I had been asked by one of my chemistry teachers to find out about an equation. The equation is:

\(3S + 4O_{2} \to SO_{2} + 2SO_{3}\)

\(4S + 5O_{2} \to 2SO_{2} + 2SO_{3}\)

How is it possible than an equation can be balanced in two ways even when it is in the simplest form ?

Also this equation cannot be balanced by P.O.A.C. method.

Please anyone help me out and enlighten me on this topic.

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## Comments

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TopNewestSulphur is going into different oxidation states in sulphur dioxide and sulphur trioxide. So in the first equation one of the sulphur atoms is getting oxidised into sulphur dioxide and the other two into sulphur trioxide. In the second one two sulphur atoms each are going into sulphur dioxide and troxide. Any such reaction could be made with some \(x\) sulphur atoms getting oxidised into dioxide and \(y\) into trioxide. Now the question is how is this possible that different number of sulphur atoms are getting oxidised in the reactions. The answer is simple. It is because the above reaction is actually a combination of two reactions - one being \( S + O_2 \rightarrow SO_2 \) and the other being \( 2 S + 3 O_2 \rightarrow 2 SO_3 \). So if the first reaction is added two the second one you would obtain the first reaction given by your teacher. Figure out what happens when you add \(2\) times the first reaction to the second reaction.

Realize the idea?

Hope this helps.

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yeah.. after doing it gives another balanced equation that too in simplest form. So, this reaction as a whole cannot be possible without breaking it into two different reactions ?

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The reaction is

actuallya linear combination of two reactions. The question isn't whether the reaction is possible by breaking it into two different reactions.Log in to reply

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Nice explanation.

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Thanks.

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Mathematically, this can be explained as follows. Suppose that you can balance the equation as follows.

\(aS+bO_2\rightarrow cSO_2+dSO_3\)

Now, comparing the number of sulfur and oxygen atoms on both sides,

\(a=c+d\) and \(2b=2c+3d\)

Now suppose you want \(a,b,c,d\) to be integers. It is clear from the second equation that \(d\) must be even. Hence \(d=2D\). So \((a,b,c,d)=(c+2D,c+3D,c,2D)\) for any pair of positive integers \(c\) and \(D\). In your first reaction you have \((c,D)=(1,1)\) and in the second reaction, you have \((c,D)=(2,1)\). You can find other solutions for other values of \(c\) and \(d\).

I don't know if there is a chemical reason behind this.

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Your explanation doesn't make it clear why this actually happens.

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Well, I find Sudeep Salgia's answer to be more insightful, especially because he divides the equation into two separate equations. My answer basically shows you how to convert this into a number theoretic problem.

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IMO, You explained it better.

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IMO ?? What it means here ?

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