Question

If the rate of disappearance of N2O5 is equal to 1.40 mol/min at a particular moment, what is the rate of appearance of NO2 at that moment? Enter your answer with 2 decimal places and no units.

Answer 1

Answer:

2.8 mol/min  is the rate of appearance of $NO_2$ at that moment.

Explanation:

$2N_2O_5\rightarrow 4NO_2+O_2$

Rate of the reaction = R

$R=\frac{-1}{2}\frac{d[N_2O_5]}{dt}=\frac{1}{4}\frac{d[NO_2]}{dt}=\frac{1}{1}\frac{d[O_2]}{dt}$

Rate of disappearance of $N_2O_5$ is $=-\frac{d[N_2O_5]}{dt}= 1.40 mol/min$

$R=\frac{-1}{2}\frac{d[NO_2]}{dt}$

$R=\frac{1}{2}\times 1.40 mol/min=0.70 mol/min$

Rate of appearance of $NO_2$ is $=\frac{1}{4}\frac{d[NO_2]}{dt}:$

$R=\frac{1}{4}\frac{d[NO_2]}{dt}$

$\frac{d[NO_2]}{dt}=4\times 0.70 mol/min=2.8 mol/min$

2.8 mol/min  is the rate of appearance of $NO_2$ at that moment.