Remember to include the coefficients in the change in concentratio, as well as the exponents. I get
If
$color(white)("From the reaction, "Deltan_"gas" = (1+3) - (2) = 2,)$
$color(white)("since there are 3 mols H"_2, "1 mol N"_2, "and 2 mols NH"_3.)$
$color(white)("So,")$
$color(white)(K_p = 1.69 cdot ("0.08206 L"cdot"atm/mol"cdot"K" cdot "298 K")^((1+3)-(2)))$
$color(white)(~~ 1010" in implied units of atm")$
Let's first find the concentrations, because
$"4.0 mols NH"_3/("2.0 L") = "2.0 M"$
$"2.0 mols NH"_3/("2.0 L") = "1.0 M"$
The ICE table uses
$color(red)(2)"NH"_3(g) rightleftharpoons "N"_2(g) + color(red)(3)"H"_2(g)$
$"I"" "2.0" "" "" "" "0" "" "" "0$
$"C"" "-color(red)(2)x" "" "+x" "" "+color(red)(3)x$
$"E"" "2.0-color(red)(2)x" "" "x" "" "color(red)(3)x$
Remember that the coefficients go into the change in concentration and exponents.
Now, the
$K_c = (x(color(red)(3)x)^color(red)(3))/(2.0 - color(red)(2)x)^color(red)(2)$
$= (27x^4)/(2.0 - 2x)^2$
But like mentioned, we know
$2.0 - 2x = "1.0 M"$ .
Therefore:
$x = "0.5 M"$ .
As a result,
$color(blue)(K_c) = (27(0.5)^4)/(2.0 - 2(0.5))^2$
$= color(blue)(1.69)$