Browns Chemistry The Central Science,15.8 Exercise 1

Brown's Chemistry The Central Science,15.8 Exercise 1 SAT / ACT Prep Online Guides and Tips This posts contains aTeaching Explanation. You can buyChemistry: The Central Sciencehere. Why You Should Trust Me:I’m Dr. Fred Zhang, and I have a bachelor’s degree in math from Harvard. I’ve racked up hundreds and hundreds of hours of experienceworking withstudents from 5thgradethroughgraduate school, and I’m passionate about teaching. I’ve read the whole chapter of the text beforehand and spent a good amount of time thinking about what the best explanation is and what sort of solutions I would have wanted to see in the problem sets I assigned myself when I taught. Exercise: 15.8 Practice Exercise 1 Question: †¦ When 9.2g of frozen $N_2O_4$ is added to a .50L reaction vessel †¦ [What is the value of $K_c$] Part 1: Approaching the Problem The question is asking for an equilibrium constant ($K_c$). We want to know$K_c$. Generally, we can know the equilibrium constant ONLY IF we can figure out the equilibrium concentrations of the species (nitrous oxide and dinitrogen tetraoxide): $$K_c = [NO_2]^2/[N_2O_4]$$ Thus, the entire game to figuring out the equilibrium constant here is to figure out the equilibrium concentrations. We are already given that in equilibrium, the concentration of $[N_2O_4]$=.057 molar. So we have half the puzzle: $$K_c = [NO_2]^2/.057$$ The other half of the puzzle if figuring out the equilibrium concentration $[NO_2]$. Sadly, the question doesn’tjust give us this. But we have a piece of information nearly as good, which is the starting (initial) amount of$[N_2O_4]$. Because we know the reaction equation, thekey now is to go from initial amount of$[N_2O_4]$ to the final (equilibrium) concentration $[NO_2]$. Part 2: Converting Grams to Molar We are given that the reaction started out with 9.2g of $N_2O_4$ in a 0.50L reaction vessel. For equilibrium calculations, we generally want to know concentrations of types molecules, instead of actual mass or volume. We apply stoichiometry here and convert grams per liter to molarity using molar mass. We use the periodic table to look up the molar mass of$N_2O_4$ is 92.01 grams per mole. We get that: $$(9.2g N_2O_4)/(0.50L) *(1 mol)/(92.01 g N_2O_4) = (0.100mol)/L = 0.200 molar$$ Thus the initial concentration of$N_2O_4$is 0.200 molar, and written as [$N_2O_4$]=.200 Part 3: Running the Reaction Now that we know the starting concentration, we want to get to final concentrations. The algebraic equation that links the two is the equation of reaction: $$N_2O_4 (g) ↔ 2 NO_2 (g)$$ This means that for every molecule of$N_2O_4$ we get two molecules of $NO_2$. As the reaction goes forward, when$N_2O_4$ decreases by $x$ molar,$NO_2$ increases by $2x$ molar. The concentration table is then: $N_2O_4 (g)$ $2 NO_2 (g)$ Initial Concentration (M) 0.200 0 Change in Concentration (M) -x +2x Equilibrium Concentration (M) 0.200-x 2x Part 4: Calculating the Equilibrium We are given that the equilibrium concentration of[$N_2O_4$]=.057 molar. The concentration table above gives the equilibrium concentration of[$N_2O_4$]=0.200-x, so we just equate the two and solve for x. 0.200-x = 0.057 x = .143 Now that we know x, 2x = .268 Or that in equilibrium, $[NO_2]=.268$ To calculate the equilibrium constant Kc, we plug in the information above: $$K_c = [NO_2]^2/[N_2O_4]=.268^2/.057= 1.43$$ Therefore, the right answer is d) 1.4 Video Solution Get full textbook solutions for just $5/month. PrepScholar Solutions has step-by-step solutions that teach you critical concepts and help you ace your tests. With 1000+ top texts for math, science, physics, engineering, economics, and more, we cover all popular courses in the country, including Stewart's Calculus. Try a 7-day free trial to check it out.