Question

Examine the graph below and determine the value of the energy of activation for the chemical process. 0 y=-11624x + 32.055 R2 = 1 2 In k -3 -5 -6 0.0028 0.0029 0.0032 0.0033 0.003 0.0031 1/T (1/K) -96654 J/mol +96654 J/mol - 11624 J/mol +11624 J/mol -32.055 J/mol +32.055 J/mol

Answer 1

Answer-

The equation for the line on the graph is given as:

y = -11624x + 32.055

And the slope of this equation is -11624.

Now arhenius equation is :

k = Ae^-E_a/RT

Take ln both side-

lnK = -E_a / RT + lnA

This equation resembles the straight line equation => y = -mx + c

Where, y = lnK , x = 1 / T

c = lnA

and m = -E_a / R

This means that slope of this equation is -E_a / R

Now we can equate this value to the value of the slope of the graph.

$\small \Rightarrow$

-E_a / R = -11624

$\small \Rightarrow$

E_a = 11624 x R

Now put R = 8.314 J / mol·K

$\small \Rightarrow$

E_a = 11624 x 8.314 = 96654 J/mol

So the right option is:

b) 96654 J/mol