Pressure can be used only for gaseous reactants? Does it mean all the reactants have to be gaseous?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Pressure can be used only for gaseous reactants? Does it mean all the reactants have to be gaseous?

Chemistry
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

For example, C(s) + O2 --> CO2 (g) Can pressure affect this heterogeneous reaction or do all the reactants need to be gaseous?
\[C(s) + O_{2}(g) -> CO_{2}(g) \]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Yes, as we know we have two words in chemistry. Total pressure of the system or partial pressure of a gas.
well for a gaseous reaction it depends on the partial pressure exerted by each gas
couple of things here to explain
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 \[Gas_{A} \frac{ 1 }{ 2 }*1atm + Gas_{B}\frac{ 2 }{ 3 }*1atm = total pressure \]
Now second point
do you know what the equilibrium constant expression is?
wel, \[\frac{ [C]^{a}[D]^{b}}{ [A]^{c}[B]^{d} } = K_{p}\]
Actually my teacher have not started that yet. She taught my calss kinetic reactions and kinetic energy
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|
@Photon336 , so that means the partial pressure of O2 (g) will affect the reaction rate of the heterogeneous reaction?
I believe that's correct; equilibrium would be different though, I think
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.
Exactly, same number of moles of gas.
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.
Ok, thank you. You are a very good chem tutor.

Not the answer you are looking for?

Search for more explanations.

Ask your own question