Question

C2H5OH + 3O2 ⇌ 2CO2 + 3H2O
If 60% of forwarding reaction completed in 15 minutes and 40% of backward reaction completed in 8 minutes, then what would be the Kc of the above reaction at room temperature (not the Qc 'position equilibrium'

Answer 1

Answer:

the chemical reaction C2H5OH(ℓ) + 3O2 (g) → 2CO2(g) + 3H2O (ℓ).

from above chemical equation,

[ where ∆ng is the difference of number of gaseous compound of products to gaseous compound of reactants ]

given, change in internal energy, ∆U = 1368 KJ/mol

so, change in enthalpy , ∆H = ∆U + ∆RT

= 1368 - RT < 1368 KJ/mole .

hence, it is clear that the change in enthalpy will be less than 1368.