12.1. Nitrogen and sulfur
A subsection of Chemistry, 9701, through 12. Nitrogen and sulfur
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Ammonium nitrate is an important fertiliser made by the acid-base reaction between ammonia and nitric acid. Write an equation for the production of ammonium nitrate from ammonia and nitric acid. The ammonia for this reaction is produced by the Haber process and the nitric acid is produced by oxidation of ammonia. The Haber process involves a reaction between nitrogen and hydrogen at a temperature of 450 °C and a pressure of 20 000 kPa. At a higher reaction temperature, the rate of production of ammonia would be greater. N2+ 3H22NH3 ∆H = –92 kJ mol–1 The Boltzmann distribution curve shows the distribution of energies in a mixture of nitrogen and hydrogen at 450 °C. number of molecules energy Ea Sketch a second line onto the axes above to show the distribution of energies in the same mixture of gases at a higher temperature. With reference to the two curves, explain why the rate of production of ammonia would be greater at a higher temperature. Add a suitable label to the horizontal axis above and use it to explain why a catalyst is used in the Haber process. Explain why a higher temperature is not used despite the fact that it would increase the rate of production of ammonia. The first stage in the production of nitric acid involves the reaction of ammonia with oxygen to form nitrogen monoxide, NO, and water. Suggest an equation for this reaction and use oxidation numbers to show that it is a redox reaction. Draw a dot-and-cross diagram of the ammonium ion. Show the outer electrons only. Use the following code for your electrons. • electrons from nitrogen × electrons from hydrogen State the shape of an ammonium ion and give the H–N–H bond angle. shape bond angle State and explain the problems that arise from the overuse of ammonium nitrate fertiliser when the excess is washed into rivers.
9701_s16_qp_23
THEORY
2016
Paper 2, Variant 3
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Nitrobenzene, C6H5NO2, can be reduced to phenylamine, C6H5NH2, in acid solution in a two step process. Balance the half-equation for this reaction to work out how many moles of electrons are needed to reduce one mole of nitrobenzene. C6H5NO2 + e– + H+ → C6H5NH2 + H2O The reducing agent normally used is granulated tin and concentrated hydrochloric acid. In the first step, the reduction of nitrobenzene to phenylammonium chloride can be represented by the equation shown. Use oxidation numbers or electrons transferred to balance this equation. You might find your answer to useful. C6H5NO2 + HCl + Sn → C6H5NH3Cl + SnCl 4 + H2O When 5.0 g of nitrobenzene was reduced in this reaction, 4.2 g of phenylammonium chloride, C6H5NH3Cl, was produced. Calculate the percentage yield. percentage yield of phenylammonium chloride = % Following the reaction in , an excess of NaOHwas added to liberate phenylamine from phenylammonium chloride. Calculate the mass of phenylamine, C6H5NH2, produced when 4.20 g of phenylammonium chloride reacts with an excess of NaOH. mass of phenylamine = g The final volume of the alkaline solution of phenylamine in was 25.0 cm3. The phenylamine was extracted by addition of 50 cm3 of dichloromethane. After the extraction, the dichloromethane layer contained 2.68 g of phenylamine. Use the data to calculate the partition coefficient, K partition, of phenylamine between dichloromethane and water. K partition = How does the basicity of phenylamine compare to that of ethylamine? Explain your answer. Phenol can be synthesised from phenylamine in two steps. NH2 OH + N2 E step 1 step 2 State the reagents and conditions for steps 1 and 2. step 1 step 2 Draw the structure of the intermediate compound E in the box above.
9701_s16_qp_41
THEORY
2016
Paper 4, Variant 1
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Nitrobenzene, C6H5NO2, can be reduced to phenylamine, C6H5NH2, in acid solution in a two step process. Balance the half-equation for this reaction to work out how many moles of electrons are needed to reduce one mole of nitrobenzene. C6H5NO2 + e– + H+ → C6H5NH2 + H2O The reducing agent normally used is granulated tin and concentrated hydrochloric acid. In the first step, the reduction of nitrobenzene to phenylammonium chloride can be represented by the equation shown. Use oxidation numbers or electrons transferred to balance this equation. You might find your answer to useful. C6H5NO2 + HCl + Sn → C6H5NH3Cl + SnCl 4 + H2O When 5.0 g of nitrobenzene was reduced in this reaction, 4.2 g of phenylammonium chloride, C6H5NH3Cl, was produced. Calculate the percentage yield. percentage yield of phenylammonium chloride = % Following the reaction in , an excess of NaOHwas added to liberate phenylamine from phenylammonium chloride. Calculate the mass of phenylamine, C6H5NH2, produced when 4.20 g of phenylammonium chloride reacts with an excess of NaOH. mass of phenylamine = g The final volume of the alkaline solution of phenylamine in was 25.0 cm3. The phenylamine was extracted by addition of 50 cm3 of dichloromethane. After the extraction, the dichloromethane layer contained 2.68 g of phenylamine. Use the data to calculate the partition coefficient, K partition, of phenylamine between dichloromethane and water. K partition = How does the basicity of phenylamine compare to that of ethylamine? Explain your answer. Phenol can be synthesised from phenylamine in two steps. NH2 OH + N2 E step 1 step 2 State the reagents and conditions for steps 1 and 2. step 1 step 2 Draw the structure of the intermediate compound E in the box above.
9701_s16_qp_43
THEORY
2016
Paper 4, Variant 3
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