AP Chemistry Questions

<p>Could someone give me a thorough explanation of this question?</p>

<p>Which combination of solutions of HCl and NaOH would
produce the largest ∆T?
(A) 50 mL of 1 M HCl with 50 mL of 1 M NaOH
(B) 50 mL of 2 M HCl with 50 mL of 2 M NaOH
(C) 100 mL of 1 M HCl with 50 mL of 2 M NaOH
(D) 100 mL of 1 M HCl with 100 mL of 1 M NaOH</p>

<p>Another question:</p>

<p>Reaction: 2NO(g) + O2(g) = 2NO2(g)
Which would increase the partial pressure of NO2(g) at
equilibrium?
(A) decreasing the volume of the system
(B) adding a noble gas to increase the pressure of the
system
(C) removing some NO(g) from the system
(D) adding an appropriate catalyst</p>

<p>Why is the answer A and not anything else?</p>

<p>For the second question, volume and pressure are inversely proportionate so decreasing the volume of the system would increase the pressure.</p>

<p>Thanks but why wouldn’t the other options work?</p>

<p>B) Adding another gas would not increase the partial pressure of the NO2. It would have a partial pressure of its own. It would increase the pressure of the system, however.
C) Removing NO(g) from the system would decrease the pressure of the system.
D) Adding a catalyst isn’t needed.</p>

<p>First question, answer is (B), most concentrated</p>

<p>Is it that simple? Or is there more to it HarveyMuddLove?</p>

<p>1) B is the most concentrated and has the lowest volume, which means least amount of water. ΔH, or enthalpy (heat energy) change depends on the number of moles of reactants. C and D both have either the same or a greater number of moles of reactant, so ΔH would be equal or higher for C and D. HOWEVER, remember that water has a high specific heat, so the more water there is, the less temperature change there will be. Thus, B will have the most ΔT since it has the most moles of reactant per volume of water.</p>

<p>2) A is correct because decrease in volume increases the total pressure. According to Le Chatelier’s Principle, when the pressures of the gases in the system at equilibrium is increased, the reaction will shift to favor the side with fewer moles of gas (in order to maintain a constant Kp). Since NO2 is on the products side, and there are fewer total moles of gas on the products side (3 mol reactant gas, 2 mol product gas), increasing the total pressure will cause the reaction to shift so that more NO2 is produced.</p>

<p>B is incorrect because the noble gas has no effect on the individual gases at equilibrium, so the reaction would be unaffected.</p>

<p>C is incorrect because removing NO would cause the reaction to shift to the left to produce more NO, thus keeping the Kp constant.</p>

<p>D is incorrect because a catalyst will only speed up the rate of reaction in both directions. It does not have any effect on Kp, so it does not cause the reaction to change.</p>

<p>^Beautiful, thank you.</p>