Question

18. Butane gas burns according to the following exothermic reaction:

C4H10 (g) + 13/2 O2 (g) → 4 CO2 (g) + 5 H2O (g) ∆H°rxn = - 2877.1 kJ

a) If 25.0 g of butane were burned, how much energy would be released?

b) If the reaction of 25.0 g of butane produced a volume change of 15.4 L against an external pressure of 748 mmHg, calculate the work done (in J).

c) Calculate the change in internal energy (∆E) for the reaction of 25.0 g of butane.

Answer 1

a)

Molar mass of C4H10 = 4*MM(C) + 10*MM(H)

= 4*12.01 + 10*1.008

= 58.12 g/mol

mass of C4H10 = 25.0 g

we have below equation to be used:

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(25.0 g)/(58.12 g/mol)

= 0.43014 mol

Since delta H is negative, heat is released

when 1 mol of C4H10 reacts, heat released = 2877.1 KJ

So,

for 0.4301 mol of C4H10, heat released = 0.43014*2877.1/1 KJ

= 1237.6 KJ

Answer: 1237.6 KJ

b)

W = -Pext(Vf-Vi)

P= 748.0 mm Hg

= (748.0/760) atm

= 0.9842 atm

Vi = 0.0 L

Vf = 15.4 L

put values in above expression

W = -Pext(Vf-Vi)

W = -0.9842 atm (15.4L - 0.0L)

W = -15.157 atm.L

W = -15.157 *101.33 J

W = -1535.8 J

Answer: -1535.8 J

c)

Q = -1237.6 KJ

W = -1535.8 J

W = -1.5358 KJ

we have below equation to be used:

deltaE = Q + W

deltaE = (-1237.6 KJ) + (-1.5358 KJ)

deltaE = -1239.1 KJ

Answer: -1239.1 KJ