NO_2 required for a reaction is produced by decomposition of N_2 O_5 in CCl_4 as by equation 2 N_2 O_5( g) -> 4 NO_2( g)+ O_2( g) The initial concentration of N_2 O_5 is 3 mol L^-1 and it is 2.75 mol L^-1 after 30 minutes. The rate of formation of NO_2 is x x 10^-3 mol L^-1 min^-1, value of x is _________. (nearest integer)

Rate of reaction (ROR) =-(1 / 2) fracDelta[ N_2 O_5]Delta t=(1 / 4) frac[ NO_2]Delta t=fracDelta[ O_2]Delta t \ ROR=-(1 / 2) fracDelta[ N_2 O_5]Delta t=-(1 / 2) ((2.75-3) / 30) mol L^-1 min^-1 \ ROR=-(1 / 2) ((-0.25) / 30) mol L^-1 min^-1 \ ROR =(1 / 240) mol L^-1 min^-1 \ Rate of formation of NO_2=fracDelta[ NO_2]Delta t=4 x ROR =(4 / 240)=16.66 x 10^-3 molL^-1 min^-1 = 17 x 10^-3 .

NO2 required for a reaction is produced by decompositio... | null (Chemistry)

Rate of reaction (ROR)

\begin{aligned} & =-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=\frac{1}{4} \frac{\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta \mathrm{t}} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=-\frac{1}{2} \frac{(2.75-3)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{(-0.25)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \text { ROR }=\frac{1}{240} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \end{aligned}

Rate of formation of

\mathrm{NO}_2=\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=4 \times \mathrm{ROR}

=\frac{4}{240}=16.66 \times 10^{-3} \mathrm{molL}^{-1} \mathrm{~min}^{-1} \simeq 17 \times 10^{-3} \text {. }

Explanation

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