Question

The decomposition of \[{N_2}{O_5}\] according to the equation $2{N_2}{O_{5(g)}} \to 4N{O_{2(g)}} + {O_{2(g)}}$ is a first order reaction. After $30$ min, from the start of the decomposition in a closed vessel, the total pressure developed is found to be $284.5{\text{mmHg}}$ . On complete decomposition, the total pressure is ${\text{584}}{\text{.5mmHg}}$. The rate constant of the reaction is:

Answer 1

Hint: We have to know the chemical decomposition, or substance breakdown, is the interaction or impact of working on a solitary compound element into at least two sections. Compound deterioration is normally viewed, and characterized as the specific inverse of substance union. To put it plainly, the substance response where at least two items are shaped from a solitary reactant is known as a disintegration response.

Complete step by step answer:
We have to know the balanced given chemical reaction, that has to be shown,
$2{N_2}{O_{5(g)}} \to 4N{O_{2(g)}} + {O_{2(g)}}$
At initial time $(t = 0)$ ,
$\begin{array}{*{20}{c}}
  a&0&0
\end{array}$
At time $(t = 30{\text{ mins)}}$ ,
$\begin{array}{*{20}{c}}
  {a - X}&{2X}&{\dfrac{X}{2}}
\end{array}$
Completion of the reaction,
$\begin{array}{*{20}{c}}
  0&{2a}&{\dfrac{a}{2}}
\end{array}$
At the point when number of mole whenever is straightforwardly, corresponding to the pressing factor created around then,
In this way,
The equation $1$ has to given
$a \propto {{\text{P}}_0}$ at $(t = 0)$ ,
The equation $2$ has to given
$a + \left( {\dfrac{{3X}}{2}} \right) \propto 284.5$ at $t = 0$
The equation $3$ has to be given,
$\left( {\dfrac{{5a}}{2}} \right) \propto 584.5$ at $t = \infty $
From equation, ${\text{1}}$ and $3$ , we get equation $4$, that has to be given below,
$a \propto {\text{233}}{\text{.8}}$
When equation $4$ is substituted in equation $2$ , we get equation $5$ ,
${\text{X}} \propto {\text{33}}{\text{.8}}$ .
Now, we have to calculate the rate constant by using the following expression,
$K = \dfrac{{2.303}}{t}\log \dfrac{a}{{(a - X)}}$
Applying all the values in the above expression,
$K = \dfrac{{2.303}}{{30}}\log \dfrac{{233.8}}{{200}}$
Therefore,
$K = 5.206 \times {10^{ - 3}}{\min ^{ - 1}}$

Note: We have to know the decay response is a response where a compound separates into at least two less complex substances. The overall type of a decay response is $AB \to A + B$ . Most deterioration responses require a contribution of energy as warmth, light, or power. The deterioration responses are additionally, called investigation responses since they are very significant in scientific strategies. Models incorporate mass spectrometry, gravimetric examination, and thermo-gravimetric investigation.