Respuesta :
Answer : The value of [tex]\Delta E[/tex] of the reaction is, -553.8 KJ
Explanation :
Formula used :
[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 = -184.6 KJ/mole = -184600 J/mole
The balanced chemical reaction is,
[tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]
when the moles of [tex]H_2\text{ and }Cl_2[/tex] are 3 moles then the reaction will be,
[tex]3H_2(g)+3Cl_2(g)\rightarrow 6HCl(g)[/tex]
From the given balanced chemical reaction we conclude that,
[tex]\Delta n_g[/tex] = change in the moles of the reaction = Moles of product - Moles of reactant = 6 - 6 = 0 mole
R = gas constant = 8.314 J/mole.K
T = temperature = [tex]25^oC=273+25=298K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta E=(-184600J/mole\times 3mole)-(0mole\times 8.314J/mole.K\times 298K)[/tex]
[tex]\Delta E=-553800J[/tex]
[tex]\Delta E=-553.8KJ[/tex]
Therefore, the value of [tex]\Delta E[/tex] of the reaction is, -553.8 KJ
The enthalpy of the reaction = The change in internal energy of the
reaction
The change in the internal energy for the reaction, ΔU is -553.8 kJ/mol.
Reasons:
The given parameters are;
Chemical reaction;
H₂(g) + Cl₂(g) → 2HCl(g)
ΔH for the reaction = -184.6 kJ/mol
Number of moles of H₂ = 3.00 moles
Number of moles of Cl₂ = 3.00 moles
Required:
ΔU (in kJ) for the reaction
Solution:
The pressure and temperature for the reaction = 1.0 atm, and 25°C, which
are the standard temperature and pressure.
The reaction is presented as follows;
3H₂(g) + 3Cl₂(g) → 6HCl(g)
The number of moles in the reactant and product are multiplied by 3,
therefore, ΔH = 3 × -184.6 kJ/mol = -553.8 kJ/mol
The change in the number of moles of gas, [tex]\Delta n_g[/tex], is given as follows;
[tex]\Delta n_g[/tex] = [tex]n_{final} - n_{initial}[/tex]
The number of moles of gas in the reactants, [tex]n_{initial}[/tex] = 6 moles
The number of moles of gas in the final product, [tex]n_{final}[/tex] = 6 moles
∴ [tex]\Delta n_g[/tex] = 6 - 6 = 0
ΔH = ΔU + [tex]\Delta n_g[/tex]·R·T
Which gives;
ΔH = ΔU + 0×R×T = ΔU
ΔH = -553.8 kJ/mol. = ΔU
ΔU = -553.8 kJ/mol.
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