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3 years ago

12

When considering free energy change, biochemists usually define a standard state, the biochemical standard state, which is modif

ied from the chemical standard state to fit biochemical applications. Determine which of the phrases describe the biochemical standard state, the chemical standard state, or both
a. ΔG (Some text use ΔG)
b. ΔG
c. Temperature is 25
d. Constant value of Mg2+
e. PH7
f. Intial concntration of reactants and products is 1M
g. Presurre is 1 atm.

1 answer:

kodGreya [7K]3 years ago

7 0

Answer:

The conditions for biochemical, chemical and both standard states are shown below

Explanation:

Chemical standard state:

Temperature is 25°

Intial concntration of reactants and products is 1M

g. Presurre is 1 atm.

PH7

Biochemical standard state:

Temperature is 25°

PH7

Constant value of Mg2+

Both:

Intial concntration of reactants and products is 1M

g. Presurre is 1 atm.

Temperature is 25°

PH7

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Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

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The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

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b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

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Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

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At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

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M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

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