ext{The data for the reaction } A+B \longrightarrow \mathrm{C} ext{, is}
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\begin{array}{lccc}
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Question

ext{The data for the reaction } A+B \longrightarrow \mathrm{C}	ext{, is}
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\begin{array}{lccc}
	ext{Exp.} & [A]_{0} & [B]_{0} & 	ext{Initial rate} \
1 & 0.012 & 0.035 & 0.10 \
2 & 0.024 & 0.070 & 0.80 \
3 & 0.024 & 0.035 & 0.10 \
4 & 0.012 & 0.070 & 0.80 \
\end{array}
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	ext{The rate law corresponds to the above data is}\

Professor PC Thomas Classes · Answer

(c) : $A+B ightarrow C$ Let rate $=k[A]^{x}[B]^{y}$\ where order of reaction is $(x+y)$.\ Putting the values of exp. 1,2 , and 3 , we get following equations.\ $0.10=k[0.012]^{x}[0.035]^{y}$\ $0.80=k[0.024]^{x}[0.070]^{y}$\ $0.10=k[0.024]^{x}[0.035]^{y}$\ Dividing (ii) by (iii), we get\ $\frac{0.80}{0.10}=\left(\frac{0.070}{0.035}ight)^{y} \Rightarrow 2^{y}=8 \Rightarrow y=3$\ Keeping $[A]$ constant, $[B]$ is doubled, rate becomes 8 times.\ Dividing eq. (iii) by eq. (i), we get\ $\frac{0.10}{0.10}=\left(\frac{0.024}{0.012}ight)^{x} \Rightarrow 2^{x}=1 \Rightarrow x=0$\ Keeping $[B]$ constant, $[A]$ is doubled, rate remains unaffected. Hence, rate is independent of $[A]$.\ rate $\propto[B]^{3}$.\