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Sir, In zero order reaction we write the slope to be m=-k But according to the eqn we get i.e., t=1/k([Ro]-[R]) Therefore the slope must be m=1/k Why dont we write it so.

Sir,

In zero order reaction we write the slope to be

m=-k

But according to the eqn we get i.e., t=1/k([Ro]-[R])

Therefore the slope must be

m=1/k

Why dont we write it so.

Grade:

2 Answers

Pankaj
askIITians Faculty 131 Points
9 years ago

Integrated rate law for a zero order reaction is :

[R]=-kt +[R]0

When we compare the equation with the equation of a straight line i.e. y = mx + c,

we get slope of line = m = -k and Y intercept = [R]0

Slope is equal to –k for the plot [R] vs t

Same equation can also be written as

t = 1/k([Ro]-[R])

When compared with the equation of straight line y = mx + c

The slope would be 1/k for the plot t vs ([Ro]-[R]) and the y intercept would be 0 as c = 0

Hope it’s clear to you.


Thanks & Regards

Pankaj Singh

askIITians Faculty

Pankaj
askIITians Faculty 131 Points
9 years ago

Integrated rate law
for a zero order reaction is :

[R]=-kt+[R]0

When we compare the equation with the equation of a straight
line i.e. y = mx + c,

we get slope of line =
m = -k and Y intercept =[R]0

Slope is equal to –k for the plot [R] vs t

Same equation can also be written as

t = 1/k([Ro]-[R])

When
compared with the equation of straight liney = mx + c

The slope would be 1/k

for the plott vs ([Ro]-[R]) and the y intercept would be 0 as c = 0

Thanks and Regards

Pankaj Singh

askIITians Faculty

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