When are Kp and Kc values equal?

[![Test screenshot](https://i.stack.imgur.com/KhEP7.png)](https://i.stack.imgur.com/KhEP7.png)


The picture shows the correct answers to this question.


I'm not sure why these answers are correct. $K\_p$ only counts with gases, while $K\_c$ only counts with aqueous solution + gases. I assumed that any reactions with only gases, or a mix of gases and solids, will have an equal $K\_p$ and $K\_c$ value, but that is clearly not the case.




Disclaimer: The following information has been extracted from my own Chemistry lecture notes which was built on contributions from many other general chemistry textbooks. It is a stereotypical example of relating $K\_c$ and $K\_p$.


**$K\_c$ and $K\_p$ relationship**


For a simple reaction involving gas phase substances


$$\ce{aA(g) <=> bB(g)}$$


The equilibrium constant can be written in terms of molar concentrations or partial pressures as


$$K\_c = \frac{[\ce{B}]^b}{[\ce{A}]^a} \qquad\text{or}\qquad K\_p = \frac{(P\_\ce{B})^b}{(P\_\ce{A})^a}$$


Assuming that the gas behaves ideally (a common and convenient assumption for general cases), we can write the ideal gas law for each gas such that


$$ PV = nRT \quad\to\quad P\_\ce{A}V = n\_\ce{A}RT \quad\to\quad P\_\ce{A} = \left(\frac{n\_\ce{A}}{V}\right)RT$$


$$ PV = nRT \quad\to\quad P\_\ce{B}V = n\_\ce{B}RT \quad\to\quad P\_\ce{B} = \left(\frac{n\_\ce{B}}{V}\right)RT$$


We note that
$$ \frac{n\_A}{V} \equiv M $$


where $M$ is simply molarity (moles/unit volume). Since $[\ce{X}] = M\_x$, we get


$$P\_\ce{A} = [\ce{A}]RT \qquad\text{and}\qquad P\_\ce{B} = [\ce{B}]RT$$


Substitute these terms into the $K\_p$ equation to give


$$K\_p = \frac{(P\_\ce{B})^b}{(P\_\ce{A})^a} = \frac{([\ce{B}]RT)^b}{([\ce{A}]RT)^a}$$


and recognize that 


$$\frac{[\ce{B}]^b}{[\ce{A}]^a} = K\_c$$


Plug this relationship into the $K\_p$ expression to give


$$K\_p = K\_c (RT)^{b-a} \quad\to\quad K\_p = K\_c (RT)^{\Delta n}$$


Where $\Delta n$ is simply the difference of the sum of the stoichiometric coefficients of the products and reactants from a balanced chemical equation like so


$$\Delta n = \sum (\mathrm{products}) - \sum (\mathrm{reactants})$$


It should be clear from this that the only way $K\_c = K\_p$ within the context of an ideal gas is when $\Delta n = 0$.


$$\begin{align\*} 
K\_p = K\_c(RT)^{\Delta n} \quad\to\quad K\_p &= K\_c(RT)^{0} \\
&= K\_c (1) \\
&= K\_c
\end{align\*}$$


By visual inspection of your image, it is clear why these are the correct answers based upon the derived relationship offered here.


 11

Kc=Kp when the mols of gas on the left side of the equation is equal to the number of mols of gas on the right side of the equation.


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