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)

Question

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) (A) 100 (B) -100 (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 5.25 kJ/mol and in the absence of catalyst is 8.314 kJ $m o l^{- 1}$ . What is the slope of the plot of lnk vs $\frac{1}{T}$ in the absence of catalyst. $(R = 8.314 J k^{- 1} m o l^{- 1})$

100

-100

-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 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 T1​ in the absence of catalyst. (R=8.314Jk−1mol−1)

100

-100

-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 5.25 kJ/mol and in the absence of catalyst is 8.314 kJ $m o l^{- 1}$ . What is the slope of the plot of lnk vs $\frac{1}{T}$ in the absence of catalyst. $(R = 8.314 J k^{- 1} m o l^{- 1})$

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

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