1) when 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and

Question

1) when 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and its normal boiling point of 80.1 C 169.5kJ are absorbed and P X delta V the vaporization process is equal to 145kJ. calculate the delta E and deltaH



2) When 0.455g of anthracene, C14H10 is combusted in a bomb calorimeter hat has a water jacket contanining 500.0g of water, the temp of the water increases by 8.63C Assming that the specific heat of water is 4.18J/(gXC) and that the heat absorption by the calorimeter is begligible, estimate the enthalpy of combustion per mole of anthracene

Explanation / Answer

1.Delta H = Delta E + (P*(delta V)), where Delta H is the enthalpy change (in j or kj) and delta E is the change in internal energy (in j or kj), P is pressure (typically in atm), and V is the volume.

Since (-)P*(delta V) = work, then we can rewrite the equation as

Delta H = Delta E - work

Delta E also equals the heat transferred (q) plus the work done (w)
So therefore, Delta H = q + w - w = q.

2. Q = 500.0 x 4.28 x 8.63 = 18468.2 J

moles C14H10 = 0.455 g /178.234 g/mol=0.00255

0.00255 : 18468.2 = 1 : x
x =7.24 x 10^6 J => 7240 kJ heat released by one mole

enthalpy = - 7060 kJ/mol

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