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Topic: **Use of maths coursework****Question:**
I need help please for my chemistry A level coursework i am planning to investigate how activation energy changes with different acids
the reaction will be between acids and metals (probably magnesium) and i will look at monobasic and dibasic acids (HCl and H2SO4)
I'm not really sure how to investigate this whether just to time how long it takes for the metal to dissolve or if there is a better method to use any ideas will be helpful
and i know i need to use k = Ae^-Ea/RT and plot a graph of lnk = -Ea/RT + LnA but i dont know what A is
sorry if this is a bit long i need to get this sorted before i go back to school and i am totally stuck
how do you work out what it is though

June 16, 2019 / By Sachie

I think your program is too ambitious. You intend to enter the field of chemical kinetics of very fast reactions. A German chemist, Manfred Eigen ( see:http://en.wikipedia.org/wiki/Manfred_Eig... , was the first to determine such rates and reaction laws with all parameters. He had to employ high energy pulses to acidic solution and brought system out of equilibrium for a short time and observed how the system reached equilibrium again. The method he employed is nowadays called relaxation method. This requires a lot of materials and a excellent knowledge of math. Look for a slower reaction you can handle.

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I think your program is too ambitious. You intend to enter the field of chemical kinetics of very fast reactions. A German chemist, Manfred Eigen ( see:http://en.wikipedia.org/wiki/Manfred_Eig... , was the first to determine such rates and reaction laws with all parameters. He had to employ high energy pulses to acidic solution and brought system out of equilibrium for a short time and observed how the system reached equilibrium again. The method he employed is nowadays called relaxation method. This requires a lot of materials and a excellent knowledge of math. Look for a slower reaction you can handle.

properly on the grounds that i'm a extensive fan of calculators i will clarify the thank you to do it with a Ti-80 3 calculator. Press "stat", then "a million:edit." the place it says "L1" it is the place you will enter your x values. enter the values for lnk into the the place it says "L1" and enter the values for a million/T into the place it says "L2". i will provide you an occasion. For the 1st one ok=4.80 one×10^-4 T=282K so which you will possibly enter ln(4.81x10^-4) into "L1" and a million/282 into "L2" try this for all the ok values and their corresponding temperatures. The slope is -Ea/RT, the place R=8.31 J/mol*ok. once you have entered all those factors, press "2nd" then "mode/stop." pass to "y=" and press enter the place it says "plot a million." Then pass to "stat," "calc" and scroll right down to "4:LinReg(ax+b)". Press enter as quickly as then press "vars," "y-vars" and then press "a million:function." Press enter two times and you will possibly desire to get a cost for a and b and you will possibly be able to get an r and r² cost yet those are not too substantial. a=-Ea/R b=lnA From there you in basic terms use some uncomplicated algebra to be certain something. in case you decide on help with any of this you are able to pass to my profile and e mail me.

properly on the grounds that i'm a extensive fan of calculators i will clarify the thank you to do it with a Ti-80 3 calculator. Press "stat", then "a million:edit." the place it says "L1" it is the place you will enter your x values. enter the values for lnk into the the place it says "L1" and enter the values for a million/T into the place it says "L2". i will provide you an occasion. For the 1st one ok=4.80 one×10^-4 T=282K so which you will possibly enter ln(4.81x10^-4) into "L1" and a million/282 into "L2" try this for all the ok values and their corresponding temperatures. The slope is -Ea/RT, the place R=8.31 J/mol*ok. once you have entered all those factors, press "2nd" then "mode/stop." pass to "y=" and press enter the place it says "plot a million." Then pass to "stat," "calc" and scroll right down to "4:LinReg(ax+b)". Press enter as quickly as then press "vars," "y-vars" and then press "a million:function." Press enter two times and you will possibly desire to get a cost for a and b and you will possibly be able to get an r and r² cost yet those are not too substantial. a=-Ea/R b=lnA From there you in basic terms use some uncomplicated algebra to be certain something. in case you decide on help with any of this you are able to pass to my profile and e mail me.

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A is the pre-exponential factor, or constant, and is experimentally determined or calculated knowing the other factors.

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Since you want to graph this, I'll have x be the number of chairs and y be the cost. The fact that all of this occurs in one day doesn't affect any of the calculations, so I'll leave that out of the work. The standard slope-intercept formula for a line is: y = mx + b where m is the slope and b is an unknown constant. m can also be expressed as the change in y divided by the change in x. In this problem that is the change in cost divided by the change in the number of chairs. m = (4800 - 2200) / (300 - 100) m= 2600/ 200 = 13 Now the equation looks like this: y = 13x + b. To solve for b I'm going to substitute x = 100, y = 2200 into that equation. 2200 = (13 * 100) + b...........I am using * as the multiplication symbol. 2200 = 1300 + b 2200 - 1300 = b 900 = b b = 900 Summarizing: 1) The equation is y = 13x + 900 2) The slope was already determined to be 13. That represents the change in y divided by the change in x. In this case that would be the change in cost divided by the change in the number of chairs manufactured. 3) The y-intercept is the point at which x = 0 for this equation. It would be the cost of making no chairs. It really does not have any useful meaning in this problem, but to solve for it I'll substitute x = 0 into the equation. y = (13 * 0) + 900 y = 900 The y-intercept is (0, 900) <<<<<

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