Question

Calculate the partial pressure of carbon monoxide form the following:

$Ca{CO}_3\left(s\right)\underrightarrow{\Delta }CaO\left(s\right)+{CO}_2\uparrow ;K_p=8\times {10}^{-2}$

${CO}_2\left(s\right)+C\left(s\right)⟶2CO\left(s\right);K_p=2$

• A. 0.2
• B. 0.4
• C. 1.6
• D. 4

Answer 1

The correct option is B 0.4

Solution:- (B) $0.4$

As we know that,

$K_p=\frac{\text{Partial pressure of product}}{\text{Partial pressure of reactant}}$

Given:-

${CaC{O}_3}_{\left(s\right)}⟶{CaO}_{\left(s\right)}+{C{O}_2}_{\left(g\right)};K_p=8\times {10}^{-2}$

From the above reaction-

$K_p=P_{C{O}_2}=8\times {10}^{-2}.....\left(1\right)$

$C_{\left(s\right)}+{C{O}_2}_{\left(g\right)}⟶2{CO}_{\left(g\right)};K_p=2$

From the above reaction-

$K_p=\frac{{\left(P_{CO}\right)}^2}{P_{C{O}_2}}$

$\Rightarrow \frac{P_{CO}^2}{P_{C{O}_2}}=2.....\left(2\right)$

Now, from ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\frac{{\left(P_{CO}\right)}^2}{8\times {10}^{-2}}=2$

$\Rightarrow {\left(P_{CO}\right)}^2=16\times {10}^{-2}$

$\Rightarrow P_{CO}=\sqrt{0.16}=0.4$

Hence the partial preeure of carbon monoxide is $0.4$.