Question

A Uonsanh The following reaction is in equilibrium in a closed system: In one experiment, 6g of lodine (la) and 50g of potassium iodide (KI) were dissolved in 1000 mL of water. After the solution was allowed to reach equilibrium it was found to have an iodide (I) concentration of 0.278 mol dm Calculate a value for the equilibrium constant, Ke, for this reaction. tbdum

Answer 1

Solution

Mass of I₂ = 6 g

Molar mass of I₂ = 253.81 g/mol

Volume of solution= 1000 mL (i.e. = 1 L)

Number of moles of I₂ = given mass / molar mass

Number of moles of I₂ = 6g / 253.81 g/mol

Number of moles of I₂ = 0.0236 mole

Molarity of I₂ = 0.0236 mole /1 L =0.0236 M

Mass of KI = 50 g

Molar mass of KI = 166 g/mol

Number of moles of KI = 50g / 166 g/mol

Number of moles of KI = 0.301 moles

Molarity of KI = 0.301 mole /1 L = 0.301 M

Initial chemical reaction is

I_2(aq) + I^-_(aq) = I^-_3(aq)

From reaction molar ratio between I₂ & I^-₃ is 1:1 & also I^-& I^-₃ is 1:1 hence we have 0.301 mole of I^-and 0.278 moles remains so 0.0230 moles of I^- must be consumed to produce 0.0230 mole of I^-_3. Also 0.230 moles of I₂ get consumed to produce 0.0230 mole of I^-₃

	I2	I-	I-3
Initial	0.0236	0.301	0
Change	-0.0230	-0.0230	+0.0230
Final	0.0006	0.278	0.0230

We know the expression for Kc

Kc = [I^-₃] / [I₂] [I^-]

Kc = [0.0230] / [0.0006] [0.278]

Kc = [0.0230] / [0.0006] [0.278]

Kc = 137.89