When a graph is plotted between log K and ( (1/T) ), the slope of line obtained represents: (K = rate constant, T = temperature)

Question

When a graph is plotted between log K and ( (1/T) ), the slope of line obtained represents: (K = rate constant, T = temperature) (A)  (Ea/2.303R)  (B)  -(Ea/R)  (C)  -(Ea/2.303R)  (D)  log A

Tardigrade Admin · Accepted Answer

The Arrhenius equation can be written as log K= log A-(E/2.303RT) On comparing this equation with general equation of a straight line, y=mx+c We get, y= log K, x=(1/T), m=-(E/2.303R), c= log A i.e., if we plot a graph between log K (at Y-axis) and (1/T) (at X-axis), then the slope of the line obtained will be equal to -(e/2.303 R).

Q. When a graph is plotted between log K and $(\frac{1}{T}),$ the slope of line obtained represents: (K = rate constant, T = temperature)

$\frac{E _{a}}{2.303 R}$

$- \frac{E _{a}}{R}$

$- \frac{E _{a}}{2.303 R}$

$lo g A$

Q. When a graph is plotted between log K and (T1​), the slope of line obtained represents: (K = rate constant, T = temperature)

2.303REa​​

−REa​​

−2.303REa​​

logA

Q. When a graph is plotted between log K and $(\frac{1}{T}),$ the slope of line obtained represents: (K = rate constant, T = temperature)

$\frac{E _{a}}{2.303 R}$

$- \frac{E _{a}}{R}$

$- \frac{E _{a}}{2.303 R}$

$lo g A$