Ammonia and the Haber process

Solutions to work

 

1) Below are the three steps of the synthesis of nitric acid from ammonia known as the Ostwald process. Give the oxidation states of nitrogen before and after reaction and identify the oxidant and reductant in each.
step 1) 4 NH3 (g) + 5O2 (g) → 4 NO (g) + 6 H2O (g) nitrogen changes oxidation state from -3 to +2 .
The oxidant is oxygen the reductant is ammonia

step 2) 2 NO (g) + O2 (g) → 2 NO2 (g) nitrogen changes oxidation state from .+2 to +4
The oxidant is oxygen gas the reductant is

step 3) 3 NO2 (g) + H2O (l) → 2 HNO3 (aq) + NO (g) nitrogen changes oxidation state from +4 in NO2 to +5 in . HNO3 and +2 in NO
The oxidant is NO2 the reductant is also NO2

 

2) 2NH3 (l) <=> NH4+ (aq) + NH2- (aq) This reaction is: an acid base reaction . In this reaction ammonia is acting as an amphiprotic substance

 

3) 2NH3(g) + H2SO4(l)=> (NH4)2SO4(s) This reaction is an acid/base . In this reaction ammonia is acting a base.

 

4) The Haber process can be summarised by the following equation 3H2(g) + N2(g) → 2NH3(g) (ΔH = –92 kJ/mol)
Which of the following conditions will maximise yield? high pressure , high temperature , a catalyst makes no difference to the yield.

 

5) High pressure in the Haber process will cause an increase to the rate of the reaction, and will casue a higher yield

 

6) During the Haber process formation of ammonia occurs when reactant molecules are passed over four beds of a porous catalyst. This has the effect of increasing the rate of the reaction by increasing hte surface area of the catalyst.

 

7) Compromises have to be made in industry during chemical synthesis. The greatest yield of ammonia is best achieved by:
- low temperatures and high pressures , however, temperatures between 400 and 450 oC are used because the reaction can proceed at a high enough rate to be ecnonomically viable and also produce an economically viable yield.

 

8) High pressure would drive the following reaction 3H2(g) + N2(g) → 2NH3(g) to the right . In a commercial environment the pressure is kept at a moderately high pressure, as opposed to a very high pressure due to the expnse of maintaining high pressure vessels and pipes.

 

9) During the Haber process ammonia is constantly removed. Removal of ammonia drives the reaction to the right and increases yield.

 

10) Excess reactants are sometimes used to increase the yield. However in this case a volume ratio of nitrogen to hydrogen of 1:3 is used. Why?

 

11) A 15% efficiency rate is achieved in the Haber process. Although this rate is economically viable, suggest what role temperature plays in such a low rate and why does a catalyst not increase the efficiency?

 

12) A great deal of heat is produced during the Haber process. What is done with this heat? It is used to heat the incoming reaction mixture of gases.

 

13) What conditions would result in the equilibrium below being established most rapidly? Any condition that increases the rate of the reaction, such as, high temperature, and or a catalyst . As a result a catalyst will have no impact on the yield but temperature will decrease the yield.

3H2(g) + N2(g) → 2NH3(g)

 

14) Ammonia dissolves readily in water.
i) Explain why. Ammonia has hydrogen bonding as does water and hence is able to interact well with the water molecules.

ii) A solution of ammonia is labeled 0.520 % w/v. Calculate the molarity of this solution to the right number of significant figures is 0.305M

 

15) Ammonia can be used as a fuel as was the case during WWII when the German military was starved of fossil fuels.

4NH3(g)  +  3O2(g) =>  2N2(g)   +   6H2O(g) (ΔH = –1267.20 kJ/mol)

a) If 0.017 g of ammonia was burnt completely in a bomb calorimeter to heat 6.00 g of water at 20oC. What is the final temperature of the water?

Step 1 mol of ammonia = 0.017 /.17.031 = 9.98 X 10-4 mol

Step 2 amount of energy released in J = (0.000998 / 4) X 1267.20 kJ = 316 J

Step 3 calculate temperature change.

ΔT = E / (4.18 X 6.00) = 316 /4.18= 12.6 oC

final temperature is 32.6 oC

 
 

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