At a certain temperature, 4.0 mol $N H_{3}$ $NH_3$ is introduced into a 2.0 L container, and the $N H_{3}$ $NH_3$ partially dissociates to $2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ $2NH_3(g)\rightleftharpoonsN_2(g)+3H_2(g)$ . At equilibrium, 2.0 mol $N H_{3}$ $NH_3$ remains. What is the value of $K_{c}$ $K_c$ ?

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At a certain temperature, 4.0 mol $N H_{3}$ $NH_3$ is introduced into a 2.0 L container, and the $N H_{3}$ $NH_3$ partially dissociates to $2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ $2NH_3(g)\rightleftharpoonsN_2(g)+3H_2(g)$ . At equilibrium, 2.0 mol $N H_{3}$ $NH_3$ remains. What is the value of $K_{c}$ $K_c$ ?

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At a certain temperature, 4.0 mol $N H_{3}$ $NH_3$ is introduced into a 2.0 L container, and the $N H_{3}$ $NH_3$ partially dissociates to $2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ $2NH_3(g)\rightleftharpoonsN_2(g)+3H_2(g)$ .

At equilibrium, 2.0 mol $N H_{3}$ $NH_3$ remains. What is the value of $K_{c}$ $K_c$ ?

Note: I started an ICE table but I don't know what to put in the $N_{2}$ $N_2$ and $3 H_{2}$ $3H_2$ columns.

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Answer 1 · 2018-06-27T21:19:37

Remember to include the coefficients in the change in concentratio, as well as the exponents. I get $K_{c} = 1.69$ $K_c = 1.69$ .

If $K_p = K_c(RT)^(Deltan_"gas")$ , what would $K_p$ be if this reaction has been occurring at $"298 K"$ ? $R = "0.08206 L"cdot"atm/mol"cdot"K"$ . Highlight below to see.

$color(white)("From the reaction, "Deltan_"gas" = (1+3) - (2) = 2,)$
$color(white)("since there are 3 mols H"_2, "1 mol N"_2, "and 2 mols NH"_3.)$

$color(white)("So,")$

$color(white)(K_p = 1.69 cdot ("0.08206 L"cdot"atm/mol"cdot"K" cdot "298 K")^((1+3)-(2)))$

$color(white)(~~ 1010" in implied units of atm")$

Let's first find the concentrations, because $K_c$ is in terms of concentrations.

$"4.0 mols NH"_3/("2.0 L") = "2.0 M"$

$"2.0 mols NH"_3/("2.0 L") = "1.0 M"$

The ICE table uses $"NH"_3$ having the initial concentration of $"2.0 M"$ and final of $"1.0 M"$ . Therefore, we should be able to find $x$ already, but let's assume we don't know $x$ yet.

$color(red)(2)"NH"_3(g) rightleftharpoons "N"_2(g) + color(red)(3)"H"_2(g)$

$"I"" "2.0" "" "" "" "0" "" "" "0$
$"C"" "-color(red)(2)x" "" "+x" "" "+color(red)(3)x$
$"E"" "2.0-color(red)(2)x" "" "x" "" "color(red)(3)x$

Remember that the coefficients go into the change in concentration and exponents.

Now, the $K_c$ would be:

$K_c = (x(color(red)(3)x)^color(red)(3))/(2.0 - color(red)(2)x)^color(red)(2)$

$= (27x^4)/(2.0 - 2x)^2$

But like mentioned, we know $x$ . This is because

$2.0 - 2x = "1.0 M"$ .

Therefore:

$x = "0.5 M"$ .

As a result,

$color(blue)(K_c) = (27(0.5)^4)/(2.0 - 2(0.5))^2$

$= color(blue)(1.69)$

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