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Standard XII

Mathematics

Higher Order Derivatives

N 2 O 5 disso...

Question

$N_{2} O_{5}$ dissociates as: $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ If concentration of 4 mol $L^{- 1}$ reduces to 2.5 mol $L^{- 1}$ in 3 minutes, what is the rate of production of $N O_{2}$

0.5 m o l L^{- 1} m i n^{- 1}

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1.0 m o l L^{- 1} m i n^{- 1}

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1.5 m o l L^{- 1} m i n^{- 1}

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1.85 m o l L^{- 1} m i n^{- 1}

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Solution

The correct option is B $1.0 m o l L^{- 1} m i n^{- 1}$
$\frac{1}{4} \frac{d [N O_{2}]}{d t} = \frac{- 1}{2} \frac{d [N O_{5}]}{d t}$
$\frac{d [N O_{2}]}{d t} = 2 \times \frac{4 - 2.5}{3}$
$= 2 \times \frac{1.5}{3} = 1$

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N2O5 dissociates as: 2N2 O5→4NO2+O2 If concentration of 4 mol L−1 reduces to 2.5 mol L−1 in 3 minutes, what is the rate of production of NO2

The correct option is B 1.0 mol L−1min−114d[NO2]dt=−12d[NO5]dt d[NO2]dt=2×4−2.53 =2×1.53=1

$N_{2} O_{5}$ dissociates as: $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ If concentration of 4 mol $L^{- 1}$ reduces to 2.5 mol $L^{- 1}$ in 3 minutes, what is the rate of production of $N O_{2}$

The correct option is B $1.0 m o l L^{- 1} m i n^{- 1}$
$\frac{1}{4} \frac{d [N O_{2}]}{d t} = \frac{- 1}{2} \frac{d [N O_{5}]}{d t}$
$\frac{d [N O_{2}]}{d t} = 2 \times \frac{4 - 2.5}{3}$
$= 2 \times \frac{1.5}{3} = 1$