91影视

Given balance chemical equation is shown below:

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

Given data:${\text{K}}_{\text{C}}{\text{= 1.7脳10}}^{\text{8}}$

${\text{P}}_{{\text{SO}}_{\text{3}}}\text{= 300.atm}$

${\text{P}}_{{\text{O}}_{\text{2}}}\text{= 100.atm}$

To find the partial pressure of SO₂, equation in terms of Kp is defined as,

${\mathbf{K}}_{\mathbf{P}}\mathbf{=}\frac{{\left[{\mathbf{P}}_{{\mathbf{SO}}_{\mathbf{3}}}\right]}^{\mathbf{2}}}{{\left[{\mathbf{P}}_{{\mathbf{SO}}_{\mathbf{2}}}\right]}^{\mathbf{2}}\left[{\mathbf{P}}_{{\mathbf{O}}_{\mathbf{2}}}\right]}$ (1)

We need to find the value of kp using Kc given above:

${\mathbf{K}}_{\mathbf{P}}\mathbf{=}{\mathbf{K}}_{\mathbf{c}}{\left(\mathbf{RT}\right)}^{\mathbf{鈭�}\mathbf{n}}$ (2)

Were,$鈭�n=2-3=-1$

${\text{K}}_{\text{P}}{\text{= 1.7脳10}}^{\text{8}}\text{脳}{\left(\text{0.082脳1600}\right)}^{\text{- 1}}$

${\text{= 3.451脳10}}^6\text{atm}$

Substituting the value of Kp in equation 1 we get,

${\text{3.451脳10}}^{\text{6}}\text{atm =}\frac{{\left[\text{300.atm}\right]}^{\text{2}}}{{\left[{\text{P}}_{{\text{SO}}_{\text{2}}}\right]}^{\text{2}}\left[\text{100.atm}\right]}$

${\left[{\text{P}}_{{\text{SO}}_{\text{2}}}\right]}^{\text{2}}\text{=}\frac{{\left[\text{300.atm}\right]}^{\text{2}}}{{\text{3.451脳10}}^{\text{6}}\text{补迟尘脳}\left[\text{100.atm}\right]}$

$\text{= 2.6079 atm}$

$$\begin{gathered} \left[{\text{P}}_{{\text{SO}}_{\text{2}}}\right]\text{=}\sqrt{\text{2.6079}} \\ \text{= 1.6149 atm} \end{gathered}$$

Equilibrium constant, ${\text{K}}_{\text{C}}\text{=}\frac{{\left[{\text{SO}}_{\text{3}}\right]}^{\text{2}}}{{\left[{\text{SO}}_{\text{2}}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]}$

$\text{=}\frac{{\left(\text{0.0020}\right)}^{\text{2}}}{{\left(\text{0.0040}\right)}^2\left(\text{0.0028}\right)}$

${\text{= 7.0脳10}}^{\text{4}}$

Thus, Kc is calculated, and it is${\text{7.0脳10}}^{\text{4}}$

Partial pressure ossf the reactant SO₂ can be determined using the ideal gas law given below:

$$\begin{gathered} \text{PV = nRT} \\ \text{ie, P =}\frac{\text{n}}{\text{v}}\text{RT} \end{gathered}$$

${\text{P}}_{{\text{SO}}_{\text{2}}}\text{= 0.0040}\frac{\text{mol}}{\text{L}}\text{脳0.0821}\frac{\text{L. atm}}{\text{mol. k}}\text{脳1000碍}$

$\text{= 0.3284 atm}$

Hence partial pressure of SO₂ is 0.3284atm