Answer to Question #104060 in General Chemistry for James Okike

Question

The equilibrium constant for the gaseous reaction
H2 + I2 <=> 2HI is 50.2 at 448°c
Calculate the number of grams of HI that are in equilibrium with 1.25 mole of H2 and 63.5g of iodine at this temperature

Answer 1

The constant for the equilibriumH2+I2↔2HI is given by the following expression:

$K=(HI) ^2/(H2) (I2)$ =50.2.

The amount of moles of iodine is:$n(I2)=m(I2)/M(I2)$ = 63.5(g)/254 (g/mole) = 0.25 moles.

 Using expression for the equilibrium constant one can find the amount of moles of HI that are in the equilibrium with given amounts of H2 and I2:

 n^2(HI) / (125*0.25) =50.2.

50.2 =n^2(HI)/0.3215

N^2(HI)=50.2*0.3125=15.7

N(HI)=3.96mol.

 The mass of the HI can be calculated in the following way by multiplying the amount of moles by the molar mass:

$m(HI)=n(HI)×M(HI)$ = 3.96 (mole) × 128 (g/mole) = 507 g.