George Lagalle I’m subbing in the y and x values for the blue point and the gradient to solve for A. Remember that in the equation for a line we multiply the gradient by the x value. Gradient is -13500 and x (1/T) is 3.10 x 10^-3. Make sense?
For the lnA question at the end, when you substituted 1/T as 3.10 x 10-3 in to the equation without converting it in to Kelvin by 1/T, was that because you used y= mx + c instead of lnk= -Ea/R. 1/t + lnA? :)
Sir I don't understand why we need to plot a graph can't we just use (x1,y1) and (x2,y2) to find the answer? If we draw a graph how big shall our triangle be?
Please could you run through the new OCR 2017 papers that’re now on the website? Thank you so much!
Thank you!!
Can we use Arrhenius to find temperature? Will ocr ask us to do that?
Never seen that!
MaChemGuy thank you!!
I don’t understand why you multiply the gradient by 3.10x10^-3, wouldn’t the gradient be -13500 x 8.314?
George Lagalle I’m subbing in the y and x values for the blue point and the gradient to solve for A. Remember that in the equation for a line we multiply the gradient by the x value. Gradient is -13500 and x (1/T) is 3.10 x 10^-3. Make sense?
For the lnA question at the end, when you substituted 1/T as 3.10 x 10-3 in to the equation without converting it in to Kelvin by 1/T, was that because you used y= mx + c instead of lnk= -Ea/R. 1/t + lnA? :)
Hi Mairead
well look who it is
hope you are not cramming ;)
does it give you the equations in AQA A level?
Sir I don't understand why we need to plot a graph can't we just use (x1,y1) and (x2,y2) to find the answer?
If we draw a graph how big shall our triangle be?
Part of the assessment is to show you can draw the graph and then use it to calculate the gradient. Triangle as big as possible
@@MaChemGuy thank you. You are a legend 👑
Sir, why is Ink against I/T graph is a downward straight line graph?
because the gradient is negative (-Ea/R)