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 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)# .)
#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)#
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)# .
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)#