Question

For $0.1mole$ of ${N_2}{O_4}\left( g \right)$ was sealed in a tube under one atmosphere conditions at ${25^0}C$ . calculate the number of $N{O_2}\left( g \right)$ present , if the equilibrium .
${N_2}{O_4} \rightleftharpoons N{O_2}\left( {{K_P} = 0.14} \right)$ is reached after some time ;
A.$1.8*{10^2}$
B.$2.8*{10^2}$
C.$0.034$
D.$2.8*{10^{2 - }}$

Answer 1

Hint: Chemical equilibrium arises in reversible reactions in which no net change in the amount of reaction and the product occur. . The mole is the unit of measurement for the amount of substance within the International System of Units of Units. A mole of a substance or a mole of particles is defined as exactly particles, which can be atoms, molecules, ions, or electrons

Complete step by step answer:As we know about chemical equilibrium , it arises when two opposite reactions go at the same rate ,hence there is no net change in the amount of substance involved .
$A + B \rightleftharpoons C + D$
For above reaction , rate constant may be given as - $K = \left[ C \right]\left[ D \right]/\left[ A \right]\left[ B \right]$
This rate constant can be represent in concentration terms , pressure , number of moles and in mole fraction also and represented as ${K_c},{K_P},{K_n},{K_x}$ respectively .rate constant is the proportionality constant in the equation that express the relation between the rate of a chemical reaction and the concentration of the reacting substance .
Now, comes to the solution part ; number of moles of ${N_2}{O_4} = 0.1$
We have to calculate the number of moles of $N{O_2}$ ,reaction involve ${N_2}{O_4} \rightleftharpoons 2N{O_2}$
Given ; ${K_p} = 0.14$ , let $\alpha $ be the degree of dissociation
${N_2}{O_4} \rightleftharpoons 2N{O_2}$

	${N_2}{O_4}$	$2N{O_2}$
At $t = 0$	$\alpha = 0.1$	$\alpha = 0$
At ${t_{eq}}$	$0.1(1 - \alpha )$	$0.1(2\alpha )$

Then total number of moles = $0.1 + x$
$\
  {k_p} = \dfrac{{{{[{P^{N{O_2}}}]}^2}}}{{{P^{{N_2}{O_4}}}}} \\
   \Rightarrow \left( {\dfrac{{{{\left( {\dfrac{{2x}}{{0.1 + x}}} \right)}^2}}}{{\left( {\dfrac{{0.1 - x}}{{0.1 + x}}} \right)}}} \right) \\
   \Rightarrow x = 0.017 \\
\ $
Then the number of $N{O_2}$=> $2x = 0.034$
Hence , the correct option is C.

Note:
Rate of reaction depends upon the concentration of reactant , pressure , temperature , catalyst .
The unit of rate constant is $mol{L^{ - 1}}{\sec ^{ - 1}}$ . For exothermic reaction k decreases with an increase in temperature but endothermic reaction k increases with an increase with an increased temperature .