Start by writing the balanced chemical equation for this equilibrium reaction
$"I"_text(2(g]) + "H"_text(2(g]) rightleftharpoons color(red)(2)"HI"_text(2(g])$
Notice that you have a
Use the molar masses of hydrogen gas and iodine to determine how many moles of each you're adding to the container
$46 color(red)(cancel(color(black)("g"))) * "1 mole I"_2/(253.81color(red)(cancel(color(black)("g")))) = "0.1812 moles I"_2$
and
$1 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.0159color(red)(cancel(color(black)("g")))) = "0.4961 moles H"_2$
You know that at equilibrium, the reaction vessel contains
$1.9 color(red)(cancel(color(black)("g"))) * "1 mole I"_2/(253.81 color(red)(cancel(color(black)("g")))) = "0.007486 moles I"_2$
This tells you that a total of
$n_"converted" = 0.1812 - 0.007486 = "0.1737 moles I"_2$
have been converted to hydrogen iodide,
According to the
$0.1737 color(red)(cancel(color(black)("moles I"_2))) * (color(red)(2)" moles HI")/(1color(red)(cancel(color(black)("mole I"_2)))) = "0.3474 moles HI"$
The
$n_(H_2) = 0.4961 - 0.1737 = "0.3224 moles H"_2$
Therefore, at equilibrium, the reaction vessel will contain - I'll leave the answers rounded to two , just for good measure
$n_(I_2) = color(green)("0.0075 moles I"_2)$
$n_(H_2) = color(green)("0.32 moles H"_2)$
$n_(HI) = color(green)("0.35 moles HI")$
By definition, the equilibrium constant for this reaction,
$K_c = (["HI"]^color(red)(2))/(["I"_2] * ["H"_2])$
As you know, is defined as moles of per liters of solution. In this case, the volume of the reaction vessel, let's say
$["HI"] = "0.35 moles"/V$
$["H"_2] = "0.32 moles"/V$
$["I"_2] = "0.0075 moles"/V$
This means that you have
$K_c = ( 0.35)^2/color(red)(cancel(color(black)(V^2))) * color(red)(cancel(color(black)(V)))/0.32 * color(red)(cancel(color(black)(V)))/0.0075 = color(green)(51)$
The relationship between
$color(blue)(K_p = K_c * (RT)^(Deltan))" "$ , where
Notice that you reaction has a total of
This means that
$K_p = K_c * (RT)^0$
$K_p = K_c$