Question

At a certain temperature, a 21.0-l contains holds four gases in equilibrium. their masses are: 3.5 g so3, 4.6 g so2, 13.5 g n2, and 0.98 g n2o. what is the value of the equilibrium constant at this temperature for the reaction of so2 with n2o to form so3 and n2 (balanced with lowest whole-number coefficients)?

Answer 1

SO2 +N2O<--->SO3+N2
Kc = concentration of product /concentration of reactant
(SO3)(N2/(SO2)(N2O)
Find the concentration of each reactant and products
concentration =number of moles/ volume in liters
moles of SO3= 3.5/80=0.044moles
SO2=4.6/64=0.072 moles
N2=13.5/28=0.483 moles
N2O=0.98/44=0.022 moles
concentration is therefore=
SO3=0.044/21= 2.095 x 10^-3
SO2=0.072 /21=3.43 x 10^-3
N2=0.483/21=0.023
N2O=0.022/22= 1.048 x 10^-3

Kc is therefore ={ (2.095 x10^-3)(0.023)}/ {(3.43 x10^-3)(1.048 x10^-3)}=13.40