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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. Photon336
    • one year ago
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    Rate law = k[a]^x[b]^y = r r is your reaction rate and k is your rate constant, which is obtained experimentally and is written based off of your slowest step in your reaction. Note that [a] and [b] are the concentrations of your reactants in your slow step. the thing is that for gases, to study how they change in concentration, one refers to the partial pressures exerted by those gases, meaning the total pressure of your system is equal to the total pressure*the mole fraction of that particular gas. changing the pressure may affect the equilibrium, but not K, b/c k depends on the temperature your system is at.

  2. Rushwr
    • one year ago
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    Hey reaction rate depends on the temperature. And also you can say that rate constant depend upon the activation energy. Because if the mechanism of a reaction is changed the rate constant also changes. Changing the mechanism is changing the activation energy. So that does matter.

  3. anonymous
    • one year ago
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    "Does the rate constant of a gaseous reaction depend on pressure? " Yes! Why? Because rate constants depend on temperature and concentration. What is concentration? It's a quantity measured in moles per liter. So look at the ideal gas equation: $$PV=nRT$$ If we divide both sides by the Volume, V we get an expression for the pressure: $$P = \frac{n}{V} RT$$ At constant temperature (that means T is just a constant like the constant R there) it means LITERALLY that Pressure \(P\) and Concentration \(\frac{n}{V}\) are the same, only multiples of each other. This should make sense. Increasing the concentration of gas particles in a can should make the pressure go up. SO recap: Reaction rate constant depends on temperature and concentration. Concentration of a gas is directly proportional to the pressure, so that means pressure has to affect it! Done! :D

  4. Photon336
    • one year ago
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    \[pV = nRT \] \[p = \frac{ nRT }{ V } \] n = mol/v = volume mol/L p = concentration since you kept it constant, they are the same.

  5. anonymous
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Photon336 n = mol/v = volume mol/L \(\color{blue}{\text{End of Quote}}\) This is not quite right. The units of n are mols and the units of V are liters. So if you use brackets like this [ ] to represent the units of that value: \[ [n] = mols \] \[ [V] = Liters \] \[ \left[ \frac{n}{V}\right] = \frac{mols}{liter} = molarity \] So molarity is the concentration, which when temperature is held constant is directly proportional to the pressure. Increase the pressure at constant temperature means you MUST have increased the concentration. Similarly this is the concept behind STP, having a standard temperature and pressure means you are holding both of these quantities constant, so the concentration is the same. \[\frac{P}{RT} = \frac{n}{V}\] And I believe they call this "Avogadro's law" but it's really just one ay of looking at this equation that applies to all ideal gasses! :D

  6. Photon336
    • one year ago
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    higher temperature implies, particles moving around at a much faster rate = hitting walls of container more = higher pressure. since temperature is kept constant, to increase pressure concentration must have gone up.

  7. Photon336
    • one year ago
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    \[k[A]^{Y}[B]^{X} = r \] K depends on concentration, so you argued that increasing concentration is synonymous with an increase in pressure at constant temperature, and must have increased k.

  8. Photon336
    • one year ago
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    Thanks for pointing that out

  9. anonymous
    • one year ago
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    Right, so k depends on temperature, and of course [A] and [B] are the concentrations. So everything's there.

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