# 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.

##### 1 Answer

I think it would be more practical to do all four questions. If you wish you can ignore

**DISCLAIMER:** *LONG ANSWER!*

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

rigidcontainer (constant-volume), preferablyclosedso nothing leaks out, andinsulatedso 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)# .)

#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 indeedequilibriumconcentrations, 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)#

In this, notice how you are given

the initial AND equilibrium concentrationsof#"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 thewith respect toreverse reaction#(2)# at thesametemperature...

Areversereaction at thesametemperature as the corresponding forward reaction has thereciprocalequilibrium 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 gotthat#K_c# from#(3)# is the same as one divided by the#K_c# in#(2)# .

This again uses the properties of equilibrium constants.

Since thecoefficientsare all multiplied by a constant, what we should expect is that theexponentsget multiplied by that constant. When you do that, youraise 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)#