Q17.30P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Calculate Kc for each of the following equilibria:

$$\begin{gathered} \left(\mathbf{a}\right){\mathbf{H}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{+}{\mathbf{I}}_{\mathbf{2}}\left(\mathbf{g}\right)\rightleftharpoons \mathbf{2}\mathbf{HI}\left(\mathbf{g}\right)\mathbf{;}{\mathbf{K}}_{\mathbf{P}}\mathbf{=}\mathbf{49}\mathbf{at}\mathbf{730}\mathbf{.}\mathbf{K} \\ \left(\mathbf{b}\right)\mathbf{2}{\mathbf{SO}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(\mathbf{s}\right)\rightleftharpoons \mathbf{2}{\mathbf{SO}}_{\mathbf{3}}\left(\mathbf{g}\right)\mathbf{;}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{at}\mathbf{500}\mathbf{.}\mathbf{K} \end{gathered}$$

Step-by-Step Solution

Answer

Answer

Kc and Kp are related to each other through the equation 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 = 49
2. K_C =  $2.5\times {10}^{10}$
   

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

Given equation is shown below:

${\text{H}}_{\text{2}}\left(\text{g}\right){\text{ + I}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{2HI}\left(\text{g}\right)$

Given

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

${\text{K}}_{\text{P}}\text{ = 49}$

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

Were,$∆n=2-2 =0$

Rearranging the above equation to find Kc then,

${\text{K}}_{\text{c}}{\text{ = K}}_{\text{p}}$

So, in this reaction value of Kp and Kc are equal.

Equilibrium constant of the given equation is  49

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

Given equation is shown below:

${\text{2SO}}_{\text{2}}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{s}\right)\rightleftharpoons {\text{2SO}}_{\text{3}}\left(\text{g}\right)$

Given $K_{p }=2.5\times {10}^{10}$

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

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

Rearranging the above equation to find Kc then,

${\text{K}}_{\text{c}}{\text{ = K}}_{\text{p}}$

So, in this reaction value of Kp and Kc are equal.

Equilibrium constant of the given equation is 2.5$\times {10}^{10}$