During the course of the reaction, the vessel is found to contain #"7.00 mols"# of #C#, #"14.0 mols"# of #H_2O#, #"3.60 mols"# of #CO#, and #"8.50 mols"# of #H_2#. What is the reaction quotient #Q#?
The following reaction was carried out in a #"2.25 L"# reaction vessel at #"1100 K"# :
#C (s) + H_2O(g) rightleftharpoons CO (g) + H_2 (g)#
The following reaction was carried out in a
#C (s) + H_2O(g) rightleftharpoons CO (g) + H_2 (g)#
1 Answer
Explanation:
The equilibrium given to you looks like this
#"C"_ ((s)) + "H"_ 2"O"_ ((g)) rightleftharpoons "CO"_ ((g)) + "H"_ (2(g))#
By definition, the reaction quotient,
Two important things to note here
- the reaction quotient can be calculated by using the concentrations of the chemical species that take part in the reaction at any given moment in the course of the reaction
- the concentration of solids and of pure liquids is excluded from the expression of the reaction quotient
In this case, the reaction quotient would take the form -- keep in mind that carbon,
#Q_c = (["CO"] * ["H"_2])/(["H"_2"O"])#
Now, use the volume of the reaction vessel to calculate the concentrations of the chemical species that are of interest here
#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/(V_"solution"[color(blue)("in liters")])color(white)(a/a)|)))#
You will have
#["H"_2"O"] = "14.0 moles"/"2.25 L" = "6.22 M"#
#["CO"] = "3.60 moles"/"2.25 L" = "1.6 M"#
#["H"_2] = "8.50 moles"/"2.25 L" = "3.78 M"#
Plug these values into the equation that gives you the reaction quotient to find
#Q_c = ("1.6 M" * 3.78 color(red)(cancel(color(black)("M"))))/(6.22color(red)(cancel(color(black)("M")))) = "0.972 M"#
The reaction quotient is usually expressed without added units, so your answer will be
#Q_c = color(green)(|bar(ul(color(white)(a/a)color(black)(0.972)color(white)(a/a)|)))#
The answer is rounded to three sig figs.