Question

Consider the reaction 2H2O(g) → 2H2(g) + O2(g)ΔH = 483.6 kJ/mol. If 2.0 moles of H2O(g) are converted to H2(g) and O2(g) against a pressure of 1.0 atm at 165°C, what is ΔU for this reaction?

Answer 1

Answer : The value of
`\Delta E` of the reaction is, 479.958 KJ/mole

Explanation :

The relation between the internal energy and enthalpy of reaction is:

`\Delta E=\Delta H-\Delta n_g* RT`

where,

`\Delta E` = internal energy of the reaction = ?

`\Delta H` = enthalpy of the reaction = 483.6 KJ/mole = 483600 J/mole

From the balanced reaction we conclude that,

`\Delta n_g` = change in the moles of the reaction = Moles of product - Moles of reactant = 3 - 2 = 1 mole

R = gas constant = 8.314 J/mole.K

T = temperature =
`165^oC=273+165=438K`

Now put all the given values in the above formula, we get:

`\Delta E=483600J/mole-(1mole* 8.314J/mole.K* 438K)`

`\Delta E=479958.468J/mole`

`\Delta E=479.958KJ/mole`

Therefore, the value of
`\Delta E` of the reaction is, 479.958 KJ/mole