Question

2. (Type III) The reaction N2(g) + 3H2(g) → 2 NH3(g) starts with 6.00 moles of N2 and 6.00 moles of H2 in a 4.00L flask. a. What are the concentrations of N2 and H2 initially? b. If, at equilibrium, the concentration of N2 is 1.14 M. what are the concentrations of H2 and NH3 at equilibrium? c. What is the value of equilibrium constant K?

Answer 1

2. N2(g) + 3 H₂ (9) 2NH3(g) the 6 moles moles of a) At initial the moles of 'N2 6 moles CH volume of the flask is is, 4 L the concentration of 16+4). motlle 1.5 und IL is 50, of N 2 15 M is the concentreation of H₂ (6+4) moll 1.5 mol/L = 15m b) H2NH3(9) N2(g) it 3 H₂. (g) 1.5 M At initial 1' 5 M 0.72.14 At equilibrium 1114M (115-1.08) 0:42M concentration of At equilibrium Ne is (1.14)M So, no dissociates (15 -1.14) = 0.36M Again, from the equation it is clear that 1. mole NA will reeaet with and produce mole zmoles H₂ Ntoi

so, the dissociated or the EN2 takes part amount amount which part in the recaction 0.36 M. and 0.36 M rNa will react with (0.36x3) 1.08 M He produce (0.36 *2)= 0.72 M NH3 C. constant equilibrium [Nita 43]² KE where [NH3] [N2] and [Hz] are the concentreation of NH3, Na and tz respectively at equilibrium (0.72 72) (114) x (0.4.23 2 Ke = 6:137