Question

H2(g)+I2(g)⇌2HI(g) A reaction mixture in a 3.71 L flask at a certain temperature initially contains 0.760 g H2 and 96.8 g I2. At equilibrium, the flask contains 90.5 g HI. Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Answer 1

first convert every thing in to moles using the formula

no of moles = weight / molar mass

no of moles of H2 = 0.760 g / 2.02 g/mol

= 0.376 mol

no f moles of I2 = 96.8 g / 253.8 g/mol

= 0.381 mol

no of moles of HI at equilibrium = 90.5 g / 127.91 g/mol

= 0.707 mol but here it is formining 2 moles sp 0.707 / 2 = 0.353

at equilibrium remaining concentratiof [H2] = 0.376 - 0.353 = 0.023 moles

at equilibrium remaining concentratiof [I2] = 0.381 - 0.353 = 0.028 moles

H2(g)+I2(g)⇌2HI(g)

equilibrium expression

Kc = [ HI ]² / ([ H² ] [ I² ])

= [0.353]2 / [0.023] x [0.028]

= 0.125 / 0.000644

= 194.1