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 30 L of propane (C3H8) at 25°C and 0.90 atm is to react, how much heat must be supplied (in KJ)? R = 0.08214 L atm mol-1 K-1
Volume (V) = 30 L
Pressure (P) = 0.90 atm
Temperature (T) =25° C = 298 K
Let n is moles
Using ideal gas equation;
PV = nRT
n = PV/RT = 0.90 * 30 / 0.08214 * 298 = 0.8259 moles of propane
From reaction;
2 mole C3H8 requires = 698 KJ
So, 0.8259 mole will require = 698*0.8259/2 = 288.24 KJ heat
So, 288.24 KJ heat must be supplied.
The reactions of hydrocarbons are often studied in the petroleum industry. One such reaction is 2C3H8(g)...
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
The reactions of hydrocarbons are often studied in the petroleum industry. One such reaction is 2C3H2(g) → CHG(1) + 5Hz(2) with AH° = 698 kJ. If 25 L of propane (CzHg) at 25°C and 0.95 atm is to react, how much heat must be supplied? R = 0.08214 L atm mol-K-1 kJ
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