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Rates of reaction?

  1. Jul 31, 2004 #1
    we have just started this topic and i dont believe that i undestand it very well.

    anyone has any usefull info for me that i should know.
    who knows of a good website for all this?
    and what are some factors that affect change in equillibrium? what, how, why?

    thanx for the help.
     
  2. jcsd
  3. Jul 31, 2004 #2
    Adding heat, increasing the concentration, or increasing the pressure (for gases) would increase the rate of a reaction. Collision theory basically says the the more that molecules run into each other the more the reaction proceeds. So adding heat for most reactions will make the rxn go faster because heat gives molecules more kinetic energy. Increasing the concentration obviously gives a greater chance that molecules will run into each because there are more available. Pressure pushes gas molecules closer together so they have a better chance of running into each other.

    As for the rate laws, it depends on what kind of reaction that you have. There are 2 different ways you will see a rate, in the differential and integrated form. They mean the same thing though. For example
    consider the reaction
    aA-------->bB (a and b are the coefficients)

    The velocity of the reaction can be described most easily in the diffential form. v=-(1/a)dA/dt=(1/b)dB/dt (A means concentration of A,B means concentration of B, t is time). You have the negative sign for A because the starting material is going away, and there is no negative sign for the product B because it is being formed. If the reaction exhibits a 0 order then the velocity of the reaction is v=-dA/dt=k where k is the rate constant By multiplying both sides by dt you have -dA=kdt. Then by using a little bit of calculus you can show that A=Ao-kt (Ao is the initial concentration). For a first order reaction you do the same thing except v=-dA/dt=kA. By the same procedure you can show that the integrated form is A=Aoe^-kt. Same thing goes for a second order reaction, v=-dA/dt=kA^2. The integrated form for this type of reaction A=Ao/1+ktAo. The order of a reaction can only be found experimentally, your teacher must tell you the order or give you data so you can figure out the order of the reaction. For other reactions such as
    A+B--->C or for an equilibrium reaction the rate laws are different. Your book should provide you with them. One other point to note is like I said before, reaction rates depend on temperature. The rate constant k is dependent on temperature. The Arrhenius equation is used to calculate the rate constant if temperature is given. k=Ae^(-Ea/RT). A is the frequency or pre-exponential factor (i.e. its just a constant),Ea is the activation energy, R the gas constant, T the temperature in K.
     
  4. Jul 31, 2004 #3

    GCT

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    If you are referring to the equilibrium constant, increasing the temperature is the only means to change it since it means changes the each of the rate constants (forward and backward) in disproportionate degrees.


    -------
    Online help with chemistry
    http://groups.msn.com/GeneralChemistryHomework
     
  5. Jul 31, 2004 #4
    thanks alot
     
  6. Jul 31, 2004 #5
    Sometimes the rate constants are given at two different temperatures (or their ratio at two different temperatures is given) and you are asked to find the activation energy, using the Arrhenius Equation (this eliminates the need to know the pre-exponential factor A).

    Cheers
    Vivek
     
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