Consider the following reaction: H2( g )+I2( g )2HI( g ) A reaction mixture in a

Question

Consider the following reaction:
H2(g)+I2(g)2HI(g)
A reaction mixture in a 3.68 L flask at a certain temperature initially contains 0.761 g H2 and 96.8 g I2. At equilibrium, the flask contains 90.3 g HI.

Part A

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Consider the following reaction:
H2(g)+I2(g)2HI(g)
A reaction mixture in a 3.68 L flask at a certain temperature initially contains 0.761 g H2 and 96.8 g I2. At equilibrium, the flask contains 90.3 g HI.

Part A

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Explanation / Answer

H2(g)+I2(g)2HI(g)

Kc = [HI]2/{[H2]*[I2]}

Molar mass of H2 = 2 g/mole

Molar mass of I2 = 254 g/mole

Molar mass of HI = 128 g/mole

Now, Initial moles of H2 = mass of H2/molar mass of H2 = 0.761/2 = 0.3805

Molar concentration of H2 = moles of H2/volume of solution in litres = 0.3805/3.68 = 0.103 M.........(1)

Initial Moles of I2 = mass of I2/molar mass of I2 = 96.8/254 = 0.381

Molar concentration of I2 = moles of I2/volume of solution in litres = 0.381/3.68 = 0.103 M.........(2)

Now, moles of HI formed = mass of HI/molar mass of HI = 90.3/128 = 0.705

Thus, molar concentration of HI = moles of HI/Volume of solution in litres = 0.705/3.68 = 0.192 ...........(3)

Let at eqb., [H2] = [I2] = (0.103-x) M ;

Thus, at eqb., [HI] = 2x M = 0.192 M

or, x = 0.096 M

Thus, at eqb., [H2] = [I2] = 0.007 M

Hence , Kc = (0.192)2/{(0.007)2} = 752.326

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