Question #7cd04

1 Answer
Apr 24, 2016

Answer:

#K_"eq"= 0.0019#

Explanation:

Your chemical equation is

#"A" ⇌ "3B" + "C"#

Remember that the numbers in an ICE table must be molarities, not moles.

We must therefore convert our numbers to molarities by using the formula

#color(blue)(|bar(ul(color(white)(a/a)"Molarity" = "moles"/"litres"color(white)(a/a)|)))" "#

I: #["A"] = "3 mol"/"2 L" = "1.5 mol/L"#

E: #["B"] = "0.6 mol"/"2 L" = "0.3 mol/L"#

Now, let's set up an ICE table in the usual way.

#color(white)(mmmmmmmm)"A"color(white)(l) ⇌color(white)(l) "3B" +color(white)(l) "C"#
#"I/mol·"^"-1":color(white)(mmm)1.5color(white)(mmm)0color(white)(mml)0#
#"C/mol·L"^"-1":color(white)(mml)"-"xcolor(white)(mm)+3xcolor(white)(l)+x#
#"E/mol·L"^"-1":color(white)(m) 1.5-xcolor(white)(mll)3xcolor(white)(mm)x#

We know that at equilibrium,

#["B"] = "0.3 mol/L" = 3xcolor(white)(l) "mol/L"#

Therefore, #x = 0.1#.

The other equilibrium concentrations must be

#["C"] = xcolor(white)(l) "mol/L" = "0.1 mol/L"#

#["A"] = (1.5 – x) color(white)(l)"mol/L" = "(1.5 – 0.1)"color(white)(l)"mol/L" = "1.4 mol/L"#

#K_"eq" = (["B"]^3["C"])/(["A"]) = (0.3^3×0.1)/1.4 = 0.0019#