The experiment data for the reaction , is :Expt.No[A][B2]Rate (mols-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 :

Rate = KAxB2yFrom Expt 11.6×10−4=K0.50x0.50y→1From Expt 23.2×10−4=K0.50x1.00y→2From Expt 33.2×10−4=K1.0x1.00y→3eq2eq1 2=2y y=1eq3eq2 1=2x 20=2x x=0

The experiment data for the reaction , is :Expt.No[A][B2]Rate (mols-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 :

Rate = KAxB2yFrom Expt 11.6×10−4=K0.50x0.50y→1From Expt 23.2×10−4=K0.50x1.00y→2From Expt 33.2×10−4=K1.0x1.00y→3eq2eq1 2=2y y=1eq3eq2 1=2x 20=2x x=0

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The experiment data for the reaction , is :
Expt.No
[A]
[B₂]
Rate (mols^-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[B₂]²

rate = k[B₂]

rate = k[A]²[B]²

rate = k[A]²[B]

answer is B.

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Detailed Solution

Rate = $K {[A]}^{x} {[B_{2}]}^{y}$

From Expt 1

$1.6 \times 10^{- 4} = K {[0.50]}^{x} {[0.50]}^{y} \to (1)$

From Expt 2

$3.2 \times 10^{- 4} = K {[0.50]}^{x} {[1.00]}^{y} \to (2)$

From Expt 3

$3.2 \times 10^{- 4} = K {[1.0]}^{x} {[1.00]}^{y} \to (3)$

$\frac{e q (2)}{e q (1)} 2 = 2^{y} y = 1$

$\frac{e q (3)}{e q (2)} 1 = 2^{x} 2^{0} = 2^{x} x = 0$

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Expt.No	[A]	[B₂]	Rate (mols^-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 experiment data for the reaction , is :Expt.No[A][B2]Rate (mols-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 :

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The experiment data for the reaction , is :
Expt.No
[A]
[B₂]
Rate (mols^-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 :