Looking for classes? Ksquare Career Institute, Bengaluru →
During the kinetic study of the reaction $2A + B \rightarrow C + D$, following results were obtained:
Run I: [A] = 0.1 mol $L^{-1}$, [B] = 0.1 mol $L^{-1}$, initial rate of formation of D = $6.0\times10^{-3}$ mol $L^{-1}min^{-1}$
Run II: [A] = 0.3, [B] = 0.2, rate = $7.2\times10^{-2}$
Run III: [A] = 0.3, [B] = 0.4, rate = $2.88\times10^{-1}$
Run IV: [A] = 0.4, [B] = 0.1, rate = $2.40\times10^{-2}$
Based on the above data which one of the following is correct?
Run I: [A] = 0.1 mol $L^{-1}$, [B] = 0.1 mol $L^{-1}$, initial rate of formation of D = $6.0\times10^{-3}$ mol $L^{-1}min^{-1}$
Run II: [A] = 0.3, [B] = 0.2, rate = $7.2\times10^{-2}$
Run III: [A] = 0.3, [B] = 0.4, rate = $2.88\times10^{-1}$
Run IV: [A] = 0.4, [B] = 0.1, rate = $2.40\times10^{-2}$
Based on the above data which one of the following is correct?
A
rate = $k[A][B]^2$
B
rate = $k[A]^2[B]$
C
rate = $k[A][B]$
D
rate = $k[A]^2[B]^2$
Detailed Solution
Let rate = $k[A]^a[B]^b$.
Runs II and III have the same [A] (0.3) while [B] changes from 0.2 to 0.4: $\frac{7.2\times10^{-2}}{2.88\times10^{-1}} = \left(\frac{0.2}{0.4}\right)^b$
$\frac{1}{4} = \left(\frac{1}{2}\right)^b$, so b = 2
Runs I and IV have the same [B] (0.1) while [A] changes from 0.1 to 0.4: $\frac{6.0\times10^{-3}}{2.40\times10^{-2}} = \left(\frac{0.1}{0.4}\right)^a$
$\frac{1}{4} = \left(\frac{1}{4}\right)^a$, so a = 1
Rate law: rate = $k[A][B]^2$
Runs II and III have the same [A] (0.3) while [B] changes from 0.2 to 0.4: $\frac{7.2\times10^{-2}}{2.88\times10^{-1}} = \left(\frac{0.2}{0.4}\right)^b$
$\frac{1}{4} = \left(\frac{1}{2}\right)^b$, so b = 2
Runs I and IV have the same [B] (0.1) while [A] changes from 0.1 to 0.4: $\frac{6.0\times10^{-3}}{2.40\times10^{-2}} = \left(\frac{0.1}{0.4}\right)^a$
$\frac{1}{4} = \left(\frac{1}{4}\right)^a$, so a = 1
Rate law: rate = $k[A][B]^2$