Comproportionation Reactions Explained
Uncover the reverse of disproportionation! In this masterclass, Abhishek Sengar Sir explains how two reactants containing the same element in different oxidation states merge to form a single unified product at an intermediate oxidation state. Learn the step-by-step redox balancing trick to tackle these equations in minutes!
Video Lecture Broadcast
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In-Depth Lecture Notes & Summary
What is a Comproportionation Reaction?
A comproportionation reaction (also known as symproportionation) is a chemical process where two reactants containing the same element in different oxidation states react to form a single product where that element is in an intermediate oxidation state.
In simple mathematical terms, if we have two starting oxidation states $O_1$ and $O_2$ (where $O_1 < O_2$), the product's oxidation state $O_p$ will satisfy the inequality:
Comproportionation vs Disproportionation
Abhishek Sengar Sir explains that comproportionation is the exact reverse of disproportionation:
Disproportionation (Split)
A single reactant containing an element in an intermediate oxidation state split-reacts into two separate products representing higher and lower oxidation states.
$$\text{Reactant (Intermediate)} \to \text{Product (Lower)} + \text{Product (Higher)}$$
Comproportionation (Merger)
Two reactants carrying the same element in different high and low oxidation states merge chemically into a single product representing an intermediate state.
$$\text{Reactant (Lower)} + \text{Reactant (Higher)} \to \text{Product (Intermediate)}$$
The Classic Sulfur Case Study
Let's balance the reaction between hydrogen sulfide ($\text{H}_2\text{S}$) and sulfur dioxide ($\text{SO}_2$) to yield elemental sulfur ($\text{S}$):
Detailed Reaction Balancing Step
Step 1: Assign Oxidation States. Sulfur in $\text{H}_2\text{S}$ is in the $-2$ oxidation state. Sulfur in $\text{SO}_2$ is in the $+4$ state. The product is elemental sulfur ($\text{S}^0$), which is in the $0$ state.
Step 2: Trace n-factors.
• Oxidation: $\text{S}^{-2} \to \text{S}^0$ (Loss of $2e^-$ per sulfur atom)
• Reduction: $\text{S}^{+4} \to \text{S}^0$ (Gain of $4e^-$ per sulfur atom)
Step 3: Equalize Electronic Shift. To balance the $2e^-$ loss with the $4e^-$ gain, we must multiply the oxidized specie ($\text{H}_2\text{S}$) by 2. This gives:
Step 4: Balance Oxygen and Hydrogen. There are 2 Oxygen atoms on the LHS and 0 on the RHS. Add $2\text{H}_2\text{O}$ to the RHS to balance oxygen:
Verification check: LHS Hydrogen: $2 \times 2 = 4$; RHS Hydrogen: $2 \times 2 = 4$. LHS Oxygen: 2; RHS Oxygen: 2. LHS Sulfur: $2 + 1 = 3$; RHS Sulfur: 3. Both atoms and net charge ($0$) are perfectly balanced!
Comproportionation Solver
Select a comproportionation template and trace how different reactant states merge step-by-step!
Computing oxidation states...
Charge: 0
Charge: 0
Thermodynamic Differences
Check why these opposite chemical mechanisms take place during standard reactions.
Single molecule splits
Two molecules merge
$$\text{Element}_{\text{LHS}(1)} \neq \text{Element}_{\text{LHS}(2)} \implies \text{Element}_{\text{RHS}}$$
Lecture Supplementary Quiz
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Doubt with Comproportionation?
If you have doubts regarding symproportionation stoichiometry, mixed-valence reactions, or reverse shifts, email Abhishek Sir directly!
Email abhishek.sengar@chemca.in →
Nice explanation, this was helpful.
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