D. B. Science College, Gondia
D. B. Science College, Gondia
D. B. Science College, Gondia
M(active) M(passive)
d ln 𝑘𝑐 ∆𝐸𝑎
= ………………………….(2)
𝑑𝑇 𝑅𝑇 2
Where, 𝑘𝑐 = Equilibrium constant
R = Gas constant
T = Absolute temperature
∆𝐸𝑎 = The change of energy content of the reaction
∆𝐸 = 𝐸1 − 𝐸2
𝑘𝐼
put equation (1) 𝑘𝑐 = in equation (2) , we get
𝑘2
d ln𝑘𝑘𝐼 ∆𝐸𝑎
2
= ( ∆𝐸 = 𝐸1 − 𝐸2 )
𝑑𝑇 𝑅𝑇 2
d ln 𝑘𝐼 d ln 𝑘2 𝐸1 −𝐸2
− = …………………….....(3)
𝑑𝑇 𝑑𝑇 𝑅𝑇 2
Van’t Hoff proposed the equation (3) can be split up into two equation
As follows,
d ln 𝑘𝐼 𝐸1
𝑑𝑇
= 𝑅𝑇 2
+ I …………………………(4)
d ln 𝑘2 𝐸
= 𝑅𝑇22 + I …………………………………….(5)
𝑑𝑇
d ln 𝑘𝐼 𝐸1
= ……………………………………….(6)
𝑑𝑇 𝑅𝑇 2
d ln 𝑘2 𝐸
𝑑𝑇
= 𝑅𝑇22 ……………………………………..(7)
𝐸
2.303 log k = 2.303log A - 𝑅𝑇𝑎
𝐸 𝑎
log k = logA - 2.303𝑅𝑇
1
Plot the graph of logKagainst temperature (𝑇) give straight lines with slope =
− 𝐸𝑎
2.303𝑅
Intercept = log A
LogK Slope =
−𝐸 𝑎
2.303𝑅
𝐸𝑎 = - (slope * 2.303R)
1
(𝑇)
………………
QUESTIONS ASKED IN UNIVERSITY EXAM
https://www.chemguide.co.uk/physical/basicrates/arrhenius
https://www.britannica.com/science/Arrhenius-equation
www.chem.tamu.edu/rgroup/hughbanks/courses/102/slides/slides
17_2.pdf