Aere Posted March 21, 2013 Share Posted March 21, 2013 Sooo, I'm studying for a test tomorrow and I need to do this old AP question for review Anybody can help? It's Chapter 14 and 19 in the AP Chem textbook... Quote Link to comment Share on other sites More sharing options...
Lord Raven Posted March 21, 2013 Share Posted March 21, 2013 I wish I paid more attention in AP Chem, all I did was sleep in class but somehow I pulled off a 3 :') But did you try looking on collegeboard? This is definitely an old AP Chem exam. Quick google search: http://www.collegeboard.com/prod_downloads/ap/students/chemistry/ap04_sg_b_chem.pdf #7 Quote Link to comment Share on other sites More sharing options...
Aere Posted March 21, 2013 Author Share Posted March 21, 2013 I know it is but I can't find the answers lol Quote Link to comment Share on other sites More sharing options...
Aere Posted March 21, 2013 Author Share Posted March 21, 2013 I know it is but I can't find the answers lol Never mind, I'm retarded. General homework thread go! Quote Link to comment Share on other sites More sharing options...
Excellen Browning Posted March 21, 2013 Share Posted March 21, 2013 Question A, I don't really have an idea what they're asking. B is straightforward. The second law of thermodynamics states that every system wants to move to more entropy, therefore the entropy favours the reactants because they have more moles of gas. C again pretty straightforward. dG = dH - dS * T, the Gibbs free energy equation. As long as dG < 0, the reaction is favoured (and spontaneous). The enthalpy change is positive, and the entropy change is negative, which turns the "-dS*T" part of the equation into a positive value. As such the Gibbs free energy of this reaction is above 0, which means this reaction favours the reactants. D I'm not sure what they're asking once again, but a way to calculate the equilibrium constant of a reaction is to measure the concentration of all reactants and products, and using the old equilibrium formula. E. The reaction favours the reactants, as we've established beforehand. As such, in the simplified equilibrium constant formula of [products] / [reactants] = Keq the concentration of the reactants will be higher than that of the products. If the concentration of the products and reactants are equal, Keq = 1, if there's more product than reactant, Keq > 1, and if there;s more reactant then Keq < 1. The reaction favours the reactants, so the concentration of reactants is higher, therefore we expect Keq < 1 Might be too late but hope it helps. Quote Link to comment Share on other sites More sharing options...
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