Question

The data below shows the change in concentration of dinitrogen pentoxide over time, at 330 K, according to the following process.
2N2O5(g) = 4NO2(g) + O2
[N2O5] Time (s)
0.100 0.00
0.066 200.00
0.044 400.00
a) Find the rate of disappearance of N2O5 from t=0 s to t=200s
b) Find the rate of appearance of NO2 from t=0 s to t =200s

Answer 1

Answer: a) $1.7\times 10^{-4}$

b) $3.4\times 10^{-4}$

Explanation:

The reaction is :

$2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)$

Rate = Rate of disappearance of $N_2O_5$ = Rate of appearance of $NO_2$

Rate =  $-\frac{d[N_2O_5]}{2dt}$ = $\frac{d[NO_2]}{4dt}$

Rate of disappearance of $N_2O_5$ = $\frac{\text {change in concentration}}{time}$ = $\frac{0.100-0.066}{200-0}=1.7\times 10^{-4}$

a) Rate of disappearance of $N_2O_5$ = $-\frac{d[N_2O_5]}{2dt}$

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

b) Rate of appearance of $NO_2$ =  $\frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}$

Answer 2

A) Find the rate of disappearance of $N_2O_5$ from t = 0 s to t = 200s

$Rate = \frac{1}{2}(\frac{-\delta N_2O_5}{\delta t})\\\\Rate = -\frac{1}{2}(\frac{0.066 - 0.100}{200 - 0})\\\\Rate = 8.5*10^{-5}$

B) Find the rate of appearance of $NO_2$ from t = 0 s to t = 200s

According to rate law,

$\frac{1}{2}(\frac{-\delta N_2O_5}{\delta t}) = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\8.5*10^{-5} = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\\frac{\delta NO_2}{\delta t} = 4 * 8.5*10^{-5}\\\\Rate = 3.4*10^{-4}$

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