Question

Two solids dissociated as follows

$\mathrm{A}\left(\mathrm{s}\right)\rightleftharpoons \mathrm{B}\left(\mathrm{g}\right)+\mathrm{C}\left(\mathrm{g}\right){\mathrm{K}}_{\mathrm{P}1}=\mathrm{x}{\mathrm{atm}}^2$

$\mathrm{D}\left(\mathrm{s}\right)\rightleftharpoons \mathrm{C}\left(\mathrm{g}\right)+\mathrm{E}\left(\mathrm{g}\right){\mathrm{K}}_{\mathrm{P}2}=\mathrm{y}{\mathrm{atm}}^2$

The total pressure when both the solids dissociate simultaneously is:

Answer 1

Step 1: Given information:

$\mathbf{A}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{B}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{C}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

• In this reaction, two gases B and C are formed.
• These gases have the same pressure as they are produced in the same amount.
• The gas B and C have Pressure ${\mathrm{P}}_1$.

$\mathbf{D}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{C}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{E}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}$

• In this reaction two gases C and E are formed.
• C and E have the pressure ${\mathrm{P}}_2$.

Step 2: Calculating Equilibrium constant

The reactant is solid and it does not have any pressure.

${\mathrm{K}}_{{\mathrm{P}}_1}=\left[{\mathrm{P}}_{\mathrm{B}}\right]\left[{\mathrm{P}}_{\mathrm{C}}\right]$

Here, ${\mathrm{P}}_{\mathrm{B}}={\mathrm{P}}_1\mathrm{and}{\mathrm{P}}_{\mathrm{C}}={\mathrm{P}}_1+{\mathrm{P}}_2$

Then, ${\mathrm{K}}_{{\mathrm{P}}_1}=\left({\mathrm{P}}_1\right)({\mathrm{P}}_1+{\mathrm{P}}_2)$

Given that ${\mathrm{K}}_{{\mathrm{P}}_1}=\mathrm{x}$

$\mathrm{x}=\left({\mathrm{P}}_1\right)({\mathrm{P}}_1+{\mathrm{P}}_2)......\left(1\right)$

Similarly

${\mathrm{K}}_{{\mathrm{P}}_2}={\mathrm{P}}_{\mathrm{c}}\cdot {\mathrm{P}}_{\mathrm{E}}$

$\mathrm{y}=({\mathrm{P}}_1+{\mathrm{P}}_2)\left({\mathrm{P}}_2\right)...\left(2\right)$

Adding $\left(1\right)\mathrm{and}\left(2\right)$

$\mathrm{x}+\mathrm{y}={({\mathrm{P}}_1+{\mathrm{P}}_2)}^2$

Step 3: Calculating Total Pressure:

${\mathrm{P}}_{\mathrm{T}}={\mathrm{P}}_{\mathrm{C}}+{\mathrm{P}}_{\mathrm{B}}+{\mathrm{P}}_{\mathrm{E}}$

${\mathrm{P}}_{\mathrm{T}}=({\mathrm{P}}_1+{\mathrm{P}}_2)+{\mathrm{P}}_1+{\mathrm{P}}_2$

${\mathrm{P}}_{\mathrm{T}}=2({\mathrm{P}}_1+{\mathrm{P}}_2)$

Since, $\mathrm{x}+\mathrm{y}={({\mathrm{P}}_1+{\mathrm{P}}_2)}^2$

Now,

${\mathrm{P}}_{\mathrm{T}}=2\sqrt{\mathrm{x}+\mathrm{y}}$

Therefore, the total pressure when both the solids dissociate simultaneously is ${\mathbf{P}}_{\mathbf{T}}\mathbf{=}\mathbf{2}\sqrt{\mathbf{x}\mathbf{+}\mathbf{y}}$.