NEET UG 2012 — Chemistry Physical Chemistry

Let the order of reaction with respect to $A$A and $B$B is $x$x and $y$y respectively. So, the rate law can be given as $R=k[A]^x[B]^y$R=k[A]x[B]y When the concentration of only $B$B is doubled, the rate is doubled, so $R_1=k[A]^x[2 B]^y=2 R$R1​=k[A]x[2B]y=2R If concentrations of both the reactants $A$A and $B$B are doubled, the rate increases by a factor of 8 , so $$\begin{aligned} R^{\prime \prime}=k[2 A]^x[2 B]^y & =8 R \\ \Rightarrow k 2^x 2^y[A]^x[B]^y & =8 R \end{aligned}$$R′′=k[2A]x[2B]y⇒k2x2y[A]x[B]y​=8R=8R​ From Eq. (i) and (ii), we get $$\begin{array}{rlrl} & \Rightarrow & \frac{2 R}{R} & =\frac{[A]^x[2 B]^y}{[A]^x[B]^y} \\ & \therefore & =2^y \\ & y & =1 \end{array}$$​⇒∴y​R2R​=2y=1​=[A]x[B]y[A]x[2B]y​ From Eq. (i) and (iv), we get $\Rightarrow \frac{8 R}{R}=\frac{2^x 2^y[A]^x[B]^y}{[A]^x[B]^y}$⇒R8R​=[A]x[B]y2x2y[A]x[B]y​ or $8=2^x 2^y$8=2x2y Substitution of the value of $y$y gives, $$\begin{aligned} & 8=2^x 2^1 \\ & 4=2^x \end{aligned}$$​8=2x214=2x​ $$\begin{aligned} & & (2)^2 & =(2)^x \\ \therefore & & x & =2 \end{aligned}$$∴​​(2)2x​=(2)x=2​ Substitution of the value of $x$x and $y$y in Eq. (i) gives, $R=k[A]^2[B]$R=k[A]2[B]