Question

8 mole of so2 and 4 mole of o2 are mixed in closed vessel at 27oc. the reaction vessel is heated to 127oc to attain equilibrium. at equilibrium 80% of the so2 enters the reaction. determine the pressure of mixture at equilibrium if initial pressure was 3 atm.

a. 3.0 atm

b. 3.93 atm

c. 2.93 atm

d. 2.0 atm

Answer 1

The reaction is depicted as:SO2+½ O2 <-------> SO3.(all species are in gaseous state )We have initially :SO2= 5 moleO2 = 5 moleSO3= 0 mole.At equilibrium :SO2= 5- 5(0.6) = 2 moleO2 = 5- 5 (0.3) = 3.5 moleSO3 = 5(0.6) = 3 mole.Now , total moles at equilibrium = 8.5 mole.Hence mole fraction of O2 = 3.5/8.5.Total pressure at equilibrium = 1 atm.Hence Partial pressure due to O2 at equilibrium will be = (1atm)×( 3.5/8.5)= 0.41 atm.solve in this way

Answer 2

Answer:

Dear Student,For the given reaction,

2 A(g) + B(g) + 2D(g)

Ang = 2 - (3)

= -1 mole

Substituting the value of AUO in the expression of AH:

ΔΗθ-Δυθ + Δ gRT

= (-10.5 kJ) – (-1) (8.314 × 10–3 kJ K

1 mol-1) (298 K)

= -10.5 kJ - 2.48 kJ

AHO = -12.98 kJ

Substituting the values of AHO and

ASe in the expression of AGe:

AGO = AHO – TASO

= -12.98 kJ – (298 K) (-44.1 J K-1)

= -12.98 kJ + 13.14 kJ

AGO = + 0.16 kJ

Since AGO for the reaction is

positive, the reaction will not occur

spontaneously.Best Of Luck.