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

Md Ariful Islam

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Md Ariful Islam

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I got $"12.19 K"$.

For ideal gases, $PV = nRT$. Note that:

  • $n/V$ is concentration in $"mol/L"$
  • $P_("N"_2"O"_4) = (nRT)/(V) = ["N"_2"O"_4]RT$, the of $"N"_2"O"_4$.
  • $P_("NO"_2) = (nRT)/(V) = ["NO"_2]RT$, the partial pressure of $"NO"_2$.

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

So, to relate $K_c$ and $K_p$, the expression would be:

$K_c = (["NO"_2]^2)/(["N"_2"O"_4])$

$bb(K_p) = (P_("NO"_2)^2)/(P_("N"_2"O"_4))$

$= (["NO"_2]RT)^2/(["N"_2"O"_4]RT)$

$= bb(RT(["NO"_2]^2)/(["N"_2"O"_4]) = K_cRT$, 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 $"K"$.

$=> RTK_c = K_p$ in $"atm"$

Thus, $RT$ must numerically equal $1$ if $K_c$ is to be equal to $K_p$.

$=> color(blue)(T = (K_p)/(RK_c))$

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

So, $color(blue)(T) = 1/0.082057 * 1/1$ $"K"$ $~~$ $color(blue)("12.19 K")$

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