🧪 TheChemSolverUSNCO / General Chemistry
Lewis Structures, Thermodynamics, Calorimetry, Hess's Law, EquilibriumFRQ

3. White phosphorus is composed of P4 molecules with a tetrahedral structure, as shown in the diagra — Lewis Structures, Thermodynamics, Calorimetry, Hess's Law, Equilibrium Chemistry Question

Problem Context

  1. White phosphorus is composed of P4 molecules with a tetrahedral structure, as shown in the diagram on the left. Each P atom is bonded to the other three P atoms by single bonds, as shown in the incomplete Lewis diagram on the right.
a.

A. In the box in part A, complete the Lewis diagram for P4 by drawing the nonbonding electrons.

Model Answer

For the correct diagram [1]. *(Note: The scoring rubric provides an
image of a P4P_4 molecule drawn as a tetrahedral structure with single bonds
between all four phosphorus atoms and one lone pair of electrons on each
phosphorus atom)* [1].

b(i).

B. The reaction of white phosphorus with oxygen to form P4O10(s) is thermodynamically favorable at 298 K. The reaction is represented by equation 1.
Equation 1: P4(s) + 5 O2(g) → P4O10(s)
i. The entropy change of the reaction, ΔS°, is negative. Using particle-level reasoning, explain why the entropy decreases as the reaction progresses.

Model Answer

Because gas particles are more dispersed (have more microstates)
than solids, the entropy decreases as the reactants (which include a gas)
convert to the solid product [2].

b(ii).

ii. The enthalpy change of the reaction, ΔH°, is also negative. A student claims that the favorability of the reaction is driven by enthalpy and not by entropy. Is the student’s claim correct? Justify your answer by using the relationship between ΔG°, ΔH°, and ΔS°.

Model Answer

Yes. Given that $\Delta G^\circ_{rxn} = \Delta H^\circ_{rxn} -
T\Delta S^\circ_{rxn},thereactionmusthave, the reaction must have \Delta G^\circ_{rxn} < 0$ to be
favorable [2]. Because the reaction is exothermic, ΔHrxn∘<0\Delta H^\circ_{rxn} < 0
and enthalpy contributes to favorability [2]. ΔSrxn∘<0\Delta S^\circ_{rxn} < 0, so
entropy does not contribute to favorability [2].

c(i).

P4O10(s) reacts exothermically with water to form phosphoric acid, as represented by equation 2.
Equation 2: P4O10(s) + 6 H2O(l) → 4 H3PO4(aq)
A chemist uses a calorimetry experiment to determine the enthalpy change for the reaction, as represented by the following diagram.

C. The chemist carries out the calorimetry experiment and records the following information.

i. Calculate the amount of heat, q, released during the experiment, in kJ. Assume that the specific heat of the solution is the same as that of water.

Model Answer

$q = mc\Delta T = (100.1 \text{ g})(4.18 \text{
J/(g}\cdot^\circ\text{C}))(22.38 ^\circ\text{C} - 22.00 ^\circ\text{C}) = 160
\text{ J} = 0.16 \text{ kJ}$ [3].

c(ii).

ii. Calculate the value of ΔH°_rxn for equation 2 in kJ/mol_rxn. Include the sign in your answer.

Model Answer

qrxn=−qsurr=−0.16 kJq_{rxn} = -q_{surr} = -0.16 \text{ kJ} [3].
$\Delta H^\circ_{rxn} = \frac{-0.16 \text{ kJ}}{0.100 \text{ g } P_4O_{10}}
\times \frac{283.9 \text{ g } P_4O_{10}}{1 \text{ mol } P_4O_{10}} = -450 \text{
kJ/mol}_{rxn}$ [3].
*(The correct final answer with sign is −450 kJ/molrxn-450 \text{ kJ/mol}_{rxn})* [3].

d.

D. The chemist weighed out 0.100 g of P4O10 and 100.0 g of H2O to perform a second trial. In the second trial, some of the solid P4O10 stuck to the weighing paper and was not transferred to the calorimeter. Given that P4O10 is the limiting reactant, would ΔT for the second trial be greater than, less than, or equal to the value in the first trial? Justify your answer.

Model Answer

Less than. If less P4O10P_4O_{10} is present, less thermal energy will
be transferred to the water during the reaction, causing the temperature
increase to be less than it was with 0.100 g0.100 \text{ g} of P4O10P_4O_{10} [3, 4].

e.

P4(s) also reacts readily with Cl2(g) to produce phosphorus trichloride, PCl3(g), which in turn reacts with Cl2(g) in an equilibrium process to produce PCl5(g). The reactions are represented by equations 3 and 4.
Equation 3: P4(s) + 6 Cl2(g) → 4 PCl3(g) ΔH°_rxn1 = -1148 kJ/mol_rxn
Equation 4: PCl3(g) + Cl2(g) ⇌ PCl5(g) ΔH°_rxn2 = -88 kJ/mol_rxn
E. Calculate the standard enthalpy of formation of PCl5(g) represented by equation 5.
Equation 5: 1/4 P4(s) + 5/2 Cl2(g) → PCl5(g) ΔH°_f = ?

Model Answer

Using Hess’s law:
ΔHrxn∘=14ΔH1∘+ΔH4∘\Delta H^\circ_{rxn} = \frac{1}{4}\Delta H^\circ_1 + \Delta H^\circ_4 [4].
ΔHrxn∘=14(−1148)+(−88)=−375 kJ/mol\Delta H^\circ_{rxn} = \frac{1}{4}(-1148) + (-88) = -375 \text{ kJ/mol} [4].

f(i).

The following particle-level diagram represents the contents of the vessel in an equilibrium mixture at 546 K involving equation 4.

F. Equation 4 for the reaction that occurs is shown.
Equation 4: PCl3(g) + Cl2(g) ⇌ PCl5(g) ΔH°_rxn2 = -88 kJ/mol_rxn
i. If each particle in the diagram represents a partial pressure of 1.00 atm, what is the value of Kp for the equilibrium mixture at 546 K?

Model Answer

$K_p = \frac{P_{PCl_5}}{P_{PCl_3}P_{Cl_2}} =
\frac{4.00}{(2.00)(6.00)} = 0.333$ [4].

f(ii).

ii. Does the value of Kp increase, decrease, or remain the same when the temperature is increased to 596 K? Justify your answer based on ΔH°.

Model Answer

Decrease. The negative value of ΔH∘\Delta H^\circ indicates that
the reaction is exothermic [4, 5]. Because exothermic reactions favor reactant
formation at higher temperature, the value of KpK_p decreases [5].

New conversation: e58311e8-5544-453d-aa46-7d2a30dcb458

💬
Still have doubts about this question?
Practice more questions like this, completely free.

Practice USNCO / General Chemistry questions like this — free

4,000+ questions across AP Chemistry, USNCO, and IChO — all free, no signup required.

Explore AP Chemistry →Try Interactive Labs