I don't understand this question? :/

Adorable98

OP

bump..

Original post by Adorable98

The natural log form of the Arrhenius equation (which you are given above) is of the same form as a straight line graph y = mx + c

lnk = -Ea/RT + lnA
y = mx + c

If the gradient is 'm' what are y and x?

Reply 3

Noj777

15

Original post by Adorable98

What's this question from?

I reckon it's choice D, as using logs with k will provide more suitable numbers for a scale, and 1/T is often used for rate (change in something per unit time)

Reply 4

Noj777

15

Original post by charco

The natural log form of the Arrhenius equation (which you are given above) is of the same form as a straight line graph y = mx + c

lnk = -Ea/RT + lnA
y = mx + c

If the gradient is 'm' what are y and x?

Yeah.... clear explanation.

Nicely done :smile:

Reply 5

Adorable98

OP

15

Original post by Noj777

What's this question from?

I reckon it's choice D, as using logs with k will provide more suitable numbers for a scale, and 1/T is often used for rate (change in something per unit time)

Edexcel A2 chemistry Jan 2013 q3
And thanks!! :smile:

Original post by charco

The natural log form of the Arrhenius equation (which you are given above) is of the same form as a straight line graph y = mx + c

lnk = -Ea/RT + lnA
y = mx + c

If the gradient is 'm' what are y and x?

I see, so y= ink K and x= 1/T

Thank you!! :smile:

I don't understand this question? :/

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