Question

H2(g)+I2(g)⇌2HI(g) A reaction mixture in a 3.75 L flask at a certain temperature initially contains 0.760 g H2 and 96.9 g I2. At equilibrium, the flask contains 90.4 g HI. Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Answer 1

V = 3.75 L

m = 0.76 g of H2

mol H2 = mass/MW = 0.76/2 = 0.38

[H2] = mol/V = 0.38/3.75 = 0.101333

m = 96.6 g of I2

mol I2 = mass/MW = (96.6/253.80894 ) = 0.38060

[I2] =0.38060/3.75 = 0.10149

in equilibrium

[HI] = mass/MW/V = 90.4/(127.911*3.75) = 0.18846

then; initially

[I2] = 0.10149

[H2] = 0.101333

[HI] = 0

in equilibrium

[I2] = 0.10149 - x

[H2] = 0.101333 - x

[HI] = 0 +2x

an we know that

[HI] = 0 +2x = 0.18846

x = 0.18846 /2 =0.09423

then

[I2] = 0.10149 - 0.09423 = 0.007103

[H2] = 0.101333 -0.09423 = 0.00726

then

K = [HI]^2 / ([H2][I2])

K = (0.18846 ^2)/(0.00726*0.007103) = 688.747

K = 688.747