Nitrogen pentoxide decomposes by a first-order process yielding N2O4 and oxygen: 2N2O5 -> 2N2O4 + O2 At a given temperature, the half-life of N2O5 is 0.85 hr. What is the first-order rate constant for N2O5 decomposition? 2.26 x 10^-4 s^-1
Added by Mary E.
Step 1
Given: Half-life of N2O5 = 0.85 hr 1 hr = 3600 seconds Therefore, half-life of N2O5 = 0.85 hr * 3600 seconds/hr = 3060 seconds ** Show more…
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Nitrogen pentoxide decomposes by a first-order process yielding N2O4 and oxygen. 2N2O5 -> 2N2O4 + O2. At a given temperature, the half-life of N2O5 is 0.85 hr. What is the first-order rate constant for N2O5 decomposition?
Maitreya E.
The rate constant for the 1st order decomposition of N2O5 in the reaction 2N2O5 (g) → 4NO2 (g) + O2 (g) is k = 3.38 x 10^-5 s^-1 at 25°C. a. What is the half-life of N2O5? b. What will be the total pressure, initially 88.3 kPa for the pure N2O5 vapor, (1) 10 s and (2) 10 min after initiation of the reaction? (Hint: partial pressure is proportional to concentration).
Penny R.
For the decomposition of gaseous dinitrogen pentoxide, $$ 2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ the rate constant is $k=2.8 \times 10^{-3} \mathrm{s}^{-1}$ at $60^{\circ} \mathrm{C} .$ The initial concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ is 1.58 $\mathrm{mol} / \mathrm{L}$ . (a) What is $\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$ after 5.00 $\min ?$ (b) What fraction of the $\mathrm{N}_{2} \mathrm{O}_{5}$ has decomposed after 5.00 $\mathrm{min} ?$
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