A) \[rate=k{{[A]}^{2}}[B]\]
B) \[rate=k[A][B]\]
C) \[rate=k{{[A]}^{2}}{{[B]}^{2}}\]
D) \[rate=k[A]{{[B]}^{2}}\]
Correct Answer: D
Solution :
Let the order of reaction with respect to A is x and with respect to B is y. Thus, (I) rate\[=k{{(.1)}^{x}}{{(0.1)}^{y}}=6.0\times {{10}^{-3}}\] (II) rate \[=k{{(0.3)}^{x}}{{(0.2)}^{y}}=7.2\times {{10}^{-2}}\] (III) rate \[=k{{(0.3)}^{x}}{{(0.30)}^{y}}=2.88\times {{10}^{-1}}\] (IV) rate\[=k{{(0.4)}^{x}}{{(0.1)}^{y}}=2.40\times {{10}^{-2}}\] On dividing Eq. (I) by (IV), we get \[{{\left( \frac{0.1}{0.4} \right)}^{x}}{{\left( \frac{0.1}{0.1} \right)}^{y}}=\frac{6.0\times {{10}^{-3}}}{2.4\times {{10}^{-2}}}\] or \[{{\left( \frac{1}{4} \right)}^{x}}={{\left( \frac{1}{4} \right)}^{1}}\] \[\therefore \] \[x=1\] On dividing Eq. (II) by (III), we get \[{{\left( \frac{0.3}{0.3} \right)}^{x}}{{\left( \frac{0.2}{0.4} \right)}^{y}}=\frac{7.2\times {{10}^{-2}}}{2.88\times {{10}^{-1}}}\] or \[{{\left( \frac{1}{2} \right)}^{x}}=\frac{1}{4}\] or \[{{\left( \frac{1}{2} \right)}^{y}}={{\left( \frac{1}{2} \right)}^{2}}\] \[\therefore \] \[y=2\] Thus, rate law is, rate \[=k{{[A]}^{1}}{{[B]}^{2}}\]or \[=k[A]{{[B]}^{2}}\]
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