Q17.29P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Calculate Kc for each of the following equilibria:

$$\begin{gathered} \left(\mathbf{a}\right)\mathbf{CO}\left(\mathbf{g}\right)\mathbf{+}{\mathbf{Cl}}_{\mathbf{2}}\left(\mathbf{g}\right)\rightleftharpoons {\mathbf{COCl}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{;}{\mathbf{K}}_{\mathbf{P}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{at}\mathbf{1000}\mathbf{.}\mathbf{K} \\ \left(\mathbf{b}\right){\mathbf{S}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{+}\mathbf{C}\left(\mathbf{s}\right)\rightleftharpoons {\mathbf{CS}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{;}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{28}\mathbf{.}\mathbf{5}\mathbf{at}\mathbf{500}\mathbf{.}\mathbf{K} \end{gathered}$$

Step-by-Step Solution

Answer

Answer

 Kc and Kp are related to each other through the equation given shown below:

$K_p=K_c{\left(RT\right)}^{∆n}$

Number of moles in product side and reactant side may vary with different chemical reactions. So, substituting value of in above equation gives following Kc values.

1. K_c = 3.2019
2. K_c = 28.5

Step 1: Determining Kc for the given equation (a)

$∆n =1 -2 =-1$Given equation is shown below:

$\text{CO}\left(\text{g}\right){\text{ + Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{COCl}}_{\text{2}}\left(\text{g}\right)$

Given $K_{p }=3.9\times {10}^{-2}$

Relation between Kp and Kc can be given in the equation,

${\mathbf{K}}_{\mathbf{P}}\mathbf{=}{\mathbf{K}}_{\mathbf{c}}{\left(\mathbf{RT}\right)}^{\mathbf{\Delta n}}$

Were,$∆n=1-2=-1$

So, above equation will be${\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{ - 1}}$

$$\begin{gathered} {\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\left(\text{RT}\right)}^1 \\ {\text{=3.9×10}}^{-2}\text{×}\left(0.0821\times 1000\right) \\ \text{=3.2019} \end{gathered}$$

Equilibrium constant of the given equation is 3.2019

Step 2: Determining Kc for the given equation (b)

Given equation is shown below:

${\text{S}}_{\text{2}}\left(\text{g}\right)\text{ + C}\left(\text{s}\right)\rightleftharpoons {\text{CS}}_{\text{2}}\left(\text{g}\right)$

Given K_p =28.5

Relation between Kp and Kc can be given in the equation,

${\text{K}}_{\text{P}}{\text{ = K}}_{\text{c}}{\left(\text{RT}\right)}^{∆n}$

Were,$∆n=1-1=0$

Rearranging the above equation to find Kc then,

(RT term turns to zero)

So, value of Kc is same for Kp

Equilibrium constant of the given equation is 28.5