Given the following, how do you calculate #K_c# for the reaction #2C Cl_4(g) + O_2(g) rightleftharpoons 2COCI_2(g) + 2Cl_2(g)#?
Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. #C Cl_4(g) + 1/2O_2(g) rightleftharpoons COCI_2(g) + Cl_2(g)#
#K_c = 4.4 xx 10^9# for this reaction with these coefficients at #"1000 K"#
EDIT: The product should be #COCl_2# .
- Truong-Son
Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.
EDIT: The product should be
- Truong-Son
1 Answer
#1.9 xx 10^(19)# at#"1000 K"# and whatever pressure this was for the first reaction.
You are just supposed to write each
(By the way,
Reaction
#C Cl_4(g) + 1/2O_2(g) rightleftharpoons COCl_2(g) + Cl_2(g)#
#K_c = (["COCl"_2(g)]["Cl"_2(g)])/(["CCl"_4(g)]["O"_2(g)]^"1/2")# (recall that the coefficients for each reaction participant are also their respective exponents.)
Reaction
#2C Cl_4(g) + O_2(g) rightleftharpoons 2COCl_2(g) + 2Cl_2(g)# therefore having:
#K_c' = (["COCl"_2(g)]^2["Cl"_2(g)]^2)/(["CCl"_4(g)]^2["O"_2(g)])#
If we look at these equilibrium constants...
#[(["COCl"_2(g)]["Cl"_2(g)])/(["CCl"_4(g)]["O"_2(g)]^"1/2")]^2 = (["COCl"_2(g)]^2["Cl"_2(g)]^2)/(["CCl"_4(g)]^2["O"_2(g)])#
Therefore:
#K_c^2 = K_c'#
And so, since we have
#color(blue)(K_c') = (4.4 xx 10^9)^2 = color(blue)(1.9 xx 10^(19))#