Question

The initial pressure of a mixture of C6H6 and an excess of H2 in a rigid vessel is 1.21 atm. A catalyst is introduced. After the reaction reaches completion, the temperature is restored to its initial value. The final pressure in the vessel is 0.839 atm. What was the mole fraction of C6H6 in the original mixture

Answer 1

mole fraction of C6H6 = 0.613 atm

Step-by-step explanation:

The equation for this reaction is :

`C_6H_6 _((g)) + 3H_2_((g)) \to C_6H_(12)_((g))`

Initial P₁ P₂ 0

Final 0 P₂ -P₁/2 P₁

After completion of the reaction;

P₁ + P₂ = 1.21 atm ----- (1)

P₂ - P₁/2 + P₁ = 0.839 atm

P₂ + P₁/2 = 0.839 atm ----- (2)

Subtracting (2) from (1); we have:

P₁/2 = 0.371

P₁ = 0.742 atm

From(1)

P₁ + P₂ = 1.21 atm

0.742 atm + P₂ = 1.21 atm

P₂ = 1.21 atm - 0.742 atm

P₂ = 0.468 atm

Thus, the partial pressure of C6H6 = 0.742 atm

∴

Partial pressure = Total pressure × mole fraction of C6H6

mole fraction of C6H6 = Partial pressure / Total pressure

mole fraction of C6H6 = 0.742 atm / 1.21 atm

mole fraction of C6H6 = 0.613 atm