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Question: At \(700{\text{K}}\), the equilibrium constant, \[{{\text{K}}_{\text{p}}}\] ​ for the reaction \[...

At 700K700{\text{K}}, the equilibrium constant, Kp{{\text{K}}_{\text{p}}} ​ for the reaction
2SO3(g)⇌2SO2(g) + O2(g)  {\text{2S}}{{\text{O}}_{\text{3}}}\left( {\text{g}} \right) \rightleftharpoons {\text{2S}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right){\text{\;}} is 1.8×10−3 kPa1.8 \times {10^{ - 3}}{\text{ kPa}}. What is the numerical value of Kc{{\text{K}}_{\text{c}}}

Explanation

Solution

To answer this question, you should recall the concept of dissociation. Calculate the change in the number of atoms of the constituents in the reaction and use the relation between Kp{{\text{K}}_{\text{p}}}and Kc{{\text{K}}_{\text{c}}}to find the answer to this question.

The formula used: Kp = Kc(RT)\Deltang{{\text{K}}_{\text{p}}}{\text{ = }}{{\text{K}}_{\text{c}}}{\left( {{\text{RT}}} \right)^{{\text{\Delta ng}}}} - ---(i) where \Deltan = change in gaseous moles = nproducts−nreactants{\text{\Delta n = change in gaseous moles = }}{{\text{n}}_{{\text{products}}}} - {{\text{n}}_{{\text{reactants}}}} , Kp{{\text{K}}_{\text{p}}} is pressure constant, Kc{{\text{K}}_{\text{c}}} is equilibrium constant, R{\text{R}} is the universal gas constant and T{\text{T}} is temperature.

Complete Step by step solution:
We know that equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. We know the reaction already:
2SO3⇌2SO2+O22S{O_3} \rightleftharpoons 2S{O_2} + {O_2}.
We first need to calculate the change in gaseous moles:
\Deltan = nproducts−nreactants=3−2=1{\text{\Delta n = }}{{\text{n}}_{{\text{products}}}} - {{\text{n}}_{{\text{reactants}}}} = 3 - 2 = 1.
Now we know that 1 atm = 105  Pa1{\text{ atm = }}{10^5}\;{\text{Pa}}.
Value of Kc{{\text{K}}_{\text{c}}}is given, we just need to convert it into appropriate units:
Kp=1.8×10−3KPa=1.8Pa=1.8×10−5atm{{\text{K}}_{{\text{p}}}} = 1.8 \times {10^{ - 3}}{\text{KPa}} = 1.8{\text{Pa}} = 1.8 \times {10^{ - 5}}{\text{atm}}.
Temperature is given as 700K700{\text{K}}. Universal gas constant R = 0.0821 L atm mol−1 K−1{\text{R = }}0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}.
Now finally using the formula given in equation (i)
Kp = Kc(RT)\Deltang{{\text{K}}_{\text{p}}}{\text{ = }}{{\text{K}}_{\text{c}}}{\left( {{\text{RT}}} \right)^{{\text{\Delta ng}}}}​,
Rearranging we get:
Kc = KpRT{{\text{K}}_{\text{c}}}{\text{ = }}\frac{{{{\text{K}}_{\text{p}}}}}{{{\text{RT}}}}.
Substituting the calculated known values into this relation:
Kc=1.8×10−5700×0.0821=0.031×10−5=3.1×10−7  molL - 1{{\text{K}}_{\text{c}}} = \frac{{1.8 \times {{10}^{ - 5}}}}{{700 \times 0.0821}} = 0.031 \times {10^{ - 5}} = 3.1 \times {10^{ - 7}}\;{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}.
Therefore, we can conclude that the correct answer to this question is 3.1×10−7  molL - 13.1 \times {10^{ - 7}}\;{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}

Note: You should know the change in concentration, pressure, catalyst, inert gas addition, etc. not affect equilibrium constant. According to Le Chatelier's principle the temperature, concentration, pressure, catalyst, inert gas addition can lead to a shift in equilibrium position only. We know that activation energy is the minimum energy required to start a chemical reaction. Collisions of particles lead to reactions. Only particles that collide sufficiently, can react. Now comes the important point. From kinetics of a reaction, we know that the rate of a reaction increases with increase in temperature due to more energy and more collisions. But the extent of increase in this rate depends on the “energy of activation” of the reaction which is different for both - the forward and the backward reaction. So, a given increase in temperature leads to an increase in the rate of forward and backward reactions to different extents. So, the value of the equilibrium constant changes with temperature.