From the given data for the reaction :2NO(g)+Cl2(g)⟶2NOCl(g) at 289 K [Cl2][NO]Rate of reaction mol L−1sec−110.05 M0.05 M1×10-320.15 M0.05 M3×10-330.05 M0.15 M9×10-3the rate law for reaction will be :

r=KNOaCl2b3×10−3=0.150.05b⇒b=19×10−3=0.150.05a⇒a=2

From the given data for the reaction :2NO(g)+Cl2(g)⟶2NOCl(g) at 289 K [Cl2][NO]Rate of reaction mol L−1sec−110.05 M0.05 M1×10-320.15 M0.05 M3×10-330.05 M0.15 M9×10-3the rate law for reaction will be :

r=KNOaCl2b3×10−3=0.150.05b⇒b=19×10−3=0.150.05a⇒a=2

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From the given data for the reaction :
$2 NO (g) + {Cl}_{2} (g) ⟶ 2 NOCl (g)$ at 289 K
$[{Cl}_{2}]$ $[NO]$ Rate of reaction $(mol L^{- 1} \sec^{- 1})$
1 0.05 M 0.05 M $1 \times 10^{- 3}$
2 0.15 M 0.05 M $3 \times 10^{- 3}$
3 0.05 M 0.15 M $9 \times 10^{- 3}$
the rate law for reaction will be :

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$r = k {[{Cl}_{2}]}^{2} [NO]$

$r = k [NO] [{Cl}_{2}]$

$r = k [{Cl}_{2}] {[NO]}^{2}_{}$

$r = k [{Cl}_{2}]$

answer is B.

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Detailed Solution

$r = K [{NO}^{a} {[{Cl}_{2}]}^{b}$

$\begin{array}{r} 3 \times 10^{- 3} = {(\frac{0 .15}{0 .05})}^{b} \Rightarrow b = 1 \\ 9 \times 10^{- 3} = {(\frac{0 .15}{0 .05})}^{a} \Rightarrow a = 2 \end{array}$

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	$[{Cl}_{2}]$	$[NO]$	Rate of reaction $(mol L^{- 1} \sec^{- 1})$
1	0.05 M	0.05 M	$1 \times 10^{- 3}$
2	0.15 M	0.05 M	$3 \times 10^{- 3}$
3	0.05 M	0.15 M	$9 \times 10^{- 3}$

From the given data for the reaction :2NO(g)+Cl2(g)⟶2NOCl(g) at 289 K [Cl2][NO]Rate of reaction mol L−1sec−110.05 M0.05 M1×10-320.15 M0.05 M3×10-330.05 M0.15 M9×10-3the rate law for reaction will be :

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From the given data for the reaction :
$2 NO (g) + {Cl}_{2} (g) ⟶ 2 NOCl (g)$ at 289 K
$[{Cl}_{2}]$ $[NO]$ Rate of reaction $(mol L^{- 1} \sec^{- 1})$
1 0.05 M 0.05 M $1 \times 10^{- 3}$
2 0.15 M 0.05 M $3 \times 10^{- 3}$
3 0.05 M 0.15 M $9 \times 10^{- 3}$
the rate law for reaction will be :