The rate of the reaction $2 NO + O_{2} \to 2 {NO}_{2}$ at $25^{°} C is 0.028 mol L^{- 1} S^{- 1} .$ The experimental rate is given by $r = K {[NO]}^{2} [O_{2}] .$ If the initial concentrations of the reactants are $[O_{2}] = 0.040 m o l L^{- 1} a n d [N O] = 0.01 m o l . L^{- 1},$ the rate constant of the reaction is

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$7 . 0^{'} 10^{- 4} L^{2} {mol}^{- 2} S^{- 1}$

$7.0' 10^{3} L^{2} {mol}^{- 2} S^{- 1}$

$7.0' 10^{2} L^{2} {mol}^{- 2} S^{- 1}$

$7.0' 10^{- 2} L {mol}^{- 1} S^{- 1}$

answer is D.

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Detailed Solution

From rate law expression, $Rate, r = K {[NO]}^{2} [O_{2}] \to eqn (1)$

As per given information,

$[O_{2}] = 0.04 m o l . L^{- 1} a n d [N O] = 0.01 m o l . L^{- 1}$

$And rate of reaction, r = 0.028 mol . L^{- 1} . S^{- 1}$

And then

$0.028 = k {[0.01]}^{2} [0.040]$

$K = \frac{0.028}{{[0.01]}^{2} [0.04]} = 7.0 \times 10^{3} L^{2} . {mol}^{- 2} . S^{- 1}$

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