The standard state heat of combustion (Delta H_c degree) of benzene, C_6H_6, is

Question

The standard state heat of combustion (Delta H_c degree) of benzene, C_6H_6, is -3267 kJ/mol. Balance the combustion reaction and place the Delta H_c degree on the appropriate side of the equation: C_6H_6(1) + O_2 (g) rightarrow CO_2 (g) + H_2O(I) The density of benzene, C_6H_6, is 0.874 g/ml. If you combust 200 ml of benzene liquid, what volume of oxygen is required at STP? If in a DIFFERENT combustion of benzene, 3.66 L of CO_2 (g) are produced at 127 degree and 650 mm Hg, what amount of heat would also have been produced?

Explanation / Answer

a) the balanced equation is

C6H6 (l) + 15/2 O2(g) -------> 6CO2(g) + 3H2O(l) ; delta H combustion = -3267kJ/mol

b) Weight of benzene used = volume x density

= 200mL x 0.874g/mL

From the balanced equation

one mole (78 g/mol) require 15/2 moles ( 15/2 x 22.4 L at STP) of O2 gas

hence 200x 0.874 g would require

= 2.136 L of O2 at STP

c) 3.66 L of Co2 produced at 127C and 650mm pressure. However the standard enthalpy of combustion is defined at 25 c and 1atm = 760 mm pressure.

Thus we need to convert this volume to standard conditions using equation of state

P1V1/T1 = P2V2/T2

Thus 650 x3.66/400 = 760x V/298

And Volume at standard statei =2.33L

Now when 1 mole of benzene is burnt 6 moles of CO2 is formed and 3267kJ of energy is liberated.

that is 6 moles (6x 24.45 L at standard state) gives 3267 kJ of heat

then 2.33 L at standard state gives = 2.33x3267 /(6x24.45)

= 51.89 kJ of heat

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