Respuesta :
Calculating using mean heat capacity, -13229 J of heat needs to be removed per mole of gas.
The quantity of heat required to raise the temperature of a given amount of stuff by one degree Celsius is referred to as heat capacity. The heat capacity of one gram of a material is known as its specific heat capacity (or specific heat), whereas the heat capacity of one mole is known as its molar heat capacity.
The amount of energy required to increase the temperature of 1 gram of substance by one degree Celsius is defined as an object's specific heat capacity. The molar heat capacity refers to the amount of energy necessary to raise one mole of a substance's temperature by one degree Celsius.
First find [tex]\Delta H_{298}[/tex]:
So, from Literature
[tex]\Delta H_{298}[/tex], [tex]Cl_2[/tex] = 0
[tex]\Delta H_{298}[/tex], [tex]O_2[/tex] = 0
[tex]\Delta H_{298}[/tex], [tex]H_2O[/tex] = -241818 J/mol
[tex]\Delta H_{298}[/tex], HCl= -92307 J/mol
So [tex]\Delta H_{298}[/tex] = ΣH Product – ΣH reactant
[tex]\Delta H_{298}[/tex] = 2 x (-241818) - 4 x (-92307) = -1.144x10^5 J/mol.
[tex]T_0[/tex] = 298K
T = 823K
Now, from Literature take constants for finding mean heat capacity (MDCPH):
[tex]H_2O[/tex]:
A = 3.470, B = 1.45 x [tex]10^{-3}[/tex], C = 0(10^-7 = 0), D=0.121 x [tex]10^5[/tex]
[tex]Cl_2[/tex]:
A = 4.442, B = 0·089 x [tex]10^{-3}[/tex], C = 0, D = -0.344 x [tex]10^5[/tex]
[tex]O_2[/tex]:
A = 3.639, B = 0.506 x [tex]10^{-3}[/tex], C = 0, D = -0.227x [tex]10^5[/tex]
HCl:
A = 3.156, B = 0.623 x [tex]10^{-3}[/tex], C = 0, D = 0.151 x [tex]10^5[/tex]
ΔΑ = ΣniAi
where, n = Stoichiometric Coefficient
nHCl = -4
n[tex]O_2[/tex] = -1
n[tex]H_2O[/tex] = 2
n[tex]Cl_2[/tex] = 2
Therefore,
ΔA = ΣniAi
ΔA = 2 x 3.470 + 2 x 4.442 – 1.3 x 3.639 – 4 x 3.156
ΔA = - 0.439
Similarly,
ΔB = ΣniBi = Q x [tex]10^{-5}[/tex]
ΔC = 0
ΔD = -8·23 x [tex]10^4[/tex]
[tex]T_f[/tex] = T / [tex]T_0[/tex] = 823 / 298 = 2.7617
R = 8.314J/molK
Put all value in this Equation:
MDCPH = ΔA + (ΔB / 2) [tex]T_0[/tex] ([tex]T_f[/tex] + 1) + (ΔC / 3) [tex]T_0^2[/tex] ([tex]T_f^2[/tex] + [tex]T_f[/tex] +1) + (ΔD / [tex]T_f[/tex] – [tex]T_0^2[/tex]) = -0.8194
Now,
[tex]\Delta H_{823}[/tex] = [tex]\Delta H_{298}[/tex] + MDCPH x R x (T-298)
Put all the values in this Equation, we get
[tex]\Delta H_{823}[/tex] = -1.144x[tex]10^5[/tex] - 0.8194 x 8.314 x (823-298)
[tex]\Delta H_{823}[/tex] = -117592 J/mol
Heat transfer per mol entering the reactor:
Q = [tex]\Delta H_{823}[/tex] x 0.45 / 4
Q = -13229 J
Result:
-13229 J heat must be removed per mole of gas.
Learn more about Heat capacity here:
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