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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.

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  • $\begingroup$ Wrong definitions. Kc is uses molar concentrations, whereas Kp uses the partial pressures of the gasses $\endgroup$ – MaxW Feb 5 '17 at 19:35
  • $\begingroup$ I was referring to the physical states that are involved in Kc and Kp. $\endgroup$ – Ray Ruiz Feb 5 '17 at 19:39
  • $\begingroup$ They are equal when the sum of the stoichiometric coefficients of the products equals the sum of the S.C. for the reactants (assuming that you are treating the gas as ideal which I assume you are). $\endgroup$ – LordStryker Feb 5 '17 at 19:56
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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.

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  • $\begingroup$ Very elegant proofing. Just a wandering question, is there any relationship been Kc, Kp and oxidation states of the reactants and products considered for the Kc, Kp? I wanted to shoot it as a question, but then I came across this. Looking at OP's chosen reaction stoichiometries, I get the feeling there is. $\endgroup$ – bonCodigo Jul 24 '17 at 22:08
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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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  • 1
    $\begingroup$ This is already mentioned in another answer which also elaborates on why this is the case and demonstrates the proof. $\endgroup$ – andselisk Apr 16 at 21:52

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