Question

The value of enthalpy change (∆h) for the reaction c2h5oh + 3o2 =2co2 + 3h2o at 27℃ is -1366.5 kj/mol the value of internal energy for the above reaction at this temprature will be

Answer 1

Given reaction is,

C₂H₅OH(liqu.) + 3O₂ (g) --------⇒ 2CO₂(g) + 3H₂O (liquid)

Change in the no. of moles of gases = no. of gas moles in products - no. of gaseous moles in reactant.

∴ Change in no. of moles = 2 - 3 = -1.

∴ Work = PΔV = Δn × RT

= - 1 × RT  

= -1 × 8.31 × 300  (T = 27° C = 300 K.)

= -2493 J/mole.

Now, Δ H = -1366.5 kJ/mol

Using the First Law of the Thermodynamics,  (of Chemistry)

ΔH = ΔU + Δn × RT

∴ ΔU = ΔH -  Δn × RT

∴ ΔU = -1366.5 kJ/mole + 2493 J/mole.

∴ ΔU =  -1364007 J/mole.

Hope it helps.

Answer 2

Answer:

-1364 kJ/mol... steps in attachment.