Practical investigations
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I am investigating the rate of reaction between HCl and Mg. I have worked out that the rate equation is rate = k[HCl]2. I then kept the concentration constant at 0.04 mol dm-3 and varied the temperature. I then worked out the k value for each temperature and plotted the log of this on the y axis and 1/temp (in Kelvin) on the x axis. My gradient however is about -42 and when this is multiplied by 8.314 my calculated activation energy is 0.19!! I am very confused as to why my answer is so low, have I drawn the graph wrong? Thanks!
110208
Corrie writes ....
I can't quite see how multiplying your gradient of -42 by 8.314 gives you an activation energy of 0.19 - in kJ mol-1?
The value of your gradient looks odd - a frequent mistake is to forget about the 10-3 factor on the values of 1/temp. For instance, for 25oC, 1/temp = 1/298 = 3.4 x 10-3 and for 50oC it is 1/323 = 3.1 x 10-3. Thus, when working out your gradient, the change along the x-axis here would be 0.3 x 10-3, not just 0.3.
Another possible source of error are your values of 'k'. Usually all one plots along the y-axis are the values of log (1/time), without working a value of k itself. How did you arrive at values for 'k'?
Yet another common error occurs when log is used instead of ln for the y-axis. If log is used, then the slope = -Ea/(2.303 x R), not -Ea/R. Check which you used.
Hope this helps. If not, come back to us via Re:act, not email.
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updated: 12 February 2008
