For the reaction "N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g), at what temperature does K_c numerically equal K_p?

a) $\text{0.1203 K}$ b) $\text{12.19 K}$ c) $\text{273 K}$ d) $\text{298 K}$

Aug 28, 2016

I got $\text{12.19 K}$.

For ideal gases, $P V = n R T$. Note that:

• $\frac{n}{V}$ is concentration in $\text{mol/L}$
• ${P}_{{\text{N"_2"O"_4) = (nRT)/(V) = ["N"_2"O}}_{4}} R T$, the partial pressure of ${\text{N"_2"O}}_{4}$.
• ${P}_{{\text{NO"_2) = (nRT)/(V) = ["NO}}_{2}} R T$, the partial pressure of ${\text{NO}}_{2}$.

Recall that stoichiometric coefficients become exponents in equilibrium expressions for gases and aqueous solutions.

So, to relate ${K}_{c}$ and ${K}_{p}$, the expression would be:

${K}_{c} = \left(\left[{\text{NO"_2]^2)/(["N"_2"O}}_{4}\right]\right)$

$\boldsymbol{{K}_{p}} = \left({P}_{{\text{NO"_2)^2)/(P_("N"_2"O}}_{4}}\right)$

$= \left(\left[{\text{NO"_2]RT)^2/(["N"_2"O}}_{4}\right] R T\right)$

= bb(RT(["NO"_2]^2)/(["N"_2"O"_4]) = K_cRT, with units of pressure

Since we are wondering when numerically, ${K}_{c} = {K}_{p}$ for this reaction, we use the expression above in bold to solve for $T$ in $\text{K}$.

$\implies R T {K}_{c} = {K}_{p}$ in $\text{atm}$

Thus, $R T$ must numerically equal $1$ if ${K}_{c}$ is to be equal to ${K}_{p}$.

$\implies \textcolor{b l u e}{T = \frac{{K}_{p}}{R {K}_{c}}}$

It is implied that ${K}_{p}$ is typically reported in units of $\text{atm}$ (only ${K}_{p}$ at standard conditions does not have units). Therefore, we would use $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$ (as opposed to $R = \text{0.083145 L"cdot"bar/mol"cdot"K}$, or $\text{8.314472 J/mol"cdot"K}$).

So, $\textcolor{b l u e}{T} = \frac{1}{0.082057} \cdot \frac{1}{1}$ $\text{K}$ $\approx$ $\textcolor{b l u e}{\text{12.19 K}}$

Aug 28, 2016

$\textsf{12.19 \textcolor{w h i t e}{x} K}$

Explanation:

The relationship between $\textsf{{K}_{c}}$ and $\textsf{{K}_{p}}$ is :

$\textsf{{K}_{p} = {K}_{c} {\left(R T\right)}^{\Delta n}}$

$\textsf{\Delta n}$ is the no. moles product - no. moles reactant

$\textsf{\Delta n = 2 - 1 = 1}$

The only way that $\textsf{{K}_{p}}$ can have the same numerical value as $\textsf{{K}_{c}}$ is when $\textsf{{\left(R T\right)}^{1} = 1}$

$\therefore \textsf{T = \frac{1}{R} = \frac{1}{0.082} = 12.19 \textcolor{w h i t e}{x} K}$

When the question states $\textsf{{K}_{p} = {K}_{c}}$ it should state the dimensions that are being used. If pressure was in $\textsf{{\text{N/m}}^{2}}$ then the value of 8.31 J/K/mol should be used for R which gives a different value for T.

You need to assume that pressure is normalised against 1 atmosphere and concentration is normalised against unit concentration.