Solve Kinetics Question: Zero Order Reaction at High Pressure

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The discussion centers on determining the order of a reaction for a gas decomposition process governed by the rate law Rate ∝ kP/(1+kP). At high pressure, it is established that kP becomes significantly larger than 1, leading to a simplification where the rate becomes constant, indicating a zero-order reaction. However, a participant questions the assumption that kP will always be much greater than 1 without knowing the value of k, suggesting that if k is very small, the reaction order could differ. Ultimately, it is agreed that while k is important at lower pressures, at sufficiently high pressures, the reaction order will approach zero regardless of k's specific value. The consensus is that high pressure dictates the reaction's zero-order behavior.
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Homework Statement


The question says:
The decomposition of a gas on a surface follows the rate law : Rate is proportional to ##\frac {kP}{1+kP}##, where ##P## is the pressure and ##k## is a constant. Find the order of the reaction at high pressure.

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The Attempt at a Solution


In the solution they have said that since ##P## is high therefore ##kP>>>1## due to which rate is proportional to ##\frac {kP}{kP}=1## which means the rate is a constant so it is a zero order reaction. What I say is that since they don't known the value of ##k## how can they come to the conclusion that ##kP>>>1##? If we assume k to be equal to a very very small number such that ##1>>>kP## then their assumption fails, and the rate is not a constant.
So am I right about this thing or making a mistake?
 
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If doesn't matter what the value of k is. If you take a high enough pressure, the order will approach zero.

chet
 
Chestermiller said:
If doesn't matter what the value of k is. If you take a high enough pressure, the order will approach zero.

chet
But if we leave that proven part (I. e if pressure is increased order will tend towards zero) and use only this :
Rate is proportional to $$\frac {kP}{1+kP} $$
Then isn't the value of ##k## important?
 
mooncrater said:
But if we leave that proven part (I. e if pressure is increased order will tend towards zero) and use only this :
Rate is proportional to $$\frac {kP}{1+kP} $$
Then isn't the value of ##k## important?
Sure. At pressures less than on the order of 1/k.

Chet
 
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