AP Chemistry 4.5 Stoichiometry - Exam Style questions - FRQs- New Syllabus
Question
(ii) The \( \mathrm{CO_2(g)} \) produced from the reaction at \( 20.0^\circ \mathrm{C} \) was collected and found to have a pressure of \( 1.25\ \mathrm{atm} \). Calculate the volume of \( \mathrm{CO_2(g)} \), in liters.
(ii) The rate-determining step for the overall reaction is the dissolving of the solids. Would the time required for the dissolving of the solids in the second experiment be longer than, shorter than, or the same as the time required in the first experiment? Justify your answer based on the collisions between particles.
(iii) When the reaction is complete, will the volume of \( \mathrm{CO_2(g)} \) at the end of the second experiment be greater than, less than, or equal to the volume at the end of the first experiment? Justify your answer.
| Trial | Mass of \( \mathrm{H_2C_4H_2O_4} \) (g) | Mass of \( \mathrm{NaHCO_3} \) (g) | Mass of \( \mathrm{CO_2} \) Produced (g) |
|---|---|---|---|
| 1 | \( 0.581 \) | \( 0.627 \) | \( 0.220 \) |
| 2 | \( 1.162 \) | \( 0.627 \) | \( 0.328 \) |
| 3 | \( 1.543 \) | \( 1.251 \) | \( 0.550 \) |
Most-appropriate topic codes (AP Chemistry):
• Topic \( 3.4 \) — Ideal Gas Law (Part (a)(ii))
• Topic \( 5.1 \) — Reaction Rates (Parts (b)(i), (b)(iii))
• Topic \( 5.5 \) — Collision Model (Part (b)(ii))
• Topic \( 9.2 \) — Absolute Entropy and Entropy Change (Part (d))
• Topic \( 9.3 \) — Gibbs Free Energy and Thermodynamic Favorability (Part (e))
• Topic \( 8.7 \) — pH and \( pK_a \) (Part (f))
• Topic \( 8.9 \) — Henderson-Hasselbalch Equation (Part (g))
▶️ Answer/Explanation
(a)(i) For the correct calculated value:
\( 0.0114\ \mathrm{mol\ CO_2} \times \dfrac{44.01\ \mathrm{g}}{1\ \mathrm{mol}} = 0.502\ \mathrm{g\ CO_2} \)
(a)(ii) For the correct calculated value:
\( PV = nRT \)
\( V = \dfrac{nRT}{P} = \dfrac{(0.0114\ \mathrm{mol})(0.08206\ \mathrm{L \cdot atm \cdot mol^{-1} \cdot K^{-1}})(293\ \mathrm{K})}{1.25\ \mathrm{atm}} = 0.219\ \mathrm{L} \)
(b)(i) For a correct claim:
The surface area of the solid reactants increases.
(b)(ii) For the correct answer and a valid justification:
Shorter than. The powdered solids have a larger surface area than the solid chunks, so collisions between water and the surface particles occur more frequently, resulting in a faster rate of dissolution and a shorter time to dissolve the solids.
(b)(iii) For the correct answer and a valid justification:
Equal to. Both experiments begin with the same amount of reactants, so they produce the same number of moles of \( \mathrm{CO_2(g)} \) under the same conditions of pressure and temperature; therefore, the final volume is the same.
(c) For the correct answer and a valid justification:
\( \mathrm{NaHCO_3} \) is the limiting reactant because the amount present has the smaller theoretical yield of \( \mathrm{CO_2} \).
\( 1.543\ \mathrm{g\ H_2C_4H_2O_4} \times \dfrac{1\ \mathrm{mol\ H_2C_4H_2O_4}}{116.07\ \mathrm{g}} \times \dfrac{2\ \mathrm{mol\ CO_2}}{1\ \mathrm{mol\ H_2C_4H_2O_4}} = 0.02659\ \mathrm{mol\ CO_2} \)
\( 1.251\ \mathrm{g\ NaHCO_3} \times \dfrac{1\ \mathrm{mol\ NaHCO_3}}{84.01\ \mathrm{g}} \times \dfrac{2\ \mathrm{mol\ CO_2}}{2\ \mathrm{mol\ NaHCO_3}} = 0.01489\ \mathrm{mol\ CO_2} \)
Because \( 0.01489\ \mathrm{mol} < 0.02659\ \mathrm{mol} \), \( \mathrm{NaHCO_3} \) is the limiting reactant.
(d) For a valid explanation:
The entropy change is positive because the reactants produce \( 2 \) moles of gas particles according to the balanced equation. Gas particles are much more dispersed than particles in aqueous solution, so the products have greater disorder and therefore greater entropy.
(e) For the correct answer and a valid justification:
Disagree. Because the reaction is endothermic, \( \Delta H > 0 \), and because \( \Delta S > 0 \), the reaction is thermodynamically favorable only at temperatures high enough that the magnitude of \( -T\Delta S \) is greater than that of \( \Delta H \). Thus, the reaction is not favorable at all temperatures.
(f) For the correct calculated value:
\( pK_{a2} = -\log(8.5 \times 10^{-7}) = 6.07 \)
(g) For the correct calculated value:
\( \mathrm{pH} = pK_{a2} + \log \left( \dfrac{[\mathrm{C_4H_2O_4^{2-}}]}{[\mathrm{HC_4H_2O_4^-}]} \right) \)
\( \dfrac{[\mathrm{C_4H_2O_4^{2-}}]}{[\mathrm{HC_4H_2O_4^-}]} = 10^{(\mathrm{pH} – pK_{a2})} = 10^{(7.00 – 6.07)} = 8.5 \)