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# At $700K$, the equilibrium constant, ${K_P}$ for the reaction $2S{O_3}\left( g \right) \rightleftarrows 2S{O_2}\left( g \right) + {O_2}\left( g \right)$ is $1.80 \times {10^{ - 3}}kpa$. What is the numerical value in moles per liter of ${K_c}$ for this reaction at same temperature?A.$3.09 \times {10^{ - 7}}mol{L^{ - 1}}$B.$3.09 \times {10^{ - 6}}mol{L^{ - 1}}$C.$3.09 \times {10^7}mol{L^{ - 1}}$D.$3.09 \times {10^6}mol{L^{ - 1}}$

Last updated date: 21st Jul 2024
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Hint: We have to know that moles per liter of ${K_c}$ 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 ${K_P}$ in $kpa$ to ${K_P}$ in atm. We can convert the value of $kpa$ to atm using the conversion factor. We have to then substitute the values of temperature, gas constant, change in moles and ${K_P}$ in the expression for ${K_c}$.

We are provided with following information,
The temperature is $700K$.
The value of ${K_P}$ is $1.80 \times {10^{ - 3}}kpa$.
The chemical reaction is given as,
$2S{O_3}\left( g \right) \rightleftarrows 2S{O_2}\left( g \right) + {O_2}\left( g \right)$
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 $\Delta n$.
We can write the formula to change in moles as,
$\Delta n = {n_{products}} - {n_{reac\tan ts}}$
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.
$\Delta n = {n_{products}} - {n_{reac\tan ts}}$
Now we can substitute the known values we get,
$\Delta n = 3 - 2$
On simplification we get,
$\Delta n = 1$
We have obtained the change in moles as 1.
Let us now convert the value of ${K_P}$ in atm.
We can convert the value of${K_P}$ in $kpa$ to atm using the conversion factor.
${K_p} = \dfrac{{1.8 \times {{10}^{ - 3}}kP{a_a}}}{{101.3kPaat{m^{ - 1}}}} = 1.78 \times {10^{ - 5}}atm$
We have obtained the value of ${K_P}$ in atm as $1.78 \times {10^{ - 5}}atm$.
We have to use the following expression below to calculate ${K_c}$
${K_c} = \dfrac{{{K_p}}}{{{{\left( {RT} \right)}^{\Delta n}}}}$
Let us now then substitute the values of temperature, gas constant, change in moles and ${K_P}$ in the expression for ${K_c}$.
We can calculate the value of ${K_c}$ as,
${K_c} = \dfrac{{1.78 \times {{10}^{ - 5}}atm}}{{0.082Latm{K^{ - 1}}mo{l^{ - 1}} \times 700K}}$
On simplification we get,
${K_c} = 3.09 \times {10^{ - 7}}mol{L^{ - 1}}$
We have calculated the value of ${K_c}$ as $3.09 \times {10^{ - 7}}mol{L^{ - 1}}$.

Option (A) is correct.

Note:
We have to know that while calculating the ${K_c}$, 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.