# For the reaction system:$\dpi{100} 2NO_{(g)}+O_{2(g)}\rightarrow 2NO_{2(g)}$ ,volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to $\dpi{100} O_{2}$  and second order with respect to $\dpi{100} NO_{2}$ , the rate of reaction will Option 1) diminish to one-­fourth of its initial value Option 2) diminish to one­-eighth of its initial value Option 3) increase to eight times of its initial value Option 4) increase to four times of its initial value

As we learnt in

Rate of Law = Dependence of Rate on Concentration -

The representation of rate of a reaction in terms of concentration of the reactants is known as Rate Law

or

The Rate Law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may/maynot be equal to stoichiometric of the reacting species in a balanced chemical equation

- wherein

Formula: $aA+bB\rightarrow cC+dD$

$Rate=\frac{dR}{dT}$

$=\alpha [A]^{x}.[B]^{y}$

$=k[A]^{x}.[B]^{y}$

K= rate constant

$2NO+O_{2 }\rightarrow 2NO_{2}$

$R_{1}=K\left [ NO^{} \right ]^{2}\left [O _{2} \right ]^{1} -- - -- -- - \left ( i \right )$

On reducing the volume to half its concentraction increases by two time

$R_{2}=K\left [ 2NO^{} \right ]^{2}\left [2O _{2} \right ]^{1}$

$R_{2}=K\times 8\times \left [ NO^{} \right ]^{2}\left [O _{2} \right ]^{1}$

$R_{2}=8R_{1}$

Option 1)

diminish to one-­fourth of its initial value

Incorrect Option

Option 2)

diminish to one­-eighth of its initial value

Incorrect Option

Option 3)

increase to eight times of its initial value

Correct Option

Option 4)

increase to four times of its initial value

Incorrect Option

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