Using an ICE Table, how do we calculate $K_{c}$ $K_c$ for the following reaction?

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Using an ICE Table, how do we calculate $K_{c}$ $K_c$ for the following reaction?

At $373 K$ $"373 K"$ , $0.1 mols$ $"0.1 mols"$ of $N_{2} O_{4}$ $N_2O_4$ is heated in a one liter flask and at equilibrium the amount of nitrogen dioxide is $0.12 mols$ $"0.12 mols"$ .

$N_{2} O_{4} ⇌ 2 N O_{2}$ $N_2O_4 rightleftharpoons 2NO_2$

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Answer 1 · 2016-11-12T03:09:44

$K_{c} = 0.36$ $K_c = 0.36$

For starters, we (of course) assume ideal gases (these are both gases at room temperature).

Then, we recall the definition of the concentration equilibrium constant for a two-component equilibrium:

$K_{c} = \frac{{[B]}_{B}^{ν_{B}}}{{[A]}_{A}^{ν_{A}}}$ $K_c = ([B]^(nu_B))/([A]^(nu_A))$ ,

for the equilibrium $ν_{A} A ⇌ ν_{B} B$ $nu_A A rightleftharpoons nu_B B$ .

Even though we are looking at two gases, we are given the $mol$ $bb"mol"$ s and the total vessel's volume, so we can still get initial and equilibrium concentrations as usual.

However, since the vessel is not going to change size, and ideal gases are assumed to fill the vessel completely and evenly, let us simply work in $mol$ $"mol"$ s for the ICE table, and divide by $1 L$ $"1 L"$ later.

Normally, we would write this:

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ $N_2O_4(g) rightleftharpoons 2NO_2(g)$

$I 0.1 0$ $"I"" "" "0.1" "" "" "0$
$C - x + 2 x$ $"C"" "" "- x" "+2x$
$E 0.1 - x 2 x$ $"E"" " 0.1 - x" "" "2x$

noting that the consumption of $x$ $x$ $mol$ $"mol"$ s of $N_{2} O_{4}$ $"N"_2"O"_4$ yields $2 x$ $2x$ $mol$ $"mol"$ s of ${NO}_{2} (g)$ $"NO"_2(g)$ , and not just $x$ $x$ .

But we know what $x$ $x$ is. Since at equilibrium, $n_{N O_{2}} = 0.12 mols$ $n_(NO_2) = "0.12 mols"$ , we have that $2 x = 0.12 mol$ $2x = "0.12 mol"$ s, or $x = 0.06 mols$ $x = "0.06 mols"$ . Therefore, we can already calculate $K_{c}$ $K_c$ :

$K_{c} = \frac{{[N O_{2}]}_{2}^{2}}{[N_{2} O_{4}]}$ $K_c = ([NO_2]^(2))/([N_2O_4])$

$= ((2x)^2" mols"^cancel(2)"/" cancel("1") "L"^cancel(2))/((0.1 - x)cancel"mols""/" cancel"1 L")$

$= (4x^2)/(0.1 - x)$ $"M"$

$= (4(0.06)^2)/(0.1 - 0.06)$ $"M"$

However, $K_c$ is conventionally reported without units, so:

$color(blue)(K_c = 0.36)$

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Using an ICE Table, how do we calculate $K_{c}$ $K_c$ for the following reaction?

At $373 K$ $"373 K"$ , $0.1 mols$ $"0.1 mols"$ of $N_{2} O_{4}$ $N_2O_4$ is heated in a one liter flask and at equilibrium the amount of nitrogen dioxide is $0.12 mols$ $"0.12 mols"$ .

$N_{2} O_{4} ⇌ 2 N O_{2}$ $N_2O_4 rightleftharpoons 2NO_2$

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Using an ICE Table, how do we calculate K_cKcK_c for the following reaction?

At "373 K"373 K"373 K", "0.1 mols"0.1 mols"0.1 mols" of N_2O_4N2O4N_2O_4 is heated in a one liter flask and at equilibrium the amount of nitrogen dioxide is "0.12 mols"0.12 mols"0.12 mols". N_2O_4 rightleftharpoons 2NO_2N2O4⇌2NO2N_2O_4 rightleftharpoons 2NO_2

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Using an ICE Table, how do we calculate $K_{c}$ $K_c$ for the following reaction?

At $373 K$ $"373 K"$ , $0.1 mols$ $"0.1 mols"$ of $N_{2} O_{4}$ $N_2O_4$ is heated in a one liter flask and at equilibrium the amount of nitrogen dioxide is $0.12 mols$ $"0.12 mols"$ .

$N_{2} O_{4} ⇌ 2 N O_{2}$ $N_2O_4 rightleftharpoons 2NO_2$