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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 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

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Answer #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.

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