The experiemental data for decomposition of N_{2}O_{5}

[2N_{2}O_{5}→ 4NO_{2} + O_{2}]

in gas phase at 318 k are given below:

t/s | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |

10 | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |

(i) plot N_{2}O_{5} against t

(ii) Find the half-life period for the Reaction.

(iii) Draw a graph between log[N_{2}O_{5}]and t.

(iv) what is the rate law?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from k and compare it with (ii)

(i)

(ii) initial conc.is 1.63 × 10^{-2} so half conc. is 0.815 × 10^{-2}M From given information and graph t_{1/2} is 1440 sec.

(iii)

(iv) As log [N_{2}O_{5}] vs time is straight line given reaction is first order. Hence its rate law will be, Rate = k[N_{2}O_{5}]

(v) The slope of above graph is slope = 0.000209

K = 2.303 × slope

⇒ 4.82 × 10^{-4}sec^{-1}

Now, t_{1/2} = 0.693/K.

⇒ 0.693/4.82 × 10^{-4}

⇒ t_{1/2} = 1438 sec. which is almost equal to (ii)

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