I'm learning equilibirum in class, but I am stuck on this question. DOes anyone how how to do this problem? I use the quadratic formula but end end with answers A and B, but the correct answer is A. Can someone show me step by step how to do this problem? thanks 9. At a certain temperature, T, K for the reaction below is 7.5 liters/mole. 2NO2 <===> N2O4 If 2.0 moles of NO2 are placed in a 2.0-liter container and permitted to react at this temperature, what will be the concentration of N2O4 at equilibrium? a) 0.39 moles/liter b) 0.65 moles/liter c) 0.82 moles/liter d) 7.5 moles/liter e) none of these
You did the problem correctly. You end up having to solve the quadratic equation 30.x^{2} - 31x + 7.5 = 0. You get the answers of x = .39 and x = .65. Now, think about the problem. You are starting with 1.0 M NO_{2}, correct? You should notice that the equilibrium concentration of NO_{2} will be 1.0 - 2x, and the equilibrium concentration of N_{2}O_{4} will be x. If x was .65, then 1.0 - 2x would yield a negative concentration, which is physically impossible. Therefore, the only correct answer can be [N_{2}O_{4}] = .39 M because 1.0 - 2(.39) gives you a positive concentration.