Question

What is the change in entropy when 7.61 mL of liquid benzene (C6H6, d = 0.879 g/mL) is combusted in the presence of 22.3 L of oxygen gas, measured at 298 K and 1 atm pressure? (R = 0.0821 L · atm/(K · mol)) 2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l); ΔS° = –437.7 J/K at 298 K

	a.	18.7 J/K
	b.	398 J/K
	c.	436 J/K
	d.	45.3 J/K
	e.	37.4 J/K

Answer 1

mass of C6H6   = volume * density

                         = 7.61*0.879   = 6.69g

no of moles of C6H6   = W/G.M.Wt   = 6.69/78   = 0.086 moles

P   = 1atm

V = 22.3L

T   = 298K

R = 0.0821 L · atm/(K · mol

PV = nRT

n   = PV/RT

      = 1*22.3/0.0821*298   = 0.911 moles

2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l)

2 moles of C6H6 react with 15 moles of O2

0.086 moles of C6H6 react with = 15*0.086/2   = 0.645 moles of O2 is required.

O2 is excess reagent

C6H6 is limiting reagent

2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l); ΔS° = –437.7 J/K at 298 K

2 moles of C6H6 combustion to gives -437.7J/K

0.086 moles of C6H6 combustion to gives = -437.7*0.086/2   = -18.7J/K

a. 18.7J/K >>>>>answer