The experimental data for decomposition of N 2 O 5-2 N 2 O 5- 4 NO 2- O 2- in gas phase at 318 K are given below-Nbeginarray -c-c-c-c-c-c-c-c-c-c-Hhine10-2 - N - 2 O - 5- - - 1-63 - 1-36 - 1-14 - 0-93 - 0-78 - 0-64 - 0-53 - 0-43 - 0-35mol - L -1- 11 h linelend array Plot - N 2 O 5- against t-hline10-2 - N - 2 O - 5- - - 1-63 - 1-36 - 1-14 - 0-93 - 0-78 - 0-64 - 0-53 - 0-43 - 0-35mol - L -1- hlinelend array Find the half-life period for the reaction Nbeginarray -c-c-c-c-c-c-c-c-c-c-hlinet - s - 0 - 400 - 800 - 1200 - 1600 - 2000 - 2400 - 2800 - 3200 illhline 10-2 times - N - 2 O - 5- - - 1-63 - 1-36 - 1-14 - 0-93 - 0-78 - 0-64 - 0-53 - 0-43 - 0-35mol - L -1- IV e n d a r r a y Draw a graph between log- N 2 O 5- and t-beginarray - c-c-c-c-c-c-c-c-c-c-Wlinet - s - 0 - 400 - 800 - 1200 - 1600 - 2000 - 2400 - 2800 - 3200 withline10-2 times - N - 2 O - 5- - - 1-63 - 1-36 - 1-14 - 0-93 - 0-78 - 0-64 - 0-53 - 0-43 - 0-35mol - L -1- 11 h linelend array

The plot of [N_2O_5] against time is given below. Initial concentration of N_2O_5 = 1.63 × 10^ 2M Half of initial concentration =1.63 × 10^ 2/2 =0.815 × 10^ 2 ...

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Standard XII

Chemistry

Half life

The experimen...

Question

The experimental data for decomposition of $N_{2} O_{5}$ $[2 N_{2} O_{5} \to 4 N O_{2} + O_{2}]$ in gas phase at 318 K are given below:

$\begin{matrix} t / s & 0 & 400 & 800 & 1200 & 1600 & 2000 & 2400 & 2800 & 3200 10^{- 2} \times [N_{2} O_{5}] / & 1.63 & 1.36 & 1.14 & 0.93 & 0.78 & 0.64 & 0.53 & 0.43 & 0.35 m o l L^{- 1} \end{matrix}$

Plot $[N_{2} O_{5}]$ against t

$\begin{matrix} t / s & 0 & 400 & 800 & 1200 & 1600 & 2000 & 2400 & 2800 & 3200 10^{- 2} \times [N_{2} O_{5}] / & 1.63 & 1.36 & 1.14 & 0.93 & 0.78 & 0.64 & 0.53 & 0.43 & 0.35 m o l L^{- 1} \end{matrix}$
Find the half-life period for the reaction

$\begin{matrix} t / s & 0 & 400 & 800 & 1200 & 1600 & 2000 & 2400 & 2800 & 3200 10^{- 2} \times [N_{2} O_{5}] / & 1.63 & 1.36 & 1.14 & 0.93 & 0.78 & 0.64 & 0.53 & 0.43 & 0.35 m o l L^{- 1} \end{matrix}$
Draw a graph between $l o g [N_{2} O_{5}]$ and t

$\begin{matrix} t / s & 0 & 400 & 800 & 1200 & 1600 & 2000 & 2400 & 2800 & 3200 10^{- 2} \times [N_{2} O_{5}] / & 1.63 & 1.36 & 1.14 & 0.93 & 0.78 & 0.64 & 0.53 & 0.43 & 0.35 m o l L^{- 1} \end{matrix}$
What is the rate law?

$\begin{matrix} t / s & 0 & 400 & 800 & 1200 & 1600 & 2000 & 2400 & 2800 & 3200 10^{- 2} \times [N_{2} O_{5}] / & 1.63 & 1.36 & 1.14 & 0.93 & 0.78 & 0.64 & 0.53 & 0.43 & 0.35 m o l L^{- 1} \end{matrix}$
Calculate the rate constant

$\begin{matrix} t / s & 0 & 400 & 800 & 1200 & 1600 & 2000 & 2400 & 2800 & 3200 10^{- 2} \times [N_{2} O_{5}] / & 1.63 & 1.36 & 1.14 & 0.93 & 0.78 & 0.64 & 0.53 & 0.43 & 0.35 m o l L^{- 1} \end{matrix}$
Calculate the half-life period from k and compare it with (ii)

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Solution

The plot of $[N_{2} O_{5}]$ against time is given below.

