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Ulleksa [173]
3 years ago
6

Calculate the equilibrium constants K’eq for each of the following reactions at pH 7.0 and 25oC, using the ∆Go’ values given: (a

) Glucose-6-phosphate + H2O → glucose + PI ∆Go’= -13.8 kJ/mol (b) Lactose + H2O → glucose + galactose ∆Go’= -15.9 kJ/mol (c) Malate → fumarate + H2O ∆Go’= +3.1 kJ/mol
Chemistry
1 answer:
Archy [21]3 years ago
4 0

Answer :

(a) The value of equilibrium constant is, 262.163

(b) The value of equilibrium constant is, 611.807

(c) The value of equilibrium constant is, 0.286

Solution :

The relation between the standard Gibbs free energy and equilibrium constant are,

\Delta G^o=-2.303\times RT\times \log K_{eq}      ...........(1)

where,

\Delta G^o = standard Gibbs free energy

R = universal gas constant = 8.314 J/K/mole

T = temperature = 25^oC=273+25=298K

K_{eq} = equilibrium constant

Now we have to determine the value of K_{eq}  for the following reaction.

(a) \text{Glucose}-6-\text{phosphate}+H_2O\rightarrow \text{Glucose}+\text{Phosphate}     \Delta G^o=-13.8kJ/mol

Now put all the given values in the above formula 1, we get:

-13.8kJ/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}

-13.8\times 1000J/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}

K_{eq}=262.163

(b) \text{Lactose}+H_2O\rightarrow \text{Glucose}+\text{galactose}     \Delta G^o=-15.9kJ/mol

Now put all the given values in the above formula 1, we get:

-15.9kJ/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}

-15.9\times 1000J/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}

K_{eq}=611.807

(c) \text{Malate}\rightarrow \text{fumarate}+H_2O     \Delta G^o=+3.1kJ/mol

Now put all the given values in the above formula 1, we get:

+3.1kJ/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}

+3.1\times 1000J/mole=-2.303\times (8.314J/K/mole)\times (298K)\times \log K_{eq}

K_{eq}=0.286

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7 0
3 years ago
You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr
defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t

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2) The density of the packed bed can be expressed as

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being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum  of the fractions ois equal to the total space).

Then we can rearrange

\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37

The void fraction of the bed is 0.37.

6 0
3 years ago
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Apply this theory onto salts.
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