General Chemistry Thermochemistry Quiz - Quiz 3 - Thermochemistry Test Your Knowledge for Free!

1

ΔH of the reaction 3 N₂O (g) + 2 NH₃ (g) → 4 N₂ (g) + 3 H₂O (g) is –879.6 kJ. What is ΔH_f^o for N₂O in kJ/mol, knowing that ΔH_f^o (NH₃) = -45.9 kJ/mol and ΔH_f^o (H₂O) = -241.8 kJ/mol?

+246

+82

-82

-246

The reaction can be analyzed using the given enthalpies of formation: ΔH = ΣnΔH^o_f (products) - ∑nΔH^o_f (reactants). ΔH = 4 x ΔH_f^o (N₂) + 3 x ΔH_f^o (H₂O) - 3 x ΔH_f^o (N₂O) - 2 x ΔH_f^o (NH₃) -879.6 = 4 x 0 + 3 x (-241.8) - 3 x ΔH_f^o (N₂O) - 2 x (-45.9) 3 x ΔH_f^o (N₂O) = 246 ΔH_f^o (N₂O) = 82 kJ/mol

Concepts related to this Question

Standard Enthalpies of Formation

2

What is the change in internal energy, ΔU, for a reaction that gives off 65 joules of heat and does 38 joules of work?

-103 J

-27 J

+27 J

+103 J

The change in internal energy, ΔU, is given by: ΔU = q + w q = -65 J (heat given off, so it’s negative) w = -38 J (work done by the system, so it’s negative) Thus, the change in internal energy for the reaction is ΔU = -103 J

Concepts related to this Question

Internal Energy, Heat and Work

3

Based on the given bond dissociation enthalpies (BE), what is ∆H^o_rxn for the addition of methane to ethyne to give propene? CH₄ (g) + HC≡CH (g) → CH₃CH=CH₂ (g) BE in kJ/mol: C-H = 415; C=C = 611; C-C = 345; C≡C = 837

-119 kJ/mol

-196 kJ/mol

-259 kJ/mol

-296 kJ/mol

The reaction involves breaking one C≡C bond (837 kJ/mol) and one C-H bond (415 kJ/mol) and forming one C=C bond (611 kJ/mol), one C–C bond (345 kJ/mol), and one C-H bond (415 kJ/mol). ΔH^o = ∑ BE (reactants) - ∑ BE (products) = ∑ BE (bonds broken) - ∑ BE (bonds formed) ΔH^o = 837 + 415 - 611 - 345 - 415 = -119 kJ/mol

Concepts related to this Question

Bond Enthalpies

4

The standard enthalpy of formation of hematite, Fe₂O₃ (s), is –825.5 kJ/mol and ΔH^o_{rxn 1} of CO (g) +

\frac{1}{2}

O₂ (g) → CO₂ (g) is -283 kJ/mol.
What is the standard enthalpy of reaction, ΔH^o_{rxn 2}, for the reduction of hematite with carbon monoxide to give metallic iron as shown below? Fe₂O₃ (s) + 3 CO (g) → 2 Fe (s) + 3 CO₂ (g)

-23.5 kJ/mol

-542.5 kJ/mol

-849 kJ/mol

-1674.5 kJ/mol

From CO (g) + $\frac{1}{2}$ O₂ (g) → CO₂ (g) and using Hess's Law: ΔH^o_{rxn 1} = -283 kJ/mol = ΔH^o_f(CO₂) - ΔH^o_f(CO).
Therefore, 3 x [ΔH^o_f(CO₂) - ΔH^o_f(CO)] = -849.0 kJ/mol.

Using Hess's Law: ΔH^o_{rxn 2}= [2 x ΔH^o_f(Fe) + 3 x ΔH^o_f(CO₂)] − [ΔH^o_f(Fe₂O₃) + 3 x ΔH^o_f(CO)]
ΔH^o_f(Fe) = 0 kJ/mol and ΔH^o_f(Fe₂O₃) = -825.5 kJ/mol.
ΔH^o_{rxn 2} = 3 x [ΔH^o_f(CO₂) - ΔH^o_f(CO)] - ΔH^o_f(Fe₂O₃) = -849 + 825.5 = -23.5 kJ/mol

Concepts related to this Question

Hess's Law Standard Enthalpies of Formation

5

The gas-phase bromination of propene [C₃H₆ (g) + Br₂ (g) → C₃H₆Br₂ (g)] has a standard enthalpy of reaction of –122.5 kJ/mol. What is the standard enthalpy of formation of C₃H₆Br₂ (g)?
ΔH^o_f in kJ/mol: C₃H₆ (g) = 20.4; Br₂ (g) = 30.9.

