# 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.

Jun 9, 2017

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

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 $\left(2\right)$ gives you a shortcut way to do $\left(3\right)$.)

2)

a) Presumably, you know this well enough, as it is just using the definition of ${K}_{c}$

${K}_{c} = \left({\left[{\text{HI"(g)]_(eq)^2)/(["H"_2(g)]_(eq)["I}}_{2} \left(g\right)\right]}_{e q}\right)$,

for the reaction

$\text{H"_2(g) + "I"_2(g) rightleftharpoons 2"HI} \left(g\right)$.

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 $\text{M}$.

$\implies \textcolor{b l u e}{{K}_{c}} = {\left(2.52 \times {10}^{- 2}\right)}^{2} / \left(\left(1.14 \times {10}^{- 2}\right) \left(0.12 \times {10}^{- 2}\right)\right)$

$= \textcolor{b l u e}{46.42}$

3)

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

In fact, right off the bat, if this question is written well, then you should expect that ${K}_{c} = \frac{1}{46.42} \approx 2.16 \times {10}^{- 2}$, for this is the reverse reaction with respect to $\left(2\right)$ 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:

$2 {\text{HI"(g) " "" "rightleftharpoons" "" " "H"_2(g) + "I}}_{2} \left(g\right)$

$\text{I"" ""0.0304 M"" "" "" "" ""0 M"" "" ""0 M}$
$\text{C"" "-2x" "" "" "" "" "+x" "" } + x$
$\text{E"" "(0.0304 - 2x)"M"" "x" M"" "" "x" M}$

But if you recognize that you were already given ${\left[\text{HI}\right]}_{i}$ and ${\left[\text{HI}\right]}_{e q}$...

$2 x = \text{0.0304 M" - "0.0235 M" = "0.0069 M}$

$\implies x = \text{0.00345 M}$

As a result, we know what ${K}_{c}$ will be:

$\textcolor{b l u e}{{K}_{c}} = \left({\left[\text{H"_2(g)]_(eq)["I"_2(g)]_(eq))/(["HI} \left(g\right)\right]}_{e q}^{2}\right)$

$= {\left(\text{0.00345 M" cdot "0.00345 M")/("0.0235 M}\right)}^{2}$

$= 0.02156 = \textcolor{b l u e}{2.16 \times {10}^{- 2}}$

This is a nicely physically reasonable answer, since we were hoping and got that ${K}_{c}$ from $\left(3\right)$ is the same as one divided by the ${K}_{c}$ in $\left(2\right)$.

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) ${K}_{c} = \left({\left[{\text{HI"(g)]_(eq)^2)/(["H"_2(g)]_(eq)["I}}_{2} \left(g\right)\right]}_{e q}\right)$

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"

$\implies \textcolor{b l u e}{{K}_{c} '} = {K}_{c}^{\text{1/2}} = \textcolor{b l u e}{\sqrt{{K}_{c}}}$

b) Now just use what you got in $\left(a\right)$.

$\implies \textcolor{b l u e}{{K}_{c} '} = \sqrt{46.42} \approx \textcolor{b l u e}{6.81}$