Question

Given,${{K}_{c}}$ for the reaction $S{{O}_{2}}+\dfrac{1}{2}{{O}_{2}}\overset{{}}{\leftrightarrows}S{{O}_{3}}$ at ${{600}^{o}}C$ is 61.7. Calculate ${{K}_{p}}$ . What is the unit of ${{K}_{p}}$ for the above equilibrium? (R = 0.0821 lit. atm per deg per mole)

Answer 1

All the quantities that are going to be used in the equation should be in standard units, such as temperature. ${{K}_{c}}$ is the equilibrium constant of a reaction with respect to molar concentration and ${{K}_{p}}$ is the equilibrium constant of a reaction with respect to atmospheric pressure.

Complete answer:
Let us first describe what we have been given.
${{K}_{c}}$ is the equilibrium constant with respect to molar concentration, ${{K}_{p}}$ is the equilibrium constant with respect to atmospheric pressure and R is universal gas constant.
${{K}_{c}}$ = 61.7 at temperature T = 600 degree C, which is T = 600 + 273 = 873K. Thus, the temperature is 873 kelvin.
Universal gas constant R = 0.0821 with the unit of $\dfrac{L\times atm}{K\times mol}$ .
Now, to find out the value of ${{K}_{p}}$ , we have to use the following formula.
${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$
Thus, from this formula we can calculate the value of ${{K}_{p}}$ .
$\Delta n$ represents the difference between the number of moles of a product with the number of moles of reactants.
Thus, $\Delta n$ = Number of moles of product – number of moles of reactant.
As we can see in the given reaction, one mole of sulphur trioxide is formed by the addition of one mole of sulphur dioxide and half mole of oxygen.
Therefore, $\Delta n$ = $1-(1+\dfrac{1}{2})$
$\Delta n$ = $-\dfrac{1}{2}$
As we have got all the values, let us substitute them in the equation and find the value of${{K}_{p}}$ .
Therefore, ${{K}_{p}}$ = $61.7{{(0.0821\times 873)}^{-\dfrac{1}{2}}}$
${{K}_{p}}$ = 7.29
Thus, the value of equilibrium constant with respect to atmospheric pressure is 7.29.
Now, let us find its unit.
The value of equilibrium constant with respect to atmospheric pressure, that is ${{K}_{p}}$ , is given as follows.
${{K}_{p}}=\dfrac{[{{P}_{S{{O}_{3}}}}]}{[{{P}_{S{{O}_{2}}}}]{{[{{P}_{{{O}_{2}}}}]}^{\dfrac{1}{2}}}}$
Now, as the unit of pressure in universal gas constant is taken as atm, the unit of pressure here will remain the same.
Therefore, ${{K}_{p}}=\dfrac{atm}{atm\times {{(atm)}^{\dfrac{1}{2}}}}$
Thus, the final unit that we obtain is ${{K}_{p}}=at{{m}^{-\dfrac{1}{2}}}$

Note:
Care should be taken while deciding the unit of entities which are to be put in the equation, if here the temperature was not converted from celsius to kelvin, the answer would have changed. Also, the relation between equilibrium constant with respect to molar concentration, ${{K}_{c}}$ and equilibrium constant with respect to atmosphere, ${{K}_{p}}$ should be known, otherwise the value of ${{K}_{p}}$ could not have been found.