Question

Given, 10 $lt$ box contains ${O_3}$ and ${O_2}$ at equilibrium at $2000K$. The $\Delta {G_0} = - 534.52kJ$ at 8 $atm$ equilibrium pressure. The following equilibrium is present in the container, $2{O_3}(g) \rightleftharpoons 3{O_2}(g)$. The partial pressure of ${O_3}$ will be___________. $\ln 10 = 2.3,R - 8.3Jmole - 1K - 1$
 A.$8 \times {10^{ - 6}}$
B.\[22.62 \times {10^{ - 7}}\]
C.$9.71 \times {10^{ - 6}}$
D.$9.7 \times {10^{ - 2}}$

Answer 1

Hint:In this question we will be seeing the equilibrium states of oxygen and ozone compounds, their partial pressures, and the chemical reaction that involves change in free energy.Each chemical reaction includes an adjustment in free energy, called delta G.

Complete answer:
First, let’s write the formula for the free energy change-
$\Delta {G^0} = - RT\ln K$
Now, we’ll substitute the values that are given to us-
$\Delta {G^0}$ = $ - 2.3 \times 2000 \times 8.3\log K$
$\dfrac{{534.52 \times {{10}^3}}}{{2.3 \times 8.3 \times 2000}}$ = $\log K$
\[K\] = ${10^{14}}$
Using, the given equilibrium equation in the question-
$2{O_3}(g) \rightleftharpoons 3{O_2}(g)$
We know that, The Partial Pressure of Ozone is very very less when compared to the Partial Pressure of Oxygen, That is represented by, ${P_{{O_3}}} < < < {P_{{O_2}}}$
And we also know that, ${P_{{O_2}}} + {P_{{O_3}}} = 8$
Now, ${P_{{O_3}}} = 8atm$
$K = {10^{14}} = \dfrac{{{{({P_{{O_2}}})}^3}}}{{{{({P_{{O_3}}})}^2}}} = \dfrac{{{{(8)}^3}}}{{{P_{{O_3}}}}}$
Now, after the simplification we get the partial pressure of Ozone as below;
${P_{{O_3}}} = 22.62 \times {10^{ - 7}}atm$

So, option B is the correct answer.

Additional information:
What is $\Delta {G^0}$?
Each chemical reaction includes an adjustment in free energy, called delta G. The adjustment in free energy can be determined for any framework that goes through a change, for example, a chemical reaction. To compute ∆G, deduct the measure of energy lost to entropy from the complete energy change of the framework. To ascertain$\Delta G$, deduct the measure of energy lost to entropy (meant as $\Delta S$) from the complete energy change of the framework. This complete energy change in the framework is called enthalpy and is signified as$\Delta H$. The recipe for ascertaining $\Delta G$ is as per the following, where the image T alludes to outright temperature in Kelvin (degrees Celsius + 273:$G = \Delta H - T\Delta S.$

Note:
Partial Pressure is the power applied by a gas. The total of the partial pressures of all gases in a combination approaches the total pressure. Total pressure is the sum of all individual gases in a mixture.