-71.2 kJ/mol

-102.1 kJ/mol

-142.9 kJ/mol

-173.8 kJ/mol

The enthalpy change of the reaction can be expressed in terms of the enthalpies of formation of reactants and products: ΔH^o_rxn = ∑nΔH^o_f (products) - ∑nΔH^o_f (reactants)

ΔH^o_rxn = ΔH^o_f (C₃H₆Br₂ (g)) - ΔH^o_f (C₃H₆ (g)) - ΔH^o_f (Br₂ (g))

ΔH^o_rxn = -122.5 = ΔH^o_f (C₃H₆Br₂ (g)) - 20.4 - 30.9

ΔH^o_f (C₃H₆Br₂ (g)) = -71.2 kJ/mol

Concepts related to this Question

Standard Enthalpies of Formation

6

For which reaction is ΔH^o_rxn equal to ΔH^o_f of CaSO₃ (s)?

Ca (s) + $\frac{1}{8}$ S₈ (s) + $\frac{3}{2}$ O₂ (g) → CaSO₃ (s)

8 Ca (s) + S₈ (s) + 12 O₂ (g) → 8 CaSO₃ (s)

8 Ca (s) + S₈ (g) + 12 O₂ (g) → 8 CaSO₃ (s)

CaO (s) + SO₂ (g) → CaSO₃ (s)

The standard enthalpy of formation (ΔH^o_f) of CaSO₃ (s) is defined as the enthalpy change when 1 mole of CaSO₃ (s) is formed from its constituent elements in their standard states. Reaction (A) represents the formation of CaSO₃ (s) from calcium, sulfur, and oxygen in their standard states.

Concepts related to this Question

Standard Enthalpies of Formation

7

If the average carbon-hydrogen bond dissociation enthalpy in ethane is 416 kJ mol⁻¹, what is the bond dissociation enthalpy of the carbon-carbon bond in ethane?
∆H^o_f in kJ/mol: C₂H₆ (g) = -84.7; H (g) = 217.9; C (g) = 718.4.

164 kJ/mol

333 kJ/mol

386 kJ/mol

404 kJ/mol

The dissociation reaction of ethane is: C₂H₆ (g) → 2 C (g) + 6 H (g)
The enthalpy change for the reaction is: ΔH_dissociation = [2 ΔH^o_f (C(g)) +6 ΔH^o_f (H(g))] − ΔH^o_f (C₂H₆ (g))
ΔH_dissociation = [2 x 718.4 + 6 x 217.9] - (-84.7) = 2829 kJ/mol

Determine the total bond dissociation enthalpy for all the bonds in ethane:
Ethane (C₂H₆) has 6 C-H bonds and 1 C-C bond.
The total bond dissociation enthalpy is: 6 x DE_C−H + DE_C−C

Given that the average C-H bond dissociation enthalpy, DE_C−H, is 416 kJ/mol, the equation becomes:
6 x 416 + DE_C−C = 2829 ⇒ DE_C−C = 333 kJ/mol

Concepts related to this Question

Standard Enthalpies of Formation Bond Enthalpies

8

A 2.00 g sample of solid RbClO₄ (M = 184.92) is added to 100.0 g water, both initially at 23.00 °C, in a well-insulated container. The final temperature of the solution is 21.56 °C. What is ΔH^o_solution of RbClO₄? (Assume that the specific heat capacity of the solution is the same as that of pure water and neglect the heat capacity of the insulated container).

-0.615 kJ/mol

0.615 kJ/mol

13.6 kJ/mol

55.8 kJ/mol

Calculate the heat absorbed by the water: q = msΔT
where m = mass of the solution = 100.0 g, s = specific heat capacity = 4.184J/g^oC, ΔT = change in temperature = 23.00 - 21.56 = 1.44 ^oC
q = msΔT = 100.0 x 4.184 x 1.44 = 0.602 kJ

Calculate the number of moles of RbClO₄:
n_RbClO4 = $\frac{m_{RbClO 4}}{M_{RbClO 4}}$ = $\frac{2.00 g}{184.92 g / mol}$ = 0.0108 mol

Calculate ΔH^o_solution:
ΔH^o_solution = $\frac{q}{n_{RbClO 4}}$ = $\frac{0.6025 kJ}{0.0108 mol}$ = 55.8 kJ/mol

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Calorimetry and Heat Capacity

Quiz 3 - Thermochemistry | Thermochemistry

General Chemistry 2 - Quiz 3 - Thermochemistry