Question #b23fd
1 Answer
Explanation:
You can derive an expression for
Your equilibrium reaction looks like this
#"PbCl"_text(3(g]) + "Cl"_text(2(g]) rightleftharpoons "PbCl"_text(5(g])#
By definition, what is
#K_c = ( ["PbCl"_5])/( ["PbCl"_3] * ["Cl"_2])#
The equilibrium concentrations featured in the expression of
If you take
#[PbCl_3] = n_text(PbCl_3)/V" "# and#" "[Cl_(2)] = n_(Cl_2)/V#
#[PbCl_5] = n_(PbCl_5)/V#
Now take a look at the expression for
#K_p = ( (PbCl_5))/((PbCl_3) * (Cl_2))#
The partial pressure of each component of the mixture can be expressed using the ideal gas law
#PV = nRT implies P = (nRT)/V = n/V * RT#
This means that you have
#(PbCl_5) = underbrace(n_(PbCl_5)/V)_(color(blue)(=[PbCl_5])) * RT#
#(Cl_2) = underbrace(n_(Cl_2)/V)_(color(blue)(=[Cl_2])) * RT#
#(PbCl_3) = underbrace(n_(PbCl_3)/V)_(color(blue)(=[PbCl_3])) * RT#
This means that you can write
#K_p = ( [PbCl_5] * color(red)(cancel(color(black)(RT))))/( [Cl_2] * color(red)(cancel(color(black)(RT))) * [PbCl_3] * RT)#
#K_p = overbrace(( [PbCl_5])/( [PbCl_3] * [Cl_2]))^(color(green)(=K_c)) * 1/(RT)#
#K_p = K_c * (RT)^(-1)#
If you take
#K_p = K_c * [0.082("atm" * "L")/("mol" * color(red)(cancel(color(black)(K)))) * 500color(red)(cancel(color(black)(K)))]^(-1)#
#K_p = K_c * 0.0244"mol"/("atm" * "L")#
What about the units?
Well, if you look at the expression for
#K_c = ( [PbCl_5]color(red)(cancel(color(black)("M"))))/( [PbCl_3]color(red)(cancel(color(black)("M"))) * [Cl_2]" M") = [1/"M"]#
SInce
#K_p = 1.67 color(red)(cancel(color(black)("L")))/color(red)(cancel(color(black)("mol"))) * 0.0244 color(red)(cancel(color(black)("mol")))/("atm" * color(red)(cancel(color(black)("L")))) = color(green)("0.0407 atm"^(-1))#