# Do electric charges increase activation barriers?

The activation energy governs the rate of reaction, and is thought to be equal to the energy needed to break the bonds needed to initiate the reaction. But if we have a reaction like:

$$\ce{A +B -> C}$$

And another like:

$$\ce{A+ + B+ -> C^2+}$$

Would the former have a lower activation energy since there is no repulsion between the molecules? Assume the same bonds are being broken in each case.

• The very example of two reactions related in such a way, especially with the requirement of same bonds being broken, looks highly improbable and artificial. That being said, if such an example can be found, your reasoning should apply. – Ivan Neretin Apr 27 '17 at 21:00
• What I meant in general is what is the definition of activation energy? I've only been introduced to it in the form of the energy required to break bonds, but if we consider a reaction like the one with Iodide ions and S2O8 2- ions ( I think ) , then the normal explanation given is that the reaction does not proceed without a catalyst due to the high activation energy required to overcome the repulsion of like charges. – Saad Apr 27 '17 at 21:42
• @Saad based on your comment, the example you chose seems to obscure your real question. Also, to get Ivan to see your comment, you have to include "@Ivan Neretin" in your comment (you can only include one reference like this in a comment and it will only ping someone if they have made a comment on that particular post). – Tyberius Apr 28 '17 at 0:07
• Some of the answer to your question is given here, chemistry.stackexchange.com/questions/34909/…. The rate constants are affected by the charges on the molecules and so we may assume that this is equivalent to affecting the activation energy. – porphyrin Apr 28 '17 at 7:09