First slide

Rate of chemical reaction

Question

For the reaction, $2 {NO}_{2} + \frac{1}{2} O_{2} \overset{}{\to} N_{2} O_{5}$ the rate of disappearance of ${NO}_{2}$ is $10^{- 3} mole . {lit}^{- 1} \min^{- 1} .$ Rate of appearance of $N_{2} O_{5}$ will be

Moderate

A

$8.33 \times 10^{- 6} mole . {lit}^{- 1} s^{- 1}$
B

$8.33 \times 10^{- 5} mole . {lit}^{- 1} s^{- 1}$
C

$10^{- 3} mole . {lit}^{- 1} s^{- 1}$
D

$5 \times 10^{- 4} mole . {lit}^{- 1} s^{- 1}$

Solution

$rate = - \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = - 2 \frac{d [O_{2}]}{dt} = + \frac{d [N_{2} O_{5}]}{dt} given \frac{- d [{NO}_{2}]}{dt} = 10^{- 3} mole {lit}^{- 1} \min^{- 1} \frac{1}{2} (10^{- 3}) = + \frac{d [N_{2} O_{5}]}{dt} + \frac{d [N_{2} O_{5}]}{dt} = 5 \times 10^{- 4} mole {lit}^{- 1} \min^{- 1} + \frac{d [N_{2} O_{5}]}{dt} = \frac{5 \times 10^{- 4}}{60} mole {lit}^{- 1} s^{- 1} + \frac{d [N_{2} O_{5}]}{dt} = 8.33 \times 10^{- 6} mole {lit}^{- 1} s^{- 1}$

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