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$$\ce{K2Cr2O7 + $x$\,H2SO4 + $y$\,SO2 -> Cr2(SO4)3 + K2SO4 + $z$\,H2O}$$

For $\ce{Cr}$ change in oxidation number is: $+12$ becomes $+6$

For $\ce{S}$ change in oxidation number is: $+4y$ becomes $+6$

For $\ce{H}$ change in oxidation number is: $+2$ becomes $+2z$

since $\ce{Cr}$ decreases by 6, $\ce{S}$ and $\ce{H}$ have to increase by 6, so

$$(4y+2)-(6+2z)=6$$

$$2y-z=5$$

Also using the principle of atom conservation (POAC) for $\ce{H}$ atoms from $\ce{H2SO4}$ to $\ce{H2O}$, we get:-

$$2x=2z$$

$$x=z$$

I can't find any more equations to solve this.

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  • $\begingroup$ Your equations for the changes in oxidation number are all wrong (apart from that of chromium). Hydrogen is not undergoing any redox anywhere. Why do you multiply sulfur in reactants by y, but in the products you just ignore the stoichiometric coefficient? You could just use the conservation of atoms for H, S and O, which gets you 3 equations in 3 unknowns which you can then solve. $\endgroup$ – orthocresol Nov 2 '15 at 10:15
  • $\begingroup$ POAC = Principle of Atom Conservation $\endgroup$ – MaxW Nov 4 '15 at 4:55
  • $\begingroup$ Change in oxidation number? Cr goes from +VI to +III, sulphur from +IV to +VI. I have an idea where you got your numbers from, but they are highly confusing at least to anybody who studied chemistry in my part of the world … $\endgroup$ – Jan Jan 3 '16 at 16:02
  • $\begingroup$ While calculating the change in oxidation number for $S$ why are we not considering the source from Sulphuric acid?According to me ideally why is it not $6x+2y$? $\endgroup$ – user586228 Aug 10 '19 at 19:22
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Never having heard of the Principle of Atom Conservation (POAC) before I'm curious.

Looking up the concept, I understand it simply to mean that atoms can't be created or destroyed in a chemical reaction. (So we're ignoring radioactive decay for example.) I start blindly with a variable for each reactant and product, and will then look for linear relationships in the variables.

$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $d$\,Cr2(SO4)3 + $e$\,K2SO4 + $f$\,H2O}$$

So with $a$ $\ce{K2Cr2O7}$ molecules, I must get $a$ $\ce{Cr2(SO4)3}$ molecules to balance $\ce{Cr}$, and $a$ $\ce{K2SO4}$ molecules to balance $\ce{K}$. So:

$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $a$\,Cr2(SO4)3 + $a$\,K2SO4 + $f$\,H2O}$$

Now $\ce{H}$ is only in $\ce{H2SO4}$ on the left and $\ce{H2O}$ on the right so $f=b$.

$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $a$\,Cr2(SO4)3 + $a$\,K2SO4 + $b$\,H2O}$$

To balance $\ce{S}$:

$$b + c = 4a$$

To balance $\ce{O}$:

$$7a + 4b + 2c = 16a + b$$ which reduces to $$3b +2c = 9a$$

At this point just balancing atoms leaves you with 2 linear equations and 3 unknowns. So there are not enough equations to explicitly solve the problem using linear equations. However relationships can be established.

\begin{array}{cccc}\\ 3b& +2c & = & 9a \\ -2(b& +c & = & 4a)\\ \hline & b & = &a \\ \end{array}

$\therefore c = 3a = 3b$

The final step in this problem is to realize that an integer solution (a Diophantine problem in mathematics) is desired. So $a=b=1$ and thus $c=3$. So the final chemical equation is:

$$\ce{K2Cr2O7 + H2SO4 + 3SO2 ->Cr2(SO4)3 + K2SO4 + H2O}$$


PS - Many thanks to user Mathew Mahindaratne for suggesting a cleaner solution in the comments.

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  • $\begingroup$ When I took chemistry I don't remember a high falutin name for this. I remember mostly doing this in my head to figure out where the sticking point was in balancing the equation. $\endgroup$ – MaxW Nov 4 '15 at 5:44
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    $\begingroup$ Indeed there are two linear equations in three unknowns. The solution to that "problem" is that you are free to multiply the entire chemical equation by any number you like and it would still be correct. For example, we could either write $$\ce{H2 + 1/2 O2 -> H2O}$$ or $$\ce{2H2 + O2 -> 2H2O}$$ and both would still be correct. Therefore, you have the freedom of arbitrarily choosing the value of one of the remaining three unknowns, say $a$, and finding the corresponding values of $b$ and $c$ using the last two equations. It's analogous to normalising a wavefunction in QM. $\endgroup$ – orthocresol Jan 3 '16 at 15:58
  • $\begingroup$ @orthocresol - Thanks for the comment. I had left the solution hanging. So I added a bit about needing an integer solution and finished the solution. To me the overall idea was that even using POAC you still got to a point where you more or less "guess" then check the guess. So $$\frac{1}{2}\ce{H2} + \frac{1}{4}\ce{O2 ->} \frac{1}{2}\ce{H2O}$$ might work in a math class, but not in a chemistry class. $\endgroup$ – MaxW Jan 3 '16 at 19:18
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    $\begingroup$ @MaxW: Assuming OP is not familiar with mathematics, it'd be nicer to show $a = b$ and $c=3a=3b$ by simple mathematics using your two linear equations. $\endgroup$ – Mathew Mahindaratne Apr 18 at 10:46
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    $\begingroup$ @MathewMahindaratne - Thanks for the suggestion! I changed the solution and it is a cleaner result. $\endgroup$ – MaxW Apr 18 at 18:08

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