1
$\begingroup$

In the reaction of combustion of gunpowder,

Sulfur is used to speed up the reaction and acts like a catalyst in that it increases the rate of reaction. However, unlike a catalyst, it is used up in the reaction.

I understand that transition metals are often used as catalysts but why can sulfur act like a catalyst?

$\endgroup$
3
$\begingroup$

It's not a catalyst here. It speeds up the reaction, which is why it acts like a catalyst.

Here is the reaction without sulfur:

$\ce{6 KNO3 + C7H8O -> 3 K2CO3 + 4 CO2 + 2 H2O + 3 N2}$

And here it is with sulfur:

$\ce{4 KNO3 + C7H4O + 2 S —> 2 K2S + 4 CO2 + 3 CO + 2 H2O + 2 N2}$

(Source)

Sulfur decreases the ignition temperature, as well as probably contributing a bit to the rate. Here, it is effectively turning $\ce{CO2}$ and $\ce{K2CO3}$ to $\ce{CO}$ and $\ce{K2S}$. If you take into account Le Chatelier's principle, the conversion of the products of the basic reactants speeds it up. (The decrease in concentrations/activities of the products due to their conversion means that the equilibrium shifts in the forward direction to compensate for it.)

So it is speeding up the reaction and making it easier to carry out. Not a catalyst.

A note: There's no reason why any element/compound can't act as a catalyst. Transition metals are generally catalysts in a certain type of reaction, and they generally catalyze the reaction by adsorbing the gases. Catalysts can work in many other ways—generally, they form intermediates in the reaction and present an alternative path via the intermediate. ($\ce{V2O5}$, for example, is a catalyst in many reactions involving organic compounds since it forms cyclic intermediates with various groups).

$\endgroup$
  • $\begingroup$ Can you elaborate on how Le Chateleir's principle comes in? I didn't get that part :-) $\endgroup$ – Brian Feb 17 '13 at 15:08
  • $\begingroup$ @Brian: done. I'm not really sure if it has an effect here, though. It may just be a thermodynamic phenomenon. $\endgroup$ – ManishEarth Feb 17 '13 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.