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Ainat [17]
4 years ago
10

Why are the optimum conditions for the Haber process a temperature of 450°C and pressure of 200 atmospheres?

Chemistry
2 answers:
Novay_Z [31]4 years ago
8 0

Answer:

See explanation.

Explanation:

Hello,

Haber process is defined as the widely acknowledged productive process of ammonia by the reaction:

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

Which is carried out in gaseous phase. Thus, by means of the Le Chatelier's principle, it is possible to know that its standard enthalpy of reaction is -45.90 kJ/mol (NIST webbook) for which it is an exothermic chemical reaction, for that reason less ammonia will be produced at high temperature, nonetheless, the temperature should not be too low since the reaction rate significantly decrease, therefore, the optimum found temperature is 450 °C.

Moreover, since there are more moles (3+1=4) at the reactants and less moles at the products (2), increasing the pressure of the reaction increases the yield of ammonia, nonetheless, higher pressures involve the purchasing of more expensive equipment to withstand the high-pressures, for that reason, the best found pressure has been set as 200 atm.

Best regards.

Rus_ich [418]4 years ago
6 0

Answer:

Lower temperature and higher pressure favours the forward reaction in the Haber process

Explanation:

The equation of the Haber process for the production of ammonia is;

3H2(g) + N2(g) ⇄2NH3(g)

The Haber process involves the combination of hydrogen gas and nitrogen gas in a ratio of 3:1.

In the Haber process, the forward reaction is exothermic, hence the reverse reaction is endothermic. The implication of this is, as temperature is increased, the position of equilibrium moves to the left, and the yield of ammonia decreases. Decrease in temperature will favour the forward reaction. Hence the process is operated at a lower temperature of 450°C.

Secondly, the total number of volumes on the reactant side is 4 volumes whereas the total number of volumes on the product side is 2 volumes. Increase in pressure will favour the direction producing a lesser total volume. Hence increasing the pressure to about 200 atmospheres will favour the forward reaction thus more ammonia is produced

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yawa3891 [41]

Answer:

a) 37.04%  b) 37.04%  c) 9.63%

Explanation:

The theoretical percent recovery (Tr), is the total percentage of each compound in the sample. Depending on the technique used to recovery the compounds, the percent recovery will be less than the theoretical, because no technique is 100% efficient.

So, to calculate the theoretical, it will be the mass of the compound divided by the mass of the sample multiplied by 100%.

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b) Tr = (250 mg)/(675 mg) * 100%

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4 years ago
typical room is 4.0 m long, 5.0 m wide, and 2.5 m high. What is the total mass of the oxygen in the room assuming that the gas i
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Answer:

mass of oxygen gas in Kg = 15.0Kg

Explanation:

Volume of air in the room = 4.0m*5.0m*2.5m = 50m³

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using the ideal gas equation; PV=nRT

number of moles of oxygen gas, n = PV/RT

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3 years ago
An empty Erlenmeyer flask weighs 241.3 g. When filled with water (d = 1.00 g/cm³), the flask and its contents weigh 489.1 g.
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The answer is- Mass of flask = 608.04 g

Density (d) is defined as mass of substance divided by its volume. Thus, if mass and density are known, then Volume can be determined.

What is the formula of  mass in terms of density?

  • Let mass of substance be 'm' and volume be 'V'. Thus, as per the definition of density, it is expressed as-

d = \frac{m}{V}

Thus, mass can be expressed as-

m = d * V

  • Now, mass of empty flask = 241.3 g and the mass of  (flask + Water) = 489.1 g.

Thus, mass of water = 489.1\ g- 241.3\ g = 247.8\ g.

  • Thus, mass of water = 247.8 g and density of water =1.00\ g/cm^3. Its volume is calculated as-

V = \frac{247.8\ g}{1.00\ g/cm^3}=247.8\ cm^3

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<span>This is because you haven't added any oxygen needed for the combustion, so your equation does'nt balance. Also a solution in water [aq] doesn't burn! </span>
<span>Try </span><span>C16H32O2(s) + 23O2(g) --> 16CO2(g) + 16H2O(l) 

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</span>
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