1. Tardigrade
  2. Question
  3. Chemistry
  4. The Ea of reaction in the presence of catalyst is 5.25 kJ/mol and in the absence of catalyst is 8.314 kJ mol- 1 . What is the slope of the plot of lnk vs (1/T) in the absence of catalyst. ( R =8.314 J k -1 mol -1)

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Q. The $E_{a}$ of reaction in the presence of catalyst is 5.25 kJ/mol and in the absence of catalyst is 8.314 kJ $mol^{- 1}$ . What is the slope of the plot of lnk vs $\frac{1}{T}$ in the absence of catalyst. $\left( R =8.314 J k ^{-1} mol ^{-1}\right)$

NTA AbhyasNTA Abhyas 2020Chemical Kinetics

Solution:

ln k $=lnA-\frac{E_{a}}{R}\times \frac{1}{T}$

slope $=-\frac{E_{a}}{R}=\frac{- 8.3 \times 1 0^{3}}{8.3}=-1000$