The Ea of reaction in the presence of catalyst is 4.15 KJ/mol and in the absence of catalyst is 8.3 KJ mol- 1 . What is the slope of the plot of lnk vs (1/T) in the absence of catalyst.

Question

The Ea of reaction in the presence of catalyst is 4.15 KJ/mol and in the absence of catalyst is 8.3 KJ mol- 1 . What is the slope of the plot of lnk vs (1/T) in the absence of catalyst. (A) +1 (B) -1 (C) +1000 (D) -1000

Tardigrade Admin · Accepted Answer

ln K =lnA-(Ea/R)× (1/T) slope =-(Ea/R)=(- 8.3 × 1 03/8.3)=-1000

Q. The $E_{a}$ of reaction in the presence of catalyst is 4.15 KJ/mol and in the absence of catalyst is 8.3 KJ $m o l^{- 1}$ . What is the slope of the plot of lnk vs $\frac{1}{T}$ in the absence of catalyst.

+1

-1

+1000

-1000

ln K $= l n A - \frac{E _{a}}{R} \times \frac{1}{T}$

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

Q. The Ea​ of reaction in the presence of catalyst is 4.15 KJ/mol and in the absence of catalyst is 8.3 KJ mol−1 . What is the slope of the plot of lnk vs T1​ in the absence of catalyst.

+1

-1

+1000

-1000

ln K =lnA−REa​​×T1​ slope =−REa​​=8.3−8.3×103​=−1000

Q. The $E_{a}$ of reaction in the presence of catalyst is 4.15 KJ/mol and in the absence of catalyst is 8.3 KJ $m o l^{- 1}$ . What is the slope of the plot of lnk vs $\frac{1}{T}$ in the absence of catalyst.

ln K $= l n A - \frac{E _{a}}{R} \times \frac{1}{T}$

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