Consider the kinetic data given in the following table for the reaction, $A+B+C \rightarrow$ product

Question

Consider the kinetic data given in the following table for the reaction, $A+B+C ightarrow$ product. \The rate of the reaction for $[A]=0.15 \mathrm{~mol} \mathrm{dm}^{-3}$, $[B]=0.25 \mathrm{~mol} \mathrm{dm}^{-3}$ and $[C]=0.15 \mathrm{~mol} \mathrm{dm}^{-3}$ is found to be $Y 	imes 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$. The value of $Y$ is _______.\

Professor PC Thomas Classes · Answer

(6.75) : $A+B+C \longrightarrow$ Product $$ r=k[A]^{a}[B]^{b}[C]^{c} $$ By expt. no. 1 and $2, b=0$\ By expt. no. 1 and $3, c=1$\ By expt. no. 1 and 4, $a=1$\ $	herefore \quad r=k[A]^{1}[B]^{0}[C]^{1}$\ From expt. no. $1,6 	imes 10^{-5}=k(0.2)(0.1)$\ $\Rightarrow \quad k=3 	imes 10^{-3}$\ Now, rate at given concentrations\ $r=k[A]^{1}[B]^{0}[C]^{1}$\ $=3 	imes 10^{-3} 	imes(0.15)^{1} 	imes(0.25)^{0} 	imes(0.15)^{1}$\ $=3 	imes 0.0225 	imes 10^{-3}$\ $=6.75 	imes 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$\ $\because \quad Y 	imes 10^{-5}=6.75 	imes 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$\ $	herefore \quad Y=6.75$\