The chemical equation is
#"2SO"_3 ⇌ "2SO"_2 + "O"_2#
The initial amount of #"SO"_3# is
#0.80 color(red)(cancel(color(black)("g SO"_3))) × ("1 mol SO"_3)/(80.06 color(red)(cancel(color(black)("g SO"_3)))) = "0.009 99 mol SO"_3#
The volume of the container is 1 L, so the initial concentration of #"SO"_3# is
#["SO"_3] = "0.009 99 mol"/"1 L" = "0.09 99 mol/L"#
We can setup an ICE table to solve this problem.
#color(white)(mmmmmmmm)"2SO"_3 color(white)(l)⇌ color(white)(l)"2SO"_2 + "O"_2#
#"I/mol·L"^"-1": color(white)(mm)"0.009 99"color(white)(mmm)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mmll) "-2"xcolor(white)(mmmll)"+2"xcolor(white)(mll)"+"x#
#"E/mol·L"^"-1": color(white)(m)"0.009 99 - 2"xcolor(white)(ml)2xcolor(white)(mmll)x#
We know that 80 % of the #"SO"_3# has decomposed.
∴ The concentration remaining at equilibrium is
#"0.20 × 0.009 99 = 0.002 00"#
∴ #"0.009 99 -"color(white)(l)2x = "0.002 00"#
#2x = "0.009 99 - 0.002 00" = "0.007 99"#
#x = "0.007 99"/2 = "0.004 00"#
So,
#["SO"_3]_text(eq) = "0.002 00 mol/L"#
#["SO"_2]_text(eq) = 2x color(white)(l)"mol/L" = "0.007 99 mol/L"#
#["O"_2]_text(eq) = x color(white)(l)"mol/L" = "0.004 00 mol/L"#
#K_c = (["SO"_2]^2["O"_2])/(["SO"_3]^2) = (("0.007 99")^2 × "0.004 00")/("0.002 00")^2 = 0.064#
