Another method exists that uses algebra to find the In some cases however it is hard to use this method because oxidation numbers cannot be unequivocally attributed. The classical approach uses oxidation numbers and many chemists cannot imagine any way to find the solution other than balancing half-reactions for the process. Took me 3 minutes to set up the algebraic equations and few more minutes to solve them all.Finding coefficients for chemical reactions is often complicated, especially in the case of redox processes. I mean, it will help in terms of keeping track on the number of atoms, but I’ll still need to do perform trial and error with the three atoms. It’s not going to be much better if I use Method 2 either. By the way, that’s EVERY single term on the right! I don’t know bout you, but I know I have better use of my time than to play trial and error with these 3 atoms. On the right, it’s in Fe 2(SO 4) 3, Cr 2(SO 4) 3, CO 2, H 2O, K 2SO 4 and KNO 3. On the left, it’s in K 2Cr 2O 7 and H 2SO 4. That’s considered the easier atom among the three. I mean look at K, on the left, it appears in K 4 and K 2Cr 2O 7, and on the right, I see it in K 2SO 4 and KNO 3. What’s 3 more atoms (K, S and O), right? WRONG!!! Those 3 that are left are tough and that’s because they appear in multiple terms all over the place. So far I have balanced Fe, C, N, H and Cr. Well, how about that? Is it my lucky day or what? There’s 2 on the left and 2 on the right. Well, look at that! It’s already balanced! Moving on to Cr. Gaining momentum and confidence… 3 down, 5 more to go. Piece of cake, kinda like C – 12 on the left and 1 on the right. That means I’ll place a “12” in front of CO 2.Ģ K 4 + K 2Cr 2O 7 + H 2SO 4 → Fe 2(SO 4) 3 + Cr 2(SO 4) 3 + 12 CO 2 + H 2O + K 2SO 4 + KNO 3 S looks tough since I see it in multiple terms on both sides. Great! I can do this….add a “2” on the left, in front of K 4, so now I have:Ģ K 4 + K 2Cr 2O 7 + H 2SO 4 → Fe 2(SO 4) 3 + Cr 2(SO 4) 3 + CO 2 + H 2O + K 2SO 4 + KNO 3.įe is balanced, for now. First question, how should I balance K? I know I’ll need to “double” the K counts on the right, but which term should I start with – K 2SO 4 or KNO 3? Not sure. Ok…what about on the right? Looks like a total 3 K on the right. I started with K since it’s the first atom I encounter from the left. How long do you think it will take to balance it? Which has 8 types of atoms in the equation: K, Fe, S, C, N, Cr, O and H. So, imagine if you have a real complex equation like this: K 4 + K 2Cr 2O 7 + H 2SO 4 → Fe 2(SO 4) 3 + Cr 2(SO 4) 3 + CO 2 + H 2O + K 2SO 4 + KNO 3 I mean, who wants to spend 30 minutes trying to balance ONE equation? Seriously, that 30 minutes is probably better-spent doing other things like watching YouTube, checking out latest posts in FB, Instagram, Twitter, uploading pictures to Snapchat, level up in that favorite game of yours or doing other things that will increase your happiness. There’s nothing wrong with that, but I think it gets a little daunting when it comes to having to keep track of multiple numbers of atoms (like 5 or more) and it will prolong the process unnecessarily. Most of the students I have worked with only knew Method 1 from their high school science/chemistry class. My quick google search seems to point towards Method 1 (by inspection or trial-error-method) and Method 2 (writing down atom counts) being the most popular methods when it comes to balancing equations. Last week I posted on the three methods to balance chemical equations.
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