A reactant (A) forms two products:A⟶k1B; Activation energy EalA→k2 C Activation energy Ea2If Ea/2=2Ea1, then k1 and k2 are related as - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$k_{1} = 2 k_{2} e^{E_{a 2} / R T}$
B
$k_{1} = k_{2} e^{E_{a 1} / R T}$
C
$k_{2} = k_{1} e^{E_{a 2} / R T}$
D
$k_{1} = A k_{2} e^{E_{a 1} / R T}$

Solution:

We have $k_{1} = A_{1} e^{- E_{al} / R T}$ and $k_{2} = A_{2} e^{- E_{a 2} / R T} = A_{2} e^{- 2 E_{a 1} / R T}$

Hence $\frac{k_{1}}{k_{2}} = \frac{A_{1}}{A_{2}} \frac{e^{- E_{al} / R T}}{e^{- 2 E_{a 1} / R T}} = \frac{A_{1}}{A_{2}} e^{E_{a 1} / R T}$ or $k_{1} = (\frac{A_{1}}{A_{2}}) k_{2} e^{E_{a ∣} / R T} = A k_{2} e^{E_{al} / R T}$ .

A reactant (A) forms two products: $A \overset{k_{1}}{⟶} B;$ Activation energy $E_{al}$
$A \overset{}{\overset{k_{2}}{\to}} C$ Activation energy $E_{a 2}$
If $E_{a / 2} = 2 E_{a 1}$ , then $k_{1}$ and $k_{2}$ are related as

Solution:

Related content