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anonymous

  • one year ago

Does the rate constant of a gaseous reaction depend on pressure? I know that it depends on temperature. Please help, I'm confused.

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  1. taramgrant0543664
    • one year ago
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    Temperature is the only thing that effects the rate constant. Pressure would affect the equilibrium.

  2. JFraser
    • one year ago
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    changing the pressure of a gas is equivalent to changing its concentration, which would change the overall \(rate\), but not the value of the \(rate \space constant\), k

  3. anonymous
    • one year ago
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    But in a closed container, when you increase the pressure (by injecting an inert gas), temperature also increases (according to Gas law ) that results in change in the rate constant K of a reaction. What about this phenomenon.??

  4. Photon336
    • one year ago
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    If we look at PV =nRT we can see that pressure and temperature are proportional to one another, so increasing the pressure means increasing the temperature if you keep the volume constant, so this phenomenon would result in a change in your rate constant for that reaction. To my knowledge, adding an inert gas neither affects your rate constant , nor the reaction because inert gases are largely un-reactive so they don't take part in the reaction, in essence it has no effect.

  5. anonymous
    • one year ago
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    But by adding an inert gas to the system at constant volume, that definetly increases the pressure. So, finally what may be generalisation for this type of question. What are the parameters on which rate constant depends on.??

  6. Photon336
    • one year ago
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    Adding an inert gas can affect the equilbrium, but only if you change the volume. Adding inert gas a system at equilibrium and constant volume, the total pressure will increase. Yeah, I agree with you there. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change. So when an inert gas is added to a system that's at equilibrium and constant volume there will be no effect on the equilibrium.

  7. Photon336
    • one year ago
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    I think with constant volume the pressure overall increased so I would think in that case, adding an inert gas wouldn't change the equilibrium. ( I mean the way I understood it is that the inert gas isn't participating in your reaction. The other case is at constant pressure PV = nRT When an inert gas is added to a system in equilibrium at constant pressure total volume will go up. # of moles per unit volume of various reactants and products will decrease. equilibrium will shift towards the direction in which there is increase in number of moles of gases. Side with greater # of moles.

  8. anonymous
    • one year ago
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    Thanks for the answer. But, I'm not talking about the equilibrium here, rather refering to rate constant of a irreversible reaction. I'm just rather curious to know, does pressure affects the rate constant of gaseous reaction? If yes, then in what all ways??

  9. Photon336
    • one year ago
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    you had asked about an inert gas before in a separate question, so that's why I responded with that answer. The only thing i can think about in terms of pressure affecting rate constant is through the Arrhenius equation: the ideal gas constant and temperature in the denominator. Ea/RT and PV=nRT pressure and temperature are proportional to one another, so an increase in one will increase the other, when keeping volume and n constant. (P is pressure, V is volume, n is moles, R is a constant value, and T is temperature) if you keep the moles and volume the same, an increase in P causes increases in (RT) and (in the arrhenius equation) Ea/(RT) is in the denominator in a negative fractional power, that means that as RT goes up, the rate constant goes up. so from that logic both of them can affect rate constant.

  10. anonymous
    • one year ago
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    okay thanks a lot !

  11. Photon336
    • one year ago
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    No problem any time!

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