Chemistry - Chemical Kinetics Question with Solution | TestHub

Question Explanation:

From the graph, find the average rate of formation of NO₂ between 600s and 700s , then use stoichiometry of decomposition of N₂O₅ to calculate the rate of formation of O₂ .

Concept:

Chemical Kinetics, Rate Of Reaction, Average Rate, Stoichiometric Rate Relation

Solution:

Decomposition reaction:

$$2N_2{O}_5⟶4N{O}_2+O_2$$

From the graph, during 600-700s:

$\Delta [N{O}_2]=0.003M$

So, the average rate of formation of NO₂ is:

$$\frac{\Delta [N{O}_2]}{\Delta t}=\frac{0.003}{100}=3\times 10^{-5}M{s}^{-1}$$

From stoichiometry:

$4N{O}_2:1O_2$

Hence, the rate of formation of O₂ is:

$$\frac{d[O_2]}{dt}=\frac{1}{4}\frac{d[N{O}_2]}{dt}$$

Rate of formation of O₂ = $\frac{1}{4}\times 3\times 10^{-5}$

$=0.75\times 10^{-5}=7.5\times 10^{-6}M{s}^{-1}$

Thus, $X=7.5$, $Y=-6$

$X+Y=7.5+(-6)=1.5$

Final Answer: 1.5.