The commercial production of nitric acid involves the following chemical reactions: (a) Which of these reactions are redox reactions?(b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction, (c) How many grams of ammonia must you start with to make 1000.0 L of a 0.150 M aqueous solution of nitric acid? Assume all the reactions give 100% yield.
Solution: Step1: Redox reactions are the reactions which involves both oxidation and reduction. Oxidation refers to loss of electrons by an atom, molecule or ion whereas reduction means gain of electrons by an atom, molecule or ion. Oxidation leads to increase in oxidation state whereas reduction leads to decrease in oxidation state of the atom, molecule or ion. The reactant which loses electron is said to be oxidized and the reactant which gains electron is said to be reduced. Step2: a) All the reactions involved in the commercial production of nitric acid are redox reaction as they involve both oxidation and reduction. This statement is illustrated in the succeeding steps. Step3: b) In the first reaction, 4NH + 5O ------> 4NO + 6H O, nitrogen undergoes oxidation and 3 2 2 oxygen undergoes reduction. As we have seen, the oxidation number of nitrogen changes from -3 to +2(loss of electrons) i.e., its oxidation number increases. Also, the oxidation number of oxygen decreases from 0 to -2(gain of electrons). Hence, nitrogen is oxidized and oxygen is reduced. Step4: In the second reaction, 2NO + O ----2> 2NO , nitro2n undergoes oxidation and oxygen undergoes reduction. As we have seen, the oxidation number of nitrogen changes from +2 to +4(loss of electrons) i.e., its oxidation number increases. Also, the oxidation number of oxygen decreases from 0 to -4(gain of electrons). Hence, nitrogen is oxidized and oxygen is reduced. Step5: In the third reaction, 3NO + 2 O -2---> 2HNO + NO, 3trogen undergoes oxidation and reduction both. As we have seen, the oxidation number of nitrogen changes from +4 in nitrogen dioxide to +5 in nitric acid(loss of electrons) i.e., its oxidation number increases. Also, the oxidation number of nitrogen decreases from +4 in nitrogen dioxide to +2 in nitrogen monoxide(gain of electrons). Hence, nitrogen is undergoing both oxidation and reduction. Step6: c) Given molarity of nitric acid solution = 0.150 M Volume of nitric acid solution = 1000.0 L Therefore, number of moles of nitric acid = Volume of solution in litres X Molarity = 1000 L X 0.150 M = 150 mol. Now, from the third reaction given above, it is clear that 2 moles of nitric acid is prepared by 3 moles of nitrogen dioxide. Therefore, 150 mol of nitric acid will be prepared by [(3/2) x 150 = 225] mol of nitrogen dioxide. Again, from the second reaction, it is clear that 2 mol of nitrogen dioxide is prepared from 2 mol of nitrogen monoxide. Therefore, 225 mol of nitrogen dioxide will be prepared from 225 mol of nitrogen monoxide. Then, from the first reaction, it is seen that 4 mol of nitrogen monoxide is prepared from 4 mole of ammonia. Therefore, 225 mol of nitrogen monoxide will be prepared from 225 mol of ammonia. Thus number of moles of ammonia required = 225 mol Step7: Now, 1 mol ammonia(NH ) = molar mass of ammonia 3 Therefore, 225 mol ammonia = 225 x molar mass of ammonia = 225 x 17 g [molar mass of ammonia = 17 g/mol] = 3825 g Thus, the required amount of ammonia is 3825 g. ------------------
