I need some Chemistry help, please?

Question

Hi.. I've just started a Chemistry course in school but I'm new to this subject and don't understand alot of the questions. Could someone please help me with these questions, showing the working or telling me how to do it, please, so I won't need to ask again? Thank you very much for any help!

I was given this graph;

http://i34.tinypic.com/105sjrc.jpg

and asked the following questions.

a) Calculate the time for reaction when the temperature is 20 degrees celsius (I answered 12.5 seconds here)

b) Find the temperature rise required to double the rate of reaction.

c) Calculate the rate of reaction at 50 degrees celsius.

d) Calculate the time taken for the reaction at i)10 degrees ii)20 degrees iii) 30 degrees

Answer 1

The Y axis is in units of s^-1. This means

inverse seconds = per second, or "amount of reaction per second"

a). At 20 degrees it looks like the reaction rate is about 0.09, so the total reaction time would be 1/0.09 = 11.1 seconds.

b). Since the graph is not a straight line, you can see that the temperature rise needed to double the rate changes a bit with temperature. It looks like about 11 degrees to go from 0.1 to 0.2 s^1 and about 12.5 degrees to go from 1.5s^-1 to 3 s^-1.

c). By eyeball from the chart, it's about 4.1s^-1. Another way to do it is to mathematically extrapolate the straight upper part of the chart. Going from 25 to 30 degrees, a 5 degree increase raises the reaction rate bya bit more than 0.05 s^-1. So a 25 degree increase from 25 to 50 degrees will raise the reaction rate by something over 0.25 s^-1 for a value at 50 degrees of perhaps 3.9s^1

d). Having already done 20 degrees you can just read the other values off the chart.

Comment: It puzzles me that you are asked to "calculate" things, when the chart is your only input.  What you are doing is reading it off the graph, which is not calculation. To calculate, you need to have or derive an equation for the curve. and it does not look as if you were given that or any means to get it.