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Question: At \(700K\), the equilibrium constant, \({K_P}\) for the reaction \(2S{O_3}\left( g \right) \rightle...

At 700K700K, the equilibrium constant, KP{K_P} for the reaction 2SO3(g)⇄2SO2(g)+O2(g)2S{O_3}\left( g \right) \rightleftarrows 2S{O_2}\left( g \right) + {O_2}\left( g \right) is 1.80×10−3kpa1.80 \times {10^{ - 3}}kpa. What is the numerical value in moles per liter of Kc{K_c} for this reaction at same temperature?
A.3.09×10−7molL−13.09 \times {10^{ - 7}}mol{L^{ - 1}}
B.3.09×10−6molL−13.09 \times {10^{ - 6}}mol{L^{ - 1}}
C.3.09×107molL−13.09 \times {10^7}mol{L^{ - 1}}
D.3.09×106molL−13.09 \times {10^6}mol{L^{ - 1}}

Explanation

Solution

We have to know that moles per liter of Kc{K_c} is nothing but concentration of the reaction. In order to determine the concentration, we have to first determine the number of changes in moles. After determining the change in moles, we have to change the value of KP{K_P} in kpakpa to KP{K_P} in atm. We can convert the value of kpakpa to atm using the conversion factor. We have to then substitute the values of temperature, gas constant, change in moles and KP{K_P} in the expression for Kc{K_c}.

Complete answer:
We are provided with following information,
The temperature is 700K700K.
The value of KP{K_P} is 1.80×10−3kpa1.80 \times {10^{ - 3}}kpa.
The chemical reaction is given as,
2SO3(g)⇄2SO2(g)+O2(g)2S{O_3}\left( g \right) \rightleftarrows 2S{O_2}\left( g \right) + {O_2}\left( g \right)
Let us determine the change in moles. We can calculate the change in moles by subtracting the moles of products with moles of reactant. We can represent the change in moles as Δn\Delta n.
We can write the formula to change in moles as,
Δn=nproducts−nreactan⁡ts\Delta n = {n_{products}} - {n_{reac\tan ts}}
We can see that there are two moles of sulfur dioxide and one mole of oxygen on the product side. So, the moles of the product are three.
In the reactant side, the moles of sulfur trioxide are two.
Let us now substitute these values in expression for change in moles.
Δn=nproducts−nreactan⁡ts\Delta n = {n_{products}} - {n_{reac\tan ts}}
Now we can substitute the known values we get,
Δn=3−2\Delta n = 3 - 2
On simplification we get,
Δn=1\Delta n = 1
We have obtained the change in moles as 1.
Let us now convert the value of KP{K_P} in atm.
We can convert the value ofKP{K_P} in kpakpa to atm using the conversion factor.
Kp=1.8×10−3kPaa101.3kPaatm−1=1.78×10−5atm{K_p} = \dfrac{{1.8 \times {{10}^{ - 3}}kP{a_a}}}{{101.3kPaat{m^{ - 1}}}} = 1.78 \times {10^{ - 5}}atm
We have obtained the value of KP{K_P} in atm as 1.78×10−5atm1.78 \times {10^{ - 5}}atm.
We have to use the following expression below to calculate Kc{K_c}
Kc=Kp(RT)Δn{K_c} = \dfrac{{{K_p}}}{{{{\left( {RT} \right)}^{\Delta n}}}}
Let us now then substitute the values of temperature, gas constant, change in moles and KP{K_P} in the expression for Kc{K_c}.
We can calculate the value of Kc{K_c} as,
Kc=1.78×10−5atm0.082LatmK−1mol−1×700K{K_c} = \dfrac{{1.78 \times {{10}^{ - 5}}atm}}{{0.082Latm{K^{ - 1}}mo{l^{ - 1}} \times 700K}}
On simplification we get,
Kc=3.09×10−7molL−1{K_c} = 3.09 \times {10^{ - 7}}mol{L^{ - 1}}
We have calculated the value of Kc{K_c} as 3.09×10−7molL−13.09 \times {10^{ - 7}}mol{L^{ - 1}}.

Option (A) is correct.

Note:
We have to know that while calculating the Kc{K_c}, it is mandatory to change in
pressure to atm to equate it with gas constant. If we are not converting the value in the standard unit of pressure, then we have used the value of the gas constant based on the unit of pressure that is provided. If one is kept at standard value and other is present in non-standard value, there are chances for calculative error.