Limiting reactant demonstration
Acid-catalyzed esterification is a good example of a reaction in which succinic acid can behave as a limiting reactant.
#underbrace(("CH"_2"COOH")_2)_color(red)("succinic acid") + underbrace("2CH"_3"CH"_2"OH")_color(red)("ethanol") stackrelcolor(blue)("H"^"+", Δcolor(white)(mm))(→) underbrace(("CH"_2"COOCH"_2"CH"_3)_2)_color(red)("diethyl succinate") + 2"H"_2"O"#
The reaction is an equilibrium, so you usually use a large excess of the alcohol to drive the equilibrium to the right.
Assume that you are esterifying 11.8 g of succinic acid with 23.0 g of ethanol. What is the limiting reactant?
#M_text(r): color(white)(mmmmm)118.09color(white)(mmmmml)46.07#
#color(white)(mmmmm)("CH"_2"COOH")_2 + color(white)(l)"2CH"_3"CH"_2"OH" → ("CH"_2"COOCH"_2"CH"_3)_2 + 2"H"_2"O"#
#"Mass/g:"color(white)(mmmll)11.8color(white)(mmmmmml)23.0#
#"Moles:"color(white)(mmmll)"0.099 92"color(white)(mmmmll)0.4992#
#"Divide by:"color(white)(mmml)1color(white)(mmmmmmmml)2#
#"Moles rxn:"color(white)(mll)"0.099 92"color(white)(mmmmll)0.2496#
#"Moles of SA" = 11.8 color(red)(cancel(color(black)("g SA"))) × "1 mol SA"/(118.09 color(red)(cancel(color(black)("g SA")))) = "0.099 02 mol SA"#
#"Moles of EtOH" = 23.0 color(red)(cancel(color(black)("g EtOH"))) × "1 mol EtOH"/(46.07 color(red)(cancel(color(black)("g EtOH")))) = "0.4992 mol EtOH"#
An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:
You divide the moles of each reactant by its corresponding coefficient in the balanced equation.
I did that for you in the table above.
Succinic acid is the limiting reactant because it gives the fewest moles of reaction.
Hess's Law demonstration
(a) Example problem
The enthalpy of combustion of succinic acid is -1491 kJ/mol. The standard enthalpies of formation of carbon dioxide and of liquid water are -393.5 kJ/mol and -285.8 kJ/mol, respectively. Use this information to calculate the enthalpy of formation of succinic acid.
(b) Solution
You are given three reactions:
#bb"(1)" "C"_4"H"_6"O"_4"(s)" + 7/2"O"_2"(g)" → "4CO"_2"(g)" + 3"H"_2"O(l)"; Δ_text(c)H = "-1491 kJ/mol"#
#bb"(2)" "C(s)" + "O"_2"(g)" → "CO"_2"(g)";color(white)(mll) Δ_text(f)H = "-393.5 kJ/mol"#
#bb"(3)" "H"_2"(g)" + 1/2"O"_2"(g)" → "H"_2"O(l)";color(white)(mll) Δ_text(f)H = "-285.8 kJ/mol"#
From these, you must devise the target equation
#"4C(s) + 3H"_2"(g)" + "2O"_2"(g)" → "C"_4"H"_6"O"_4"(s)"; Δ_text(f)H = ?#
The target equation has #"4C(s)"# on the left, so you start by writing equation (2) normally, but multiplied by 4.
#bb"(4)" "4C(s)" + "4O"_2"(g)" → "4CO"_2"(g)";color(white)(mll) ΔH = "-1574.0 kJ"#
The target equation has #"3H"_2"(g)"# on the left, so you write equation (3) normally, but multiplied by 3.
#bb"(5)" "3H"_2"(g)" + 3/2"O"_2"(g)" → "3H"_2"O(l)";color(white)(mll) Δ_text(f)H = "-857.4 kJ"#
The target equation has #"C"_4"H"_6"O"_4"(s)"# on the right, so you write equation (1) in reverse.
#bb"(6)" "4CO"_2"(g)" + 3"H"_2"O(l)" →"C"_4"H"_6"O"_4"(s)" + 7/2"O"_2"(g)"; ΔH = "+1491 kJ"#
When you reverse an equation, you change the sign of its #ΔH#.
Then you add equations #bb"(4)", bb"(5)"#, and #bb"(6)"# together, cancelling species that appear on opposite sides of the reaction arrows.
When you add thermochemical equations, you add their #ΔH# values.
This gives us the target equation (7):
#bb"(4)" "4C(s)" + stackrelcolor(blue)(2)(color(red)(cancel(color(black)(4))))"O"_2"(g)" → "4CO"_2"(g)";color(white)(mmmmmmmmm) Δ_text(f)H = "-1574.0 kJ"#
#bb"(5)" "3H"_2"(g)" + color(red)(cancel(color(black)(3/2)))"O"_2"(g)" → "3H"_2"O(l)";color(white)(mmmmmmmll) Δ_text(f)H = color(white)(l)"-857.4 kJ"#
#bb"(6)" "4CO"_2"(g)" + 3"H"_2"O(l)" →"C"_4"H"_6"O"_4"(s)" + color(red)(cancel(color(black)(7/2)))"O"_2"(g)"; ΔH = "+1491 kJ"#
#bb"(7)"stackrel(———————————————————)("4C(s) + 3H"_2"(g)" + "2O"_2"(g)" → "C"_4"H"_6"O"_4"(s)"); color(white)(mml)Δ_text(f)H =color(white)(ll) "-940 kJ"#
The enthalpy of formation of succinic acid is -940 kJ/mol.
