Rate Equations

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  • Created by: chunks-42
  • Created on: 09-01-16 11:55

Reaction Rate

  • The reaction rate is the change in the amount of reactants or products per unit time (normally per second). If the reactants are in solution, the rate will change in concentration per second and the units will be moldm-3.
  • If you draw a graph of the amount of reactant or product against time for a reaction, the rate at any point is given by the gradient at that point. If the graph's a curve, you have to draw a tangent to the curve and find the gradient of that.
  • A graph of the concentration of a reactant against time might look something like this:


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Rate Equation

  • Rate equations look ghastly, but all they're really telling you is how the rate is affected by the concentration of reactants. For a general reaction: A + B --> C + D, the rate equation is:      Rate = k [A]m[B]n
  • m and n are the orders of the reaction with respect to reactant A and B. m tells you how the concentration of reactant A affects the rate and n tells you the same for reactant B.

1. If [A] changes and the rate stays the same, the order of reaction with respect to A is 0.

2. If the rate is proportional to [A], then the order of reaction with respect to A is 1.

3. If the rate is proportional to [A]2, then the order of reaction with respect to A is 2.

  • The overall order to the reaction is m+n.
  • You can only find orders of reaction from experiments. You can't work them out from chemical equations.
  • k is the rate constant - the bigger it is, the faster the reaction. The rate constant is always the same for a certain reaction at a particular temperature - but if you increase the temperature, the rate constant rises too.
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Rate Equations 2

When you increase the temperature of a reaction, the rate of reaction increases - you're increasing the number of collisions between reactant molecules and also the energy of each collision. But the concentrations of the reactants and the orders of reaction stay the same. So the value of k must increase  for the rate equation to balance.

Example. The chemical equation below shows the acid-catalysed reaction between propanone and iodine. CH3COCH3 + I2 --> CH3COCH2I + H+ + I-

This reaction is first order with respect to propanone and H+ and zero order with respect to iodine. Write down the rate equation for this reaction.

The rate equation is : rate = k[CH3COCH3]1[H+]1[I2]0

But [X1] is usually written as [X], and [X] 0 equals 1 so is usually left out of the rate equation.

So you can simplify the rate equation to: rate = k[CH3COCH3][H+]

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Initial Rates Method

The initial rate of a reaction is the rate right at the start of the reaction. You can find this from a concentration - time graph by calculating the gradient of the tangent at time = 0.

Here's a quick explanation of how to use the initial rates method:

1. Repeat the experiment several times using different initial concentrations of reactants. You should usually only change one of the concentrations at a time, keeping the rest constant.

2. Calculate the initial rate for each experiment using the method above.

3. Finally, see how the initial concentrations affect the initial rates and figure out the order for each reactant. The example below shows you how to do this. One you know the orders, you can work out the rate equation.

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Rates Example

The table below shows the results of a series of initial rate experiments for the reaction:              NO + CO + O2 --> NO2 + CO2. The experiments were carried out at a constant temperature. Write down the rate equation for the reaction.

(table did not show properly, will add later on)

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Rate Example Answer

1. Look at experiments 1 and 2 - when [NO] triples 9and all the other concentrations stay constant) the rate is nine times faster, and nine =3 2. So the reaction is second order with respect [NO].

2. Look at experiments 1 and 3 - when [CO] doubles (but all the other concentrations stay constant), the rate stays the same. So the reaction is zero order with repect to [CO].

3. Look at experiments 1 and 4 - the rate of experiment 4 is four times faster than experiment 1. The reaction is second order with respect to [NO], so the rate will quadruple when you double [NO]. But in experiment 4, [O2] has also been doubled. As doubling  [O2] hasn't had any effect on the rate, the reaction must be zero order with respect [O2].

4. Now that you know the order with respect to each reactant you can write the rate equation :    rate = k[NO]2

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