Initial concentration of $N_{2} O_{5} = 1.63 \times 10^{- 2} M$
Half of initial concentration
$= \frac{1.63 \times 10^{- 2}}{2} = 0.815 \times 10^{- 2} M$
Half-life period = Time corresponding to the above amount in the graph = 1400 s (approximately)

For graph between $l o g [N_{2} O_{5}]$ and time, we first find the values of $l o g [N_{2} O_{5}]$ .
$\begin{matrix} T i m e (s) & 0 & 400 & 800 & 1200 & 1600 & 2000 & 2400 & 2800 & 3200 [N_{2} O_{5}] \times 10^{- 2} & 1.63 & 1.36 & 1.14 & 0.93 & 0.78 & 0.64 & 0.53 & 0.43 & 0.35 l o g [N_{2} O_{5}] & - 1.79 & - 1.87 & - 1.94 & - 2.03 & - 2.11 & - 2.19 & - 2.28 & - 2.37 & - 2.46 \end{matrix}$

Since, the graph between $l o g [N_{2} O_{5}]$ and time is a straight line, the reaction is of first order.
The rate law:Rate $= k [N_{2} O_{5}]$

Slope of the time $= - \frac{k}{2.303} = \frac{- 2.46 - (- 1.79)}{3200 - 0} = \frac{- 0.67}{3200} \frac{- k}{2.303} = \frac{- 0.67}{3200} k = \frac{0.67 \times 2.303}{3200}$
or $k = 4.82 \times 10^{- 4} s^{- 1}$

Half-life period $(t_{\frac{1}{2}}) = \frac{0.693}{k} = \frac{0.693}{4.82 \times 10^{- 4} s^{- 1}} = 1438 s$
Half-life period $(t_{\frac{1}{2}})$ as calculated from the formula and slope are approximately the same.

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Similar questions

The experimental data for decomposition of N₂O₅

in gas phase at 318K are given below:

t(s)	0	400	800	1200	1600	2000	2400	2800	3200
	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

(i) Plot [N₂O₅] against t.

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

(iii) Draw a graph between log [N₂O₅] 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).

Q. The experimental data for decomposition of

N_{2} O_{5}

[2 N_{2} O_{5} \to 4 N O_{2} + O_{2}]

in gas phase at

318 K

are given below:

t/s	$0$	$400$	$800$	$1200$	$1600$	$2000$	$2400$	$2800$	$3200$
$10^{2} \times [N_{2} O_{5}] / m o l L^{- 1}$	$1.63$	$1.36$	$1.14$	$0.93$	$0.78$	$0.64$	$0.53$	$0.43$	$0.35$

(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)

Q. The experimental data for decomposition of

N_{2} O_{5} [2 N_{2} O_{5} ⟶ {4 N O}_{2} + O_{2}]

in gas phase at 318 K are given below:

$t / s$	0	400	800	1200	1600	2000	2400	2800	3200
$10^{- 2} \times [N_{2} O_{5}] / m o l L^{- 1}$	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

Calculate the rate constant and the half-life period.

Q. Match the following columns and choose the correct option.

\begin{matrix} C o l u m n I & C o l u m n I I (a) Mutation breeding & 1. Laborious and expensive process to obtain gene variation (b) Selection & 2. Hybrid vigour can be maintained for several generations (c) Hybridisation & 3. Planting a plant in a new locality from its native place (d) Introduction & 4. Oldest breeding method 5. Quick method to obtain gene variation \end{matrix}

Consider a binary operation $*$ on set{1,2,3,4,5} given by the following multiplication table:
(i)Compute $(2 * 3) * 4$ and $2 * (3 * 4)$

$\begin{matrix} * & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

(ii) Is $*$ commutative?

$\begin{matrix} * & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

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$\begin{matrix} * & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

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Half life

Standard XII Chemistry

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