olu e n e , C7H8(g), is an important organic solvent and can be produced by thefollowing equilibrium equation.C7H14(g) + energy C7H8(g) + 3 H2(g)A technician placed 3.00 mol of C7H14(g) into a 1.00 L flask. The flask was then stoppered and allowed to reach equilibrium. At equilibrium, 1.20 mol of H2(g) was present in the flask.1. The value of the equilibrium constant is
3 years ago
Answered By Mandy H
To start off make an ICE (Initial, Change, Equilibrium concentrations) table for the reaction.
mol/L | C7H14(g) + energy --> C7H8 (g) + 3H2(g)
I | 3. 00 | ~ | 0 | 0 <--3.00mol/ 1.00 L of C7H14 was placed into flask; products always start out at 0
C | -1. 20 | ~ | +1.20 | +1.20 <-- At equilibrium there is 1.20 mol H2 so there has been a change of 1.20 M (1.20 M of H2 was produced)
E | 1.80 | ~ | 1.20 | 1.20 <-- 3.00- 1.20 = 1.80 M of C7H14 left at equilibrium
For the equilibrium constant, remember that it is always products/ reactants of the substances that are in gaseous or aqueous state.
Keq = [C7H8] [H2]3 / [C7H14] <-- remember that there is a coeffient of 3 for H2 (3 mols of H2 required to react) in the original equation so there is an exponent of 3 for H2 in the equilibrium constant
Keq = [1.20] [1.20]3 / [1.80] <-- substitue the value from the equilibrium row from your ICE table (hint: when typing in your calculator remember brackets/ BEDMAS to avoid making calculation errors)
Keq = 1.15 <-- Don't forget your significant digits and that there are no units for the equilibrium constant