Effect of Pressure on the Initial Rate

[Apple96]

Hi,

Let's suppose an elementary reaction of the low-pressure mixture has the initial reaction rate of x (2A + B --> A2B). What would be the initial reaction rate of the high-pressure mixture? (Imagine a mass M is placed upon the beaker filled with the mixture, and a mass 3M is placed upon another beaker).

I'm really not sure how to approach this problem. Yes, more pressure means more concentration (exactly 3 times), but I'm not sure how this affects the initial rate. Concentration can be used to compare the reaction rate of molecules involving in one equation (like 2A+ B --> A2B, the reaction rate of A will be twice as much as that of B). But, between the two different system? I'm really not sure. Please help!

Thank you,

 

[jbsteele]

The initial reaction rate of the high-pressure mixture will depend on the type of reaction that is taking place. If the reaction is a first-order reaction, then the initial reaction rate will increase with pressure (by a factor of 3 in this case). If the reaction is a second-order reaction, then the initial reaction rate will be independent of pressure and remain the same. If the reaction is a zero-order reaction, then the initial reaction rate will decrease with pressure (by a factor of 3 in this case).

 

[jbsteele]

The answer to your question depends on the reaction mechanism and whether pressure has a direct effect on the rate of reaction. If the pressure has an effect, then the initial reaction rate of the high-pressure mixture will be three times that of the low-pressure mixture. This is because when you increase the pressure, the molecules are closer together, making it easier for them to react. However, if the reaction mechanism does not involve pressure, then the initial reaction rate of the high-pressure mixture will be the same as that of the low-pressure mixture.