Short Free Response Questions

Short Free Response Questions

Question 1: Decomposition of $N<em>2O</em>5$

• The reaction is the decomposition of $N<em>2O</em>5$, with the rate law: $rate = k[N<em>2O</em>5]$.

• A sample of pure $N<em>2O</em>5$ gas is placed in an evacuated container and allowed to decompose at a constant temperature of 300 K.

• The concentration of $N<em>2O</em>5$ gas in the container is measured over time.

Part a: Determine the value of the rate constant k

• The rate law for a first-order reaction is related to the half-life ($t_{1/2}$).

• Half-life is the time required for the reactant concentration to decrease by half.

• From the data table, the concentration of $N<em>2O</em>5$ goes from 0.160 to 0.080 in 1.67 hours (a factor of two).

• In the next 1.67 hours, the concentration goes from 0.080 to 0.040 (halving again).

• Therefore, the half-life, $t_{1/2}$, is 1.67 hours.

• Using the equation $k = \frac{0.693}{t_{1/2}}$, where k is the rate constant.

• Plugging in the half-life gives $k = \frac{0.693}{1.67 \text{ hours}} = 0.415 \text{ hours}^{-1}$.

Part b: Identify the rate-determining step

• The rate-determining step should have the same rate law as the overall reaction: $rate = k[N<em>2O</em>5]$.

• Looking at the three steps, step 1 has $N<em>2O</em>5$ as the sole reactant.

• Step 1 ( $N<em>2O</em>5(g) \rightarrow NO<em>2(g) + NO</em>3(g)$) has the same rate law format as the overall rate law.

• Therefore, step 1 is the rate-determining step.

• Explanation: The rate law of elementary step 1 is $rate = k[N<em>2O</em>5]$, which matches the overall reaction rate law.

Part c: Effect of doubling the initial concentration on k

• The rate constant, k, is concentration-independent.

• K depends on temperature, but the temperature is constant in this problem.

• If the temperature doesn't change, k doesn't change.

• Answer: k will remain the same because it is independent of concentration and remains constant at a constant temperature.

Question 2: Titration of Oxalic Acid

• The balanced chemical equation is: $6H^+(aq) + 2MnO<em>4^-(aq) + 5H</em>2C<em>2O</em>4(aq) \rightarrow 10CO<em>2(g) + 8H</em>2O(l) + 2Mn^{2+}(aq)$.

• A student dissolves a 0.139 gram sample of oxalic acid ($H<em>2C</em>2O_4$) in water.

• The student titrates the oxalic acid solution with a solution of $KMnO_4(aq)$, which has a dark purple color.

Part a: Identify the species that was reduced

• Need to determine oxidation numbers for each element.

• Hydrogen's oxidation number is the same on both sides.

• Manganese goes from +7 in $MnO_4^-$ to +2 in $Mn^{2+}$ (gains 5 electrons, so it's reduced).

• Oxygen's oxidation number stays the same.

• Carbon goes from +3 to +4 (oxidized, gives up an electron).

• Since manganese becomes less positive, it is the element reduced, meaning it gained electrons.

• The species that manganese is present in on the reactant side is $MnO_4^-$.

• $MnO_4^-$ is the species reduced (gained five electrons).

Part b: Determine the volume of $KMnO_4$ added

• The burette scale has markings every 0.1 mL. Read to one digit more precise than that, i.e., to the hundredths place (0.01 mL).

• Final volume reading: 29.55 mL.

• Initial volume reading: 3.35 mL.

• Volume of $KMnO_4$ added: 29.55 mL - 3.35 mL = 26.20 mL.

• Note: Credit is lost if the burette readings aren't explicitly written down and if the correct significant figures are not used.

Part c: Calculate moles of $Mn^{2+}$ that reacted

• Molarity of $KMnO_4(aq)$ is 0.0235 M.

• Volume from part b is 26.20 mL, which is 0.02620 L.

• $Moles = Molarity \times Volume$

• $Moles = 0.0235 \frac{mol}{L} \times 0.02620 L = 0.000615 mol$

• Answer: 0.000615 moles (three significant figures).

Part d: Is a dilute titrant concentration reasonable?

• The proposed concentration of $KMnO_4$ is 0.00143 M, which is more than 10 times more dilute than the 0.0235 M used in part c.

• This would result in a titration volume more than 10 times larger than what was previously calculated (26.20 mL).

• That would mean around 262 mL, which would not fit in a 50 mL burette.

• Correct answer: No, that would not work.

• Reason: The titrant is so dilute that the titration volume would be much greater than the 50 mL capacity of the burette.