Solve Kinetics Question: Zero Order Reaction at High Pressure

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SUMMARY

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. At high pressures, where ##kP>>>1##, the rate simplifies to a constant value, indicating a zero-order reaction. The discussion highlights the importance of the pressure value in determining the reaction order, confirming that at sufficiently high pressures, the reaction order approaches zero regardless of the constant ##k##.

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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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