Question

For the decomposition reaction:
$N_2{O}_{4\left(g\right)}\to 2{NO}_{2\left(g\right)}$ ; the initial pressure of
$N_2{O}_4$ falls from 0.46 atm to 0.28 atm in 30 minute. What is the rate of appearance of ${NO}_2$?

• A. $12\times {10}^{2}atm\cdot {min}^{-1}$
  
• B. $1.2\times {10}^{2}atm\cdot {min}^{-1}$
• C. $1.2\times {10}^{-2}atm\cdot {min}^{-1}$
• D. $1.8\times {10}^{-1}atm\cdot {min}^{-1}$

Answer 1

The correct option is C $1.2\times {10}^{-2}atm\cdot {min}^{-1}$

The correct answer is 3 is correct

According to formula

$\frac{d\left[N_2{O}_4\right]}{dt}=\frac{1}{2}\frac{d\left[N{o}_2\right]}{dt}$

$\frac{1}{2}\frac{d\left[N{o}_2\right]}{dt}=\frac{(0.28-0.46)}{30}$

$\frac{1}{2}\frac{d\left[N{o}_2\right]}{dt}=6\times {10}^{-3}$

$\frac{d\left[N{o}_2\right]}{dt}=1.2\times {10}^{-2}$atm $mi{n}^{-1}$