Question

Question #6216a

Answer 1

`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`