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ad-work [718]
3 years ago
8

Ethanol, C2H5OH, is manufactured on a large scale for a wide range of uses such as alcoholic drinks, as an industrial solvent an

d as a raw material for the synthesis of many organic compounds. Ethanol, C2H5OH, is manufactured on a large scale by two methods: • Fermentation, using yeast, of sugars, such as glucose, C6H12O6. C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g) The ethanol is then distilled off. • Hydration of ethene, C2H4, with steam in the presence of an acid catalyst. C2H4(g) + H2O(g)  C2H5OH(g) Compare the sustainability of these methods of manufacturing ethanol in terms of: • availability of starting materials and energy requirements; • atom economy. In your answer, you should make clear how the atom economy of the processes links with chemical theory.
Chemistry
1 answer:
Scorpion4ik [409]3 years ago
6 0

Answer:

See explanation

Explanation:

In considering the sustainability of the two processes, we must remember that fermentation involves the use of sugars as raw materials. These sugars are grown naturally as plants from which sugars are extracted for the fermentation process. This is a sustainable process because sugar is a renewable resource.

On the other hand, ethene is produced from the cracking of fractions obtained from distillation of natural gas and oil. Natural gas and oil are not renewable resources hence the process is not sustainable. The cracking of ethene requires a lot of energy compared to the fermentation of sugar.

As regards atom economy, the cracking of ethene, C2H4(g) + H2O(g) ------> C2H5OH(g) has a better atom economy compared to the fermentation of sugar; C6H12O6(aq) -------> 2C2H5OH(aq) + 2CO2(g).

Atom economy has to do with the process of designing reactions in such a way that atoms which are in the starting materials end up in the final product and not as waste materials.

Fermentation produces a waste material which is CO2 while the cracking of ethene produces no waste material. All atoms of the reactants are converted into atoms of products.

Hence, the cracking of ethene displays 100% atom economy. This implies that all reactant atoms are found in the desired product.

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mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)

<u>Oxidation half reaction:</u> Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V

<u>Reduction half reaction:</u> Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V       ( × 2)

<u>Net cell reaction:</u> Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.799-(-0.762)=1.561V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

[Zn^{2+}]=0.125M

[Ag^{+}]=0.240M

Putting values in above equation, we get:

E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})

E_{cell}=1.551V

Hence, the potential of the given cell is 1.551 V

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What is the pressure in a 5.00 L tank with 49.00 grams of oxygen gas at 350 K? ___atm
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The right answer is "8.81 atm".

Explanation:

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By substituting the values, we get

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