The experimental data for the reaction, 2A+B2→2AB, is Expt. No.[A][B2]Rate (s−1)1.0.500.501.6×10−42.0.501.003.2×10−43.1.01.003.2×10−4The rate equation for the above data is

r=k[A]x[B]y1.6×10−4=k[0.5]x[0.5]y .....(1)3.2×10−4=k[0.5]x[1.0]y .....(2)3.2×10−4=k[1.0]x[1.0]y .....(3)(2)÷(1) 2=2y ⇒ y=1(3)÷(2) 1=2x ⇒ x=0 r=k[A]o[B]1 ⇒ r=K[B]

The experimental data for the reaction, 2A+B2→2AB, is Expt. No.[A][B2]Rate (s−1)1.0.500.501.6×10−42.0.501.003.2×10−43.1.01.003.2×10−4The rate equation for the above data is

r=k[A]x[B]y1.6×10−4=k[0.5]x[0.5]y .....(1)3.2×10−4=k[0.5]x[1.0]y .....(2)3.2×10−4=k[1.0]x[1.0]y .....(3)(2)÷(1) 2=2y ⇒ y=1(3)÷(2) 1=2x ⇒ x=0 r=k[A]o[B]1 ⇒ r=K[B]

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The experimental data for the reaction, $2 A + B_{2} \to 2 AB$ , is
Expt. No. [A] $[B_{2}]$ Rate $(s^{- 1})$
1. 0.50 0.50 $1.6 \times 10^{- 4}$
2. 0.50 1.00 $3.2 \times 10^{- 4}$
3. 1.0 1.00 $3.2 \times 10^{- 4}$
The rate equation for the above data is

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$rate = k {[A]}^{2} {[B]}^{2}$

$rate = k {[A]}^{2} [B]$

$rate = k {[B_{2}]}^{2}$

$rate = k [B]$

answer is B.

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Detailed Solution

$\begin{array}{l} r = k {[A]}^{x} {[B]}^{y} \\ 1 .6 \times 10^{- 4} = k {[0 .5]}^{x} {[0 .5]}^{y} . .... (1) \\ 3 .2 \times 10^{- 4} = k {[0 .5]}^{x} {[1 .0]}^{y} . .... (2) \\ 3 .2 \times 10^{- 4} = k {[1 .0]}^{x} {[1 .0]}^{y} . .... (3) \end{array}$