1. How do you identify the limiting reactant in a reaction?
2. How do you use Hess's Law to determine the enthalpy change of a reaction?

Question

Luna Chen · Accepted Answer

WARNING! VERY long answer! How about these?

Reactant demonstration limitation One reaction where succinic acid can act as a limiting reactant is acid-catalyzed esterification. #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"# Since the reaction is an equilibrium, moving the equilibrium to the right usually requires using a significant excess of alcohol. Suppose you have 23.0 g of ethanol and 11.8 g of succinic acid to esterify. What is the limiting reactant in this reaction? #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"# Finding the "moles of reaction" that each will yield will make it simple to determine which reactant is the limiting one. In the balanced equation, you divide the moles of each reactant by the corresponding coefficient. In the table above, I completed that for you. Since succinic acid produces the fewest moles of reaction, it is the limiting reactant. demonstration of Hess's Law (a) Sample issue The standard enthalpies of formation of carbon dioxide and liquid water are -393.5 kJ/mol and -285.8 kJ/mol, respectively. Utilize this information to calculate the enthalpy of formation of succinic acid. The enthalpy of combustion of succinic acid is -1491 kJ/mol. (b) Resolution Three responses are provided to you: #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"# You have to create the desired equation using these. #"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. We now have the goal 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"# Succinic acid has an enthalpy of formation of -940 kJ/mol.

Serenity Allen · Answer

undefined 1. To identify the limiting reactant in a reaction, calculate the amount of product that can be formed from each reactant. The reactant that produces the least amount of product is the limiting reactant.

2. To use Hess's Law to determine the enthalpy change of a reaction, first identify a series of intermediate reactions that can be combined to form the desired reaction. Then, determine the enthalpy change for each intermediate reaction and apply Hess's Law by summing up the enthalpy changes of the intermediate reactions to find the overall enthalpy change of the desired reaction.

1. How do you identify the limiting reactant in a reaction? 2. How do you use Hess's Law to determine the enthalpy change of a reaction?