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olga_2 [115]
3 years ago
10

Find the theoretical oxygen demand for the

Chemistry
1 answer:
never [62]3 years ago
7 0

Answer:

a) 213.3 mg/L

b) 62.61 mg/L

c) 0.0225 mg/L

Explanation:

Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products

a) Given:

Concentration of acetic acid,[CH3COOH] = 200 mg/L

CH3COOH + 2O2 \rightarrow 2CO2 + 2H2O

ThOD = \frac{mass\ O2}{mass\ CH3COOH} * conc. CH3COOH

Based on the reaction stoichiometry:

mass of CH3COOH = 60 g

mass of O2= 2(32) = 64 g

ThOD = \frac{64 g}{60 g} * 200mg/L = 213.3 mg/L

b) Given:

Concentration of ethanol, [C2H5OH] = 30 mg/L

C2H5OH + 3O2 \rightarrow 2CO2 + 3H2O

ThOD = \frac{mass\ O2}{mass\ C2H5OH} * conc. C2H5OH

Based on the reaction stoichiometry:

mass of C2H5OH = 46 g

mass of O2= 3(32) = 96 g

ThOD = \frac{96 g}{46 g} * 30mg/L = 62.61 mg/L

c) Given:

Concentration of sucrose, [C12H22O11] = 50 mg/L

C12H22O11 + 12O2 \rightarrow 12CO2 + 11H2O

ThOD = \frac{mass\ O2}{mass\ C22H22O11} * conc. C12H22O11

Based on the reaction stoichiometry:

mass of C12H22O11 = 342 g

mass of O2= 12(32) = 384 g

ThOD = \frac{384 g}{342 g} * 50mg/L = 0.0225 mg/L

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The information that the third quantum number of an electron gives is the direction the electron in spinning. That is option A.

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2 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
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Answer:

a) \triangle G^{0} = 7.31 kJ/mol

b) K_{-1} = 0.0594 m^{-1} s^{-1}

Explanation:

Equation of reaction:

                                     2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

b) value of k−1 at 27 °C, i.e. 300K

K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}

K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

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