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?

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Alleei Alleei · Answer 1 · 2019-07-30T09:09:32+02:00

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

Explanation :

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

[tex]\Delta E=\Delta H-\Delta n_g\times RT[/tex]

where,

[tex]\Delta E[/tex] = internal energy of the reaction = ?

[tex]\Delta H[/tex] = enthalpy of the reaction = 483.6 KJ/mole = 483600 J/mole

From the balanced reaction we conclude that,

[tex]\Delta n_g[/tex] = 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 = [tex]165^oC=273+165=438K[/tex]

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

[tex]\Delta E=483600J/mole-(1mole\times 8.314J/mole.K\times 438K)[/tex]

[tex]\Delta E=479958.468J/mole[/tex]

[tex]\Delta E=479.958KJ/mole[/tex]

Therefore, the value of [tex]\Delta E[/tex] of the reaction is, 479.958 KJ/mole