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How would we represent the reaction of #SO_2# with #K_2Cr_2O_7# to give #SO_3#?

Answer 1

Naomi Aronson

We would assume that the sulfur dioxide is oxidized to sulfur trioxide.

#stackrel(IV+)"SO"_2 +H_2O rarr stackrel(VI+)"SO"_3 +2H^(+) +2e^(-)# #(1)#

Moreover, dichromate is converted to chromic ion.

#Cr_2O_7^(2-) +14H^(+) + 6e^(-) rarr 2Cr^(3+) + 7H_2O# #(2)#

And we take #3xx(1)+(2)# to eliminate the electrons:

#Cr_2O_7^(2-) +8H^(+) + 3SO_2 rarr 2Cr^(3+) + 3SO_3 + 4H_2O#

I'm not sure if the reaction would proceed as written, but industrially, the oxidant is oxygen gas. If the answer is no, we cannot accept the reaction. Are charge and mass balanced?

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Answer 2

Cameron Andrews

4 K₂SO₄ + 2 Cr₂(SO₄)₃ + 3 H₂SO₄ = 6 K₂Cr₂O₇ + 7 SO₂ + 6 H₂O

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Answer 3

Natalie Anton

The reaction of ( \text{SO}_2 ) with ( \text{K}_2\text{Cr}_2\text{O}_7 ) to give ( \text{SO}_3 ) can be represented by the following chemical equation:

[ \text{SO}_2 + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{SO}_3 + \text{Cr}_2(\text{SO}_4)_3 + \text{K}_2\text{SO}_4 + \text{H}_2\text{O} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.