Question

Can someone help me with questions 1, 3, and 4? As soon as possible please, I have several questions regarding the properties of equilibrium constants.

Answer 1

I think it would be more practical to do all four questions. If you wish you can ignore `(2)`, but it may be helpful for you to check it out anyway.

DISCLAIMER: LONG ANSWER!

`1)`

Well, you're at high temperature, so you have fast-moving ideal gases (which are most ideal at high temperature and low pressure).

It'd be most convenient to have a rigid container (constant-volume), preferably closed so nothing leaks out, and insulated so that it's self-contained at constant temperature.

(In fact, the constant temperature facet is crucial to the claim I make that doing `(2)` gives you a shortcut way to do `(3)`.)

`2)`

`a)` Presumably, you know this well enough, as it is just using the definition of `K_c`

`K_c = (["HI"(g)]_(eq)^2)/(["H"_2(g)]_(eq)["I"_2(g)]_(eq))`,

for the reaction

`"H"_2(g) + "I"_2(g) rightleftharpoons 2"HI"(g)`.

`b)` And I presume you then plugged in the numbers. These are indeed equilibrium concentrations, so they do just go in. All the concentrations used here are in `"M"`.

`=> color(blue)(K_c) = (2.52 xx 10^(-2))^2/((1.14 xx 10^(-2))(0.12 xx 10^(-2)))`

`= color(blue)(46.42)`

`3)`

In this, notice how you are given the initial AND equilibrium concentrations of `"HI"`. Hence, this is one of the more simple scenarios where you already know what `2x` is.

In fact, right off the bat, if this question is written well, then you should expect that `K_c = 1/46.42 ~~ 2.16 xx 10^(-2)`, for this is the reverse reaction with respect to `(2)` at the same temperature...

A reverse reaction at the same temperature as the corresponding forward reaction has the reciprocal equilibrium constant.

Let's prove it. If we construct the ICE table, we would get:

`2"HI"(g) " "" "rightleftharpoons" "" " "H"_2(g) + "I"_2(g)`

`"I"" ""0.0304 M"" "" "" "" ""0 M"" "" ""0 M"`
`"C"" "-2x" "" "" "" "" "+x" "" "+x`
`"E"" "(0.0304 - 2x)"M"" "x" M"" "" "x" M"`

But if you recognize that you were already given `["HI"]_i` and `["HI"]_(eq)`...

`2x = "0.0304 M" - "0.0235 M" = "0.0069 M"`

`=> x = "0.00345 M"`

As a result, we know what `K_c` will be:

`color(blue)(K_c) = (["H"_2(g)]_(eq)["I"_2(g)]_(eq))/(["HI"(g)]_(eq)^2)`

`= ("0.00345 M" cdot "0.00345 M")/("0.0235 M")^2`

`= 0.02156 = color(blue)(2.16 xx 10^(-2))`

This is a nicely physically reasonable answer, since we were hoping and got that `K_c` from `(3)` is the same as one divided by the `K_c` in `(2)`.

`4)`

This again uses the properties of equilibrium constants.

Since the coefficients are all multiplied by a constant, what we should expect is that the exponents get multiplied by that constant. When you do that, you raise everything to the constant.

`a)` `K_c = (["HI"(g)]_(eq)^2)/(["H"_2(g)]_(eq)["I"_2(g)]_(eq))`

`K_c' = (["HI"(g)]_(eq))/(["H"_2(g)]_(eq)^"1/2"["I"_2(g)]_(eq)^"1/2") = {(["HI"(g)]_(eq)^2)/(["H"_2(g)]_(eq)["I"_2(g)]_(eq))}^"1/2"`

`=> color(blue)(K_c') = K_c^"1/2" = color(blue)(sqrt(K_c))`

`b)` Now just use what you got in `(a)`.

`=> color(blue)(K_c') = sqrt(46.42) ~~ color(blue)(6.81)`