Question

If 0.10 g of benzene (C6H6) is ignited in a 0.3 L vessel at 298K and surrounded by 25 atm of oxygen: a) Write the net reaction; b) How many moles of O2 were consumed?; c) How many moles of product were produced?; d) What is the change in moles in the reaction vessel?; e) Assuming all of the products were ideal gas what is the new pressure in the container?; f) What is the work done by this reaction?; g) If one of the products is a liquid what would be the change in pressure. Assuming the volume of liquid is very small compared to gas, how much work was done by this system? PLEASE SHOW ALL WORK! thank you.

Answer 1

a)

net reaction :

2 C6H6 + 15 O2 ----------------> 12 CO2 (g) + 6   H2O (g)

b)

number of moles of benzene = 0.10 / 78 = 1.282 x 10^-3

2 mol C6H6   ---------------> 15 mol O2

1.282 x 10^-3 mol C6H6   ------------> ??

number of moles of O2 consumed = 1.282 x 10^-3 x 15 / 2

                                                      = 9.615 x 10^-3 mol

c)

2 mol C6H6   ------------> 12 mol CO2

1.282 x 10^-3 mol C6H6 ------------> ??

moles of CO2 formed = 1.282 x 10^-3 x 12 / 2

                                       = 7.69 x 10^-3 mol

moles of water formed = 3.85 x 10^-3 mol

d)

change in moles = 18 - 17 = 1 mol