The rate constant for the first-order decomposition of N2O5 in the reaction 2 N2O5(g)→4 NO2(g)+O2(g) is kr=3.38×10−5 s−1 at 25 °C. What is the half-life of N2O5? What will be the pressure, initially 500 Torr, after (i) 50 s, (ii) 20min after initiation of the reaction?

Respuesta :

Explanation:

[tex]2 N_2O_5(g)\rightarrow 4 NO_2(g)+O_2(g)[/tex]

Rate of the reaction ,k= [tex]3.38\times 10^{-5} s^{-1}[/tex]

Half life of the [tex]N_2O_5=t_{\frac{1}{2}}[/tex]

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{3.38\times 10^{-5} s^{-1}}[/tex](first order kinetics)

[tex]t_{\frac{1}{2}}=20,502.958 seconds[/tex]

Half life of the [tex]N_2O_5[/tex] is 20,502.958 seconds.

Integrated rate equation for first order kinetics in gas phase is given as:

[tex]k=\frac{2.303}{t}\log\frac{p_o}{2p_o-p}[/tex]

p= pressure of the gas at given time t.

[tex]p_o[/tex] = Initial pressure of the gas

(i) When, t = 50 sec

[tex]p_o=500 torr[/tex]

[tex]3.38\times 10^{-5} s^{-1}=\frac{2.303}{50 s}\log\frac{500 Torr}{2(500 Torr)-p}[/tex]

p = 500.49 Torr

(ii)When, t = 20 min = 1200 sec

[tex]p_o=500 torr[/tex]

[tex]3.38\times 10^{-5} s^{-1}=\frac{2.303}{1200 s}\log\frac{500 Torr}{2(500 Torr)-p}[/tex]

p = 519.83 Torr