Question
Could someone help me with this nitrogen fixation question, please? :)
II. Nitrogen fization A mixture of one part nitrogen and three parts hydrogen molecules is heated to a temperature of 500 °C (the reaction of nitrogen "fixation," N2 + 3H2 2NH3 puts the nitrogen in a form that plants can utilize to synthesize amino acids and other important molecules). 1. Write down the chemical equilibrium condition 2. Derive the corresponding "law of mass action expressing ratios of powers of con centrations as a function of the temperature. (Usem28mp2, mH2mp mNH3c2 17mpC2-1, where 'np is the proton mass, where I is the energy freed (per ammo- nia molecule) when a nitrogen and three hydrogen molecules form two ammonia molecules Suppomethe final tpialoh nitrogen, hydrogen, and ammonia) is 400 atm. Pretend that the gases still behave as ideal despite the high pressure. What fraction of the nitrogen molecules is converted to ammonia, then? (Write your answer as an equation expressing a function of the ratio RNHs/nNg as a number times e) High pressures are required in order to produce a large enough fraction of ammonia can you explain qualitatively why?
Explanation / Answer
1.) An Equilibrium reaction do not go under completion,instead, the two reactions occurs simultaneously.The forward and reverse reactions are opposite. The favourable chemical euilibrium conditions for the reaction given of nitogen fixation requires high pressure between 2100 and 3600 psi and temperature between 300 and 550 degree celsius.At this condition NH3 spontaneously decomposes into N2 and H2 . Eventually forward and backward reactions occur at the same rate which maintains equilibrium.
2.) On applying law of mass action : Keq = [NH3]2 / [N2] [H2]3 = Kc
3.) given : pN2 + pH2 +pNH3 = 400 atm
total no. of moles = 1+ 3 + 2 = 6 moles
the mole fraction of nitrogen molecules can be calculated as = nN2 / ntoal
= 1/6