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Question: Given, 10 \(lt\) box contains \({O_3}\) and \({O_2}\) at equilibrium at \(2000K\). The \(\Delta {G_0...

Given, 10 ltlt box contains O3{O_3} and O2{O_2} at equilibrium at 2000K2000K. The ΔG0=−534.52kJ\Delta {G_0} = - 534.52kJ at 8 atmatm equilibrium pressure. The following equilibrium is present in the container, 2O3(g)⇌3O2(g)2{O_3}(g) \rightleftharpoons 3{O_2}(g). The partial pressure of O3{O_3} will be___________. ln⁡10=2.3,R−8.3Jmole−1K−1\ln 10 = 2.3,R - 8.3Jmole - 1K - 1
A.8×10−68 \times {10^{ - 6}}
B.22.62×10−722.62 \times {10^{ - 7}}
C.9.71×10−69.71 \times {10^{ - 6}}
D.9.7×10−29.7 \times {10^{ - 2}}

Explanation

Solution

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-
ΔG0=−RTln⁡K\Delta {G^0} = - RT\ln K
Now, we’ll substitute the values that are given to us-
ΔG0\Delta {G^0} = −2.3×2000×8.3log⁡K - 2.3 \times 2000 \times 8.3\log K
534.52×1032.3×8.3×2000\dfrac{{534.52 \times {{10}^3}}}{{2.3 \times 8.3 \times 2000}} = log⁡K\log K
KK = 1014{10^{14}}
Using, the given equilibrium equation in the question-
2O3(g)⇌3O2(g)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, PO3<<<PO2{P_{{O_3}}} < < < {P_{{O_2}}}
And we also know that, PO2+PO3=8{P_{{O_2}}} + {P_{{O_3}}} = 8
Now, PO3=8atm{P_{{O_3}}} = 8atm
K=1014=(PO2)3(PO3)2=(8)3PO3K = {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;
PO3=22.62×10−7atm{P_{{O_3}}} = 22.62 \times {10^{ - 7}}atm

So, option B is the correct answer.

Additional information:
What is ΔG0\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 ascertainΔG\Delta G, deduct the measure of energy lost to entropy (meant as ΔS\Delta S) from the complete energy change of the framework. This complete energy change in the framework is called enthalpy and is signified asΔH\Delta H. The recipe for ascertaining ΔG\Delta G is as per the following, where the image T alludes to outright temperature in Kelvin (degrees Celsius + 273:G=ΔH−TΔS.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.