Question

Hydrogen iodide decomposes according to the following reaction. 2 HI(g) equilibrium reaction arrow H2(g) + I2(g) A sealed 1.5 L container initially holds 0.00615 mol H2, 0.00445 mol I2, and 0.0163 mol HI at 703 K. When equilibrium is reached, the equilibrium concentration of H2(g) is 0.00364 M. What are the equilibrium concentrations of HI(g) and I2(g)?

Answer 1

the given reaction is :-

2HI(g) <-------------> H₂(g) + I₂(g)

volume of container = 1.5 L

initial moles oh hydrogen iodide = 0.0163

initial moles of hydrogen = 0..00615

initial moles of iodine = 0.00445

2HI(g) <-------------> H₂(g) + I₂(g)

0.0163 0.00615 0.00445

0.0163 -2a 0.00615 + a 0.00445 + a

given that equilibrium concentration of hydrogen = 0.00364 M

moles at equilibrium will be = 0.00364 x 1.5 = 0.00546 moles

the value of a will be = 0.00546 - 0.00615 = - 0.00069 mol

here minus ign indicates that the reaction proceedes in backward direction

so the equlibrium moles of HI = 0.0163 + 2 x 0.00069 = 0.01768 mol

concentration of HI = 0.01178 M

equilibrium moles of iodine = 0.00445 - 0.00069 = 0.00376 mol

concentration will be = 0.00376 / 1.5 = 0.00251 M