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korosh23

  • one year ago

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

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  1. korosh23
    • one year ago
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    @Photon336

  2. korosh23
    • one year ago
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    For example, C(s) + O2 --> CO2 (g) Can pressure affect this heterogeneous reaction or do all the reactants need to be gaseous?

  3. Photon336
    • one year ago
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    \[C(s) + O_{2}(g) -> CO_{2}(g) \]

  4. korosh23
    • one year ago
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    Yes, as we know we have two words in chemistry. Total pressure of the system or partial pressure of a gas.

  5. Photon336
    • one year ago
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    well for a gaseous reaction it depends on the partial pressure exerted by each gas

  6. Photon336
    • one year ago
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    couple of things here to explain

  7. Photon336
    • one year ago
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    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
    • one year ago
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    Now second point

  9. Photon336
    • one year ago
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    do you know what the equilibrium constant expression is?

  10. Photon336
    • one year ago
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    wel, \[\frac{ [C]^{a}[D]^{b}}{ [A]^{c}[B]^{d} } = K_{p}\]

  11. korosh23
    • one year ago
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    Actually my teacher have not started that yet. She taught my calss kinetic reactions and kinetic energy

  12. Photon336
    • one year ago
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    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
    • one year ago
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    @Photon336 , so that means the partial pressure of O2 (g) will affect the reaction rate of the heterogeneous reaction?

  14. Photon336
    • one year ago
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    I believe that's correct; equilibrium would be different though, I think

  15. Photon336
    • one year ago
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    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
    • one year ago
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    Exactly, same number of moles of gas.

  17. Photon336
    • one year ago
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    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
    • one year ago
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    Ok, thank you. You are a very good chem tutor.

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