# Question 6216a

Jan 7, 2017

${K}_{c} = {K}_{P} = 0.019$

#### Explanation:

For the equilibrium $\text{H"_2 + "I"_2 ⇌ "2HI}$,

${K}_{c} = \left({\left[\text{H"_2]["I"_2])/(["HI}\right]}^{2}\right)$

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"#

${K}_{p} = {K}_{c} {\left(R T\right)}^{0} = {K}_{c} = 0.019$