Question #6216a

1 Answer
Jan 7, 2017

Answer:

#K_c = K_P = 0.019#

Explanation:

For the equilibrium #"H"_2 + "I"_2 ⇌ "2HI"#,

#K_c = (["H"_2]["I"_2])/(["HI"]^2)#

If the values given are the equilibrium concentrations,

#K_c = ((2.9 × 10^"-3")(1.7 × 10^"-3"))/(1.6 × 10^"-2")^2 = 0.019#

#K_P = K_c(RT)^(Δn)#,

where #Δn# is the change in the number of moles of gas.

#Δn = "2 - 2 = 0"#

#K_p = K_c(RT)^0 = K_c = 0.019#