AP Chemistry 6.1 Endothermic and Exothermic Processes - Exam Style questions - FRQs- New Syllabus

(a)
For the correct balanced equation (state symbols not required):
Accept one of the following:

• \( \mathrm{CaCO}_{3}(s) + 2\,\mathrm{H}^{+}(aq) \rightarrow \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \)
• \( \mathrm{CaCO}_{3}(s) + 2\,\mathrm{H}_{3}\mathrm{O}^{+}(aq) \rightarrow \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{2}(g) + 3\,\mathrm{H}_{2}\mathrm{O}(l) \)

Chloride is a spectator ion, so it does not appear in the net ionic equation.

(b)
For a correct explanation:
Accept one of the following:

• Even though the concentration of \( \mathrm{HCl} \) is greater in trial \( 5 \) than in trial \( 2 \), the reaction time is significantly longer. Both trial \( 2 \) and \( 5 \) occur under otherwise identical conditions. The trend for trial \( 1 \) and \( 4 \) indicates that higher concentration of \( \mathrm{HCl} \) results in a shorter time of reaction.

• The time of reaction in trial \( 5 \), with small chunks of calcium carbonate, is longer than trial \( 6 \) with large chunks. Both trial \( 5 \) and \( 6 \) occur under otherwise identical conditions. The trend for trials \( 1 \), \( 2 \), and \( 3 \) shows that larger chunks of the solid result in longer time of reaction.

So trial \( 5 \) does not fit the pattern shown by the other trials and is therefore inconsistent.

(c)
For a correct explanation of the effect of surface area on reaction time:
The time of reaction in trial \( 2 \) is shorter than trial \( 3 \) because the calcium carbonate in trial \( 2 \) has a larger surface area \( (\text{meaning that more particles of calcium carbonate are exposed to the } \mathrm{H}^{+} \text{ particles in solution}) \).

For a correct explanation of the effect of particle collisions on reaction rate:
The larger interface between the two reacting substances means there will be more collisions between the particles in a given amount of time, and thus, a higher frequency of successful collisions in which the particles react to form the products.

In short, more exposed surface means more opportunities for acid particles to collide with the solid each second.

(d)
For the correct answer and a valid justification:
Accept one of the following:

• Disagree. If the reaction was zeroth order with respect to \( \mathrm{HCl} \), then changing the concentration of \( \mathrm{HCl} \) would not affect the rate of reaction, and the time of reaction would be the same for trials in which the only difference was \( [\mathrm{HCl}] \). The student’s data for trials \( 1 \) and \( 4 \) \( (\text{and likewise for } 3 \text{ and } 6) \) show that changing \( [\mathrm{HCl}] \) significantly alters the time of reaction.

• Disagree. The reaction appears to be first order, not zeroth order, with respect to \( [\mathrm{HCl}] \). Tripling \( [\mathrm{HCl}] \) results in a reaction time that is about \( \dfrac{1}{3} \) of that when \( [\mathrm{HCl}] = 1.00\ \mathrm{M} \), which means the reaction rate has also tripled, indicating a first-order process.

Since rate changes when concentration changes, the order with respect to \( \mathrm{HCl} \) cannot be zero.

(e)
For the correct calculated moles of \( \mathrm{HCl} \) reacted (may be implicit):
\( 1.00\ \mathrm{g\ CaCO}_{3} \times \dfrac{1\ \mathrm{mol}}{100.09\ \mathrm{g}} = 0.00999\ \mathrm{mol\ CaCO}_{3} \)

\( 0.00999\ \mathrm{mol\ CaCO}_{3} \times \dfrac{2\ \mathrm{mol\ HCl}}{1\ \mathrm{mol\ CaCO}_{3}} = 0.0200\ \mathrm{mol\ HCl\ reacted} \)

For the correct calculated \( [\mathrm{HCl}] \) remaining, consistent with the number of moles reacted:
\( 0.0500\ \mathrm{L} \times \dfrac{1.00\ \mathrm{mol\ HCl}}{1\ \mathrm{L}} = 0.0500\ \mathrm{mol\ HCl\ initially\ present} \)

\( 0.0500\ \mathrm{mol} – 0.0200\ \mathrm{mol} = 0.0300\ \mathrm{mol\ remaining} \)

\( \dfrac{0.0300\ \mathrm{mol}}{0.0500\ \mathrm{L}} = 0.600\ \mathrm{M\ HCl\ remaining} \)

Therefore, the molarity of \( \mathrm{HCl}(aq) \) after completion in trial \( 2 \) is \( \boxed{0.600\ \mathrm{M}} \).

(f)
For the correct answer and a valid justification:
Exothermic. The solution temperature increases as the reaction proceeds, rising from \( 21.20^\circ \mathrm{C} \) to \( 21.90^\circ \mathrm{C} \). This means thermal energy is released by the reaction to the surroundings.

Since the solution warms up, \( q_{\mathrm{sur}} \) is positive and the reaction itself must have \( q_{\mathrm{sys}} < 0 \).

(g)(i)
For the correct calculated value (sign not required):
\( q_{\mathrm{sur}} = mc\Delta T = (51.0\ \mathrm{g})(4.0\ \mathrm{J/(g \cdot ^\circ C)})(21.90^\circ \mathrm{C} – 21.20^\circ \mathrm{C}) = 140\ \mathrm{J} \)

Here, \( \Delta T = 0.70^\circ \mathrm{C} \), so the heat absorbed by the solution is \( \boxed{140\ \mathrm{J}} \).

(g)(ii)
For the correct calculated value, consistent with part \( \mathrm{(g)(i)} \), and the correct sign, consistent with part \( \mathrm{(f)} \):
\( q_{\mathrm{sys}} = -q_{\mathrm{sur}} = -140\ \mathrm{J} = -0.14\ \mathrm{kJ} \)

\( 1.00\ \mathrm{g\ CaCO}_{3} \times \dfrac{1\ \mathrm{mol\ CaCO}_{3}}{100.09\ \mathrm{g\ CaCO}_{3}} \times \dfrac{1\ \mathrm{mol}_{\mathrm{rxn}}}{1\ \mathrm{mol\ CaCO}_{3}} = 0.00999\ \mathrm{mol}_{\mathrm{rxn}} \)

\( \Delta H^\circ_{\mathrm{rxn}} = \dfrac{-0.14\ \mathrm{kJ}}{0.00999\ \mathrm{mol}_{\mathrm{rxn}}} = -14\ \mathrm{kJ/mol}_{\mathrm{rxn}} \)

Therefore, the enthalpy of reaction is \( \boxed{-14\ \mathrm{kJ/mol}_{\mathrm{rxn}}} \).