The composition of N 2 O 5 is a first order reaction represented by: N 2 O 5 arrow N 2 O 4+1 / 2 O 2 After 20 min the volume of O 2 produced is 10 mL and at the end of the reaction 40 mL. The rate constant is equal to

Question

The composition of N 2 O 5 is a first order reaction represented by: N 2 O 5 arrow N 2 O 4+1 / 2 O 2 After 20 min the volume of O 2 produced is 10 mL and at the end of the reaction 40 mL. The rate constant is equal to (A) (1/20) ln (30/50) (B) (1/20) ln (50/30) (C) (1/20) ln (50/40) (D) (1/20) ln (40/30)

Tardigrade Admin · Accepted Answer

Consider a condition when one of the product is estimated directly or indirectly, i.e., when

V_{0}

is not given
For example, in the decomposition of

N H_{4} N O_{2}

, the volume of

N_{2}

is directly measured at different intervals of time. Then, the formula used is

k = \frac{2.303}{t} lo g \frac{V _{\infty}}{V _{\infty} - V _{t}}

An example is the decomposition of ammonium

(N H_{4} N O_{2})

and benzene diazonium chloride

(C_{6} H_{5} N = NCl)

N H_{4} N O_{2} (s) \to 2 H_{2} O (l) + N_{2} (g)

C_{6} H_{5} - N = N - Cl (l) \to C_{6} H_{5} - Cl (l) + N_{2} (g)

The rate of both the reactions is studied (measured) in similar manner. The volume of nitrogen

(N_{2})

is collected after a regular interval of times as follows:

Time instants	$t = 0$	$t_{1}$	$t_{2}$	$t_{3}$	$t_{4}$	$t_{\infty}$
Vol of $N_{2}$	$0$	$V_{1}$	$V_{2}$	$V_{3}$	$V_{4}$	$V_{\infty}$

At

t = 0

, clearly the volume of

N_{2} = 0

Time instant

t = \infty

means the end of a reaction, i.e., when whole of

N H_{4} N O_{2}

or

C_{6} H_{5} - N = N - Cl

is decomposed

\Rightarrow

At

t = \infty, V_{\infty}

corresponds to the initial volume of

N H_{4} N O_{2}

or

C_{6} H_{5} - N = N - Cl

(Note that the ratio of stoichiometric coefficient for both

N_{2} : N H_{4} N O_{2}

or

N_{2} : C_{6} H_{5} N = NCl

is

1 : 1

)

\Rightarrow c_{0} \propto V_{t}

At

t = t_{1}, t_{2}, t_{3}, \dots

the volume of

N_{2}

corresponds to the concentration of product formed, i.e., equal to

x

\Rightarrow x \propto V_{t}

\Rightarrow c_{0} - x \propto V_{\infty} - V_{t}

Hence, from first order kinetic

k = \frac{2.303}{t} lo g \frac{V _{\infty}}{V _{\infty} - V _{t}}

or

k = \frac{1}{t} In \frac{V _{\infty}}{V _{\infty} - V _{t}}

k = \frac{1}{20} In \frac{40}{( 40 - 10 )}

= \frac{1}{20} In \frac{40}{30}

Q. The composition of $N_{2} O_{5}$ is a first order reaction represented by: $N_{2} O_{5} \to N_{2} O_{4} + 1/2 O_{2}$
After $20 min$ the volume of $O_{2}$ produced is $10 m L$ and at the end of the reaction $40 m L$ . The rate constant is equal to

$\frac{1}{20} ln \frac{30}{50}$

$\frac{1}{20} ln \frac{50}{30}$

$\frac{1}{20} ln \frac{50}{40}$

$\frac{1}{20} ln \frac{40}{30}$

Q. The composition of N2​O5​ is a first order reaction represented by: N2​O5​→N2​O4​+1/2O2​ After 20min the volume of O2​ produced is 10mL and at the end of the reaction 40mL. The rate constant is equal to

201​ln5030​

201​ln3050​

201​ln4050​

201​ln3040​

Q. The composition of $N_{2} O_{5}$ is a first order reaction represented by: $N_{2} O_{5} \to N_{2} O_{4} + 1/2 O_{2}$
After $20 min$ the volume of $O_{2}$ produced is $10 m L$ and at the end of the reaction $40 m L$ . The rate constant is equal to

$\frac{1}{20} ln \frac{30}{50}$

$\frac{1}{20} ln \frac{50}{30}$

$\frac{1}{20} ln \frac{50}{40}$

$\frac{1}{20} ln \frac{40}{30}$