C6H6(g) + 3H2(g)<-------> C6H12(g) When 1.00 moles of C6H6(g) and 3.00moles H2 are put into a 200. L container and allowed toreach equilibrium over a catalyst at elevated temperature, theresulting mixture contains 0.137 mol . What is theequilibrium amount of H2 in moles.? C6H6(g) + 3H2(g)<-------> C6H12(g) When 1.00 moles of C6H6(g) and 3.00moles H2 are put into a 200. L container and allowed toreach equilibrium over a catalyst at elevated temperature, theresulting mixture contains 0.137 mol . What is theequilibrium amount of H2 in moles.?
Explanation / Answer
C6H6(g) + 3H2(g) <-------> C6H12(g) Initial: 1.00mol/2.00L 3.00mol/2.00L 0 mol/L Change: - 0.137 mol / 2.00 L - 3(0.137 mol / 2.00L) + 0.137mol/2.00L Equilibrium:0.4315mol/L 1.2945mol/L 0.0685 mol/L Equilibrium concentration of H2 = 1.2945mol/L The equilibrium amount in moles of H2 = 1.295mol/L * 2.00 L = 2.59 mol Initial: 1.00mol/2.00L 3.00mol/2.00L 0 mol/L Change: - 0.137 mol / 2.00 L - 3(0.137 mol / 2.00L) + 0.137mol/2.00L Equilibrium:0.4315mol/L 1.2945mol/L 0.0685 mol/L Equilibrium concentration of H2 = 1.2945mol/L The equilibrium amount in moles of H2 = 1.295mol/L * 2.00 L = 2.59 mol