Question

At elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen: 2N2O5(g) → 4NO2 (g) + O2 (g) When the rate of formation of NO2 is 5.5 ⋅ 10-4 M/s, the rate of decomposition of N2O5 is ________ M/s. At elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen: 2N2O5(g) 4NO2 (g) + O2 (g) When the rate of formation of NO2 is 5.5 10-4 M/s, the rate of decomposition of N2O5 is ________ M/s. 2.8 ⋅ 10-4 1.4 ⋅ 10-4 5.5 ⋅ 10-4 10.1 ⋅ 10-4 2.2 ⋅ 10-3

Answer 1

Answer :  The rate of decomposition of $N_2O_5$ is $2.8\times 10^{-4}M/s$

Explanation : Given,

Rate of formation of $NO_2$ = $5.5\times 10^{-4}M/s$

The given rate of reaction is,

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

The expression for rate of reaction :

$\text{Rate of decomposition of }N_2O_5=-\frac{1}{2}\times \frac{d[N_2O_5]}{dt}$

$\text{Rate of formation of }NO_2=+\frac{1}{4}\frac{d[NO_2]}{dt}$

$\text{Rate of formation of }O_2=+\frac{d[O_2]}{dt}$

Now we have to calculate the rate of decomposition of $N_2O_5$.

As we know that,

$\text{Rate of reaction}=-\frac{1}{2}\times \frac{d[N_2O_5]}{dt}=+\frac{1}{4}\frac{d[NO_2]}{dt}=+\frac{d[O_2]}{dt}$

So,

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

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

$-\frac{d[N_2O_5]}{dt}=\frac{2}{4}\times (5.5\times 10^{-4}M/s)$

$-\frac{d[N_2O_5]}{dt}=2.8\times 10^{-4}M/s$

Thus, the rate of decomposition of $N_2O_5$ is $2.8\times 10^{-4}M/s$

Answer 2

Answer:

The rate of descomposition of N₂O₅ is: 2,8 M/s

Explanation:

The global reaction is:

2 N₂O₅ (g) → 4 NO₂ (g) + O₂ (g)

The problem says that rate of formation of NO₂ = 5,5x10⁻⁴ M/s where M is molarity -$\frac{mol}{L}$-

The rate of descomposition of N₂O₅ is:

5,5x10⁻⁴ $\frac{molNO_{2} }{L.s}$ × $\frac{2 mol N_{2}O_{5}}{4 mol NO_{2} }$ = 2,8 M/s

I hope it helps!