Discussion Forum



1)

The slope of the straight line obtained by plotting log10k against 1/T represents what term?


A) $-E_{a}$

B) $-2.303E_{a}/R$

C) $-E_{a}/2.303R$

D) $-E_{a}/R$

Answer:

Option C

Explanation:

According to Arrhenius equation,

  $k= Ae^{-E_{a}/RT}$

 By taking log on both sides

$\log k= \log_{e}A-\frac{E_{a}}{RT}$

$\log_{10} k= \log_{10}A-\frac{E_{a}}{2.303RT}$

 Comparing it with y=mx+c

 A plot between log10 k and ($\frac{1}{T}$) is a straight line and slope is given by

1662021652_9.PNG

  Slope = $-\frac{E_{a}}{2.303R}$

                   ( slope has negative value)