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Grade 12th passPhysical Chemistry

The heat liberated on complete combustion of 7.8 g benzene is 327 kj. This heat has been measured at constant volume and pressure and at 27* C.Calculate the heat of combustion.at constant pressure

Profile image of Joshraj Sen
9 Years agoGrade 12th pass
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2 Answers

Profile image of Piyush Kumar Behera
C6H6 + 15/2 O------> 6 CO2 + 3H2O
Heat liberated on complete combustion of 7.8 g or benzene = 327 kJ
∴Heat liberated on complete combustion of 78 g (1mol) of benzene = 327 kJ  
Δn = 6-7 ½ = -1.5  
  R = 8.3 × 10 kJ mole K  
  T = 300 K
ΔH = ΔE + ΔngRT;   
    = -327kJ + (-1.5) (8.314 × 10 ) (300) kJ mole-1                 
-327 -3.7413 = 330.7413 KJ 
So the answer is 330.7413
Profile image of Soumyadip Sengupta
Let the no. of moles of C6H6 be n.
So, n= Given mass/Molar mass  = 7.8/78 = 0.1
at constant volume the heat liberated is Cv = 327kJ
We have to find Cp!!
 
We know, Cp = Cv + nR
=> Cp = (327 * 103) + 0.1 * 8.314  = 3.27 * 105kJ
CHEERS!!

question mark

In the reaction A + 2B ⇌ 2C, if 2 moles of A, 3 moles of B and 2 moles of C are placed in a 2 L flask and
the equilibrium concentration of C is 0.5 mol/L. The equilibrium constant (KC) for the reaction is.....and please tell how you know that the reaction is going forward .. imean how to calculate the reaction quotient ???

physical chemistry

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