Question

The reactions of hydrocarbons are often studied in the petroleum industry. One such reaction is

2C₃H₈(g) → C₆H₆(ℓ) + 5H₂(g)

with ΔH° = 698 kJ. If 40 L of propane (C₃H₈) at 25°C and 0.94 atm is to react, how much heat must be supplied?

R = 0.08214 L atm mol^-1 K^-1

Answer 1

Let us consider a reaction,

2C₃H₈(g) → C₆H₆(ℓ) + 5H₂(g)

Given data,

Pressure = 0.94 atm

Volume = 30 L

Temperature = 25⁰C + 273

= 298 K

We know that, PV = nRT

0.94atm x 30L = n x 0.08214 x 298 K

n =  0.94atm x 30L / 0.08214 x 298 K

= 28.2 / 24.47

= 1.152 mol

From the reaction, 2 mol of C3H8 reacts, heat supplied = 698 KJ ( Given )

For 1.152 mol, heat supplied = ( 698 x 1.152 ) / 2

= 804.096 / 2

= 402.04 kJ of heat is supplied