For the reaction $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$ $"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)$ , at what temperature does $K_{c}$ $K_c$ numerically equal $K_{p}$ $K_p$ ?

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For the reaction $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$ $"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)$ , at what temperature does $K_{c}$ $K_c$ numerically equal $K_{p}$ $K_p$ ?

$a)$ $a)$ $0.1203 K$ $"0.1203 K"$
$b)$ $b)$ $12.19 K$ $"12.19 K"$
$c)$ $c)$ $273 K$ $"273 K"$
$d)$ $d)$ $298 K$ $"298 K"$

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Answer 1 · 2016-08-28T17:33:47

I got $12.19 K$ $"12.19 K"$ .

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

$\frac{n}{V}$ $n/V$ is concentration in $mol/L$ $"mol/L"$

$P_{N_{2} O_{4}} = \frac{n R T}{V} = [N_{2} O_{4}] R T$ $P_("N"_2"O"_4) = (nRT)/(V) = ["N"_2"O"_4]RT$ , the partial pressure of $N_{2} O_{4}$ $"N"_2"O"_4$ .

$P_{{NO}_{2}} = \frac{n R T}{V} = [{NO}_{2}] R T$ $P_("NO"_2) = (nRT)/(V) = ["NO"_2]RT$ , the partial pressure of ${NO}_{2}$ $"NO"_2$ .

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

So, to relate $K_{c}$ $K_c$ and $K_{p}$ $K_p$ , the expression would be:

$K_{c} = \frac{{[{NO}_{2}]}_{2}^{2}}{[N_{2} O_{4}]}$ $K_c = (["NO"_2]^2)/(["N"_2"O"_4])$

$K_{p} = \frac{P_{{NO}_{2}}^{2}}{P_{N_{2} O_{4}}}$ $bb(K_p) = (P_("NO"_2)^2)/(P_("N"_2"O"_4))$

$= \frac{{([{NO}_{2}] R T)}^{2}}{[N_{2} O_{4}] R T}$ $= (["NO"_2]RT)^2/(["N"_2"O"_4]RT)$

$= R T \frac{{[{NO}_{2}]}_{2}^{2}}{[N_{2} O_{4}]} = K_{c} R T$ $= bb(RT(["NO"_2]^2)/(["N"_2"O"_4]) = K_cRT$ , with units of pressure

Since we are wondering when numerically, $K_{c} = K_{p}$ $K_c = K_p$ for this reaction, we use the expression above in bold to solve for $T$ $T$ in $K$ $"K"$ .

$\Rightarrow R T K_{c} = K_{p}$ $=> RTK_c = K_p$ in $atm$ $"atm"$

Thus, $R T$ $RT$ must numerically equal $1$ $1$ if $K_{c}$ $K_c$ is to be equal to $K_{p}$ $K_p$ .

$\Rightarrow T = \frac{K_{p}}{R K_{c}}$ $=> color(blue)(T = (K_p)/(RK_c))$

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

So, $T = \frac{1}{0.082057} \cdot \frac{1}{1}$ $color(blue)(T) = 1/0.082057 * 1/1$ $K$ $"K"$ $\approx$ $~~$ $12.19 K$ $color(blue)("12.19 K")$

Answer 2 · 2016-08-28T17:56:04

$sf(12.19color(white)(x)K)$

Explanation:

The relationship between $sf(K_c)$ and $sf(K_p)$ is :

$sf(K_p=K_c(RT)^(Deltan))$

$sf(Deltan)$ is the no. moles product - no. moles reactant

$sf(Deltan=2-1=1)$

The only way that $sf(K_p)$ can have the same numerical value as $sf(K_c)$ is when $sf((RT)^1=1)$

$:.sf(T=1/R=1/0.082=12.19color(white)(x)K)$

When the question states $sf(K_p=K_c)$ it should state the dimensions that are being used. If pressure was in $sf("N/m"^2)$ 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.

A badly worded question.

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For the reaction $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$ $"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)$ , at what temperature does $K_{c}$ $K_c$ numerically equal $K_{p}$ $K_p$ ?

$a)$ $a)$ $0.1203 K$ $"0.1203 K"$
$b)$ $b)$ $12.19 K$ $"12.19 K"$
$c)$ $c)$ $273 K$ $"273 K"$
$d)$ $d)$ $298 K$ $"298 K"$

2 Answers

Explanation:

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For the reaction "N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)N2O4(g)⇌2NO2(g)"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g), at what temperature does K_cKcK_c numerically equal K_pKpK_p?

a)a)a) "0.1203 K"0.1203 K"0.1203 K" b)b)b) "12.19 K"12.19 K"12.19 K" c)c)c) "273 K"273 K"273 K" d)d)d) "298 K"298 K"298 K"

2 Answers

Explanation:

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Impact of this question

For the reaction $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$ $"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)$ , at what temperature does $K_{c}$ $K_c$ numerically equal $K_{p}$ $K_p$ ?

$a)$ $a)$ $0.1203 K$ $"0.1203 K"$
$b)$ $b)$ $12.19 K$ $"12.19 K"$
$c)$ $c)$ $273 K$ $"273 K"$
$d)$ $d)$ $298 K$ $"298 K"$