Question

Consider the following reaction: . SO2Cl2(g)⇌SO2(g)+Cl2(g) . A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M . At equilibrium, [Cl2]= 1.3×10−2M .. . Calculate the value of the equilibrium constant (Kc).

Answer 1

The equilibrium constant (Kc) is the product of the equilibrium concentrations of the products raised to their corresponding stoichiometric coefficients divided by the reactants as well. In this case the equilibrium concentration of Cl2 which also applies to SO2 is 1.3x10^-2. The final equilibrium concentration of SO2Cl2 is 9x10^-3. Kc is then equal to 0.0188.

Answer 2

Answer : The the value of equilibrium constant is, $1.87\times 10^{-2}$

Solution : Given,

Concentration of $SO_2Cl_2$ = $2.2\times 10^{-2}M$

Concentration of $Cl_2$ = $1.3\times 10^{-2}M$

The given balanced equilibrium reaction is,

                     $SO_2Cl_2(g)\rightleftharpoons SO_2(g)+Cl_2(g)$

Initially          $2.2\times 10^{-2}M$        0        0

At eqm,

The concentration of $SO_2Cl_2$ = $(2.2\times 10^{-2}-1.3\times 10^{-2})M=0.9\times 10^{-2}M$

The concentration of $Cl_2$ = $1.3\times 10^{-2}M$

The concentration of $SO_2$ = $1.3\times 10^{-2}M$

The expression for equilibrium constant will be,

$K_c=\frac{[SO_2]\times [Cl_2]}{[SO_2Cl_2]}$

Now put all the given values in this formula, we get

$K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{(0.9\times 10^{-2})}$

$K_c=1.87\times 10^{-2}$

Therefore, the value of equilibrium constant is, $1.87\times 10^{-2}$