Chemistry Coursework Rate Reaction Analysis

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If you look at the expressions in the table above, you should recognise that the initial rate is inversely proportional to the time taken.

In symbols: In experiments of this sort, you often just use 1/t as a measure of the initial rate without any further calculations.

You can then plot 1/t as a measure of rate against the varying concentrations of the reactant you are investigating.

If the reaction is first order with respect to that substance, then you would get a straight line.

If you were looking at the effect of the concentration of hydrogen peroxide on the rate, then you would have to change its concentration, but keep everything else constant.

The temperature would have to be kept constant, so would the total volume of the solution and the mass of manganese(IV) oxide.So you would convert all the values you had for rate into log(rate). I suspect that in the unlikely event of you needing it in an exam at this level, it would be given to you. You probably have to enter 2 and then press the log button, but on some calculators it might be the other way around.Convert all the values for [A] into log[A], and then plot the graph. All you need to do is find the log button on your calculator and use it to convert your numbers. If you do it the wrong way around, you will just get an error message. I wouldn't really recommend that you try to read it all in one go.Or it could be the time taken for a small measurable amount of precipitate to be formed.This could be a reaction between a metal and an acid, for example, or the catalytic decomposition of hydrogen peroxide.If you plotted the volume of gas given off against time, you would probably get the first graph below.The maths goes like this: If you have a reaction involving A, with an order of n with respect to A, the rate equation says: If you plotted log(rate) agains log[A], this second equation would plot as a straight line with slope n.If you measure the slope of this line, you get the order of the reaction. Note: Don't worry if you don't understand logs (logarithms), or how I got from the first equation to the second one!A measure of the rate of the reaction at any point is found by measuring the slope of the graph. Since we are interested in the initial rate, we would need the slope at the very beginning.If you then look at the second graph, enlarging the very beginning of the first curve, you will see that it is approximately a straight line at that point.

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