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Following data is for a reaction between reactants A and B: Rate (mol L$^{-1}$ s$^{-1}$): $2\times10^{-3}$ at [A]=0.1 M, [B]=0.1 M; $4\times10^{-3}$ at [A]=0.2 M, [B]=0.1 M; $1.6\times10^{-2}$ at [A]=0.2 M, [B]=0.2 M. The order of the reaction with respect to A and B, respectively, are
A
1, 0
B
0, 1
C
1, 2
D
2, 1
Detailed Solution
Let Rate $=k[A]^x[B]^y$. Comparing rows 1 and 2 (B constant): $\dfrac{4\times10^{-3}}{2\times10^{-3}}=\left(\dfrac{0.2}{0.1}\right)^x\Rightarrow2=2^x\Rightarrow x=1$. Comparing rows 2 and 3 (A constant): $\dfrac{1.6\times10^{-2}}{4\times10^{-3}}=\left(\dfrac{0.2}{0.1}\right)^y\Rightarrow4=2^y\Rightarrow y=2$. So order w.r.t. A is 1 and w.r.t. B is 2.