## korosh23 one year ago Pressure can be used only for gaseous reactants? Does it mean all the reactants have to be gaseous?

1. korosh23

@Photon336

2. korosh23

For example, C(s) + O2 --> CO2 (g) Can pressure affect this heterogeneous reaction or do all the reactants need to be gaseous?

3. Photon336

$C(s) + O_{2}(g) -> CO_{2}(g)$

4. korosh23

Yes, as we know we have two words in chemistry. Total pressure of the system or partial pressure of a gas.

5. Photon336

well for a gaseous reaction it depends on the partial pressure exerted by each gas

6. Photon336

couple of things here to explain

7. Photon336

the first is that whenever you have a container of gases the pressure exerted by each individual gas will be proportional to the mole fraction of that gas. say you have 1 mole of A and 2 moles of B both gases first we find out how many total moles of gas we have that's 1+2 = 3 moles of gas. A + 2B --> AB_{2} and the pressure is 1 atm $Gas_{A} \frac{ 1 }{ 2 }*1atm + Gas_{B}\frac{ 2 }{ 3 }*1atm = total pressure$

8. Photon336

Now second point

9. Photon336

do you know what the equilibrium constant expression is?

10. Photon336

wel, $\frac{ [C]^{a}[D]^{b}}{ [A]^{c}[B]^{d} } = K_{p}$

11. korosh23

Actually my teacher have not started that yet. She taught my calss kinetic reactions and kinetic energy

12. Photon336

what this expression tells us is that we take the concentration of something mol/L for each reactant and raise it to the power of the stoichiometric coefficient that means the number in front of our reactants and products in your balanced reaction. now this is what i believe you need to know: the fact here is that solids and pure liquids aren't included in this expression, because their concentrations don't change very much over time. So gases will always be included in this expression, and I guess maybe some aqueous solutions too. so for our reaction |dw:1442777278804:dw|

13. korosh23

@Photon336 , so that means the partial pressure of O2 (g) will affect the reaction rate of the heterogeneous reaction?

14. Photon336

I believe that's correct; equilibrium would be different though, I think

15. Photon336

This is an entirely different concept but like say if your reaction were at equilibrium, Increasing/decreasing the pressure wouldn't do anything to the equilibrium. C(s) + O2 --> CO2(g) why, because there's the same number of moles on both sides and C(s) doesn't matter.

16. korosh23

Exactly, same number of moles of gas.

17. Photon336

Yes @korosh23 exactly Last thing/different concept if you were asked for the rate constant or to determine the rate law something along the lines like k[A][B] = r you cant do this because you would need a chart for the concentrations of A, B. too. hope this helped.

18. korosh23

Ok, thank you. You are a very good chem tutor.