9.1 Practice Problems
Attempt these problems as if they were real exam questions in an exam environment.
Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem. Having to look up information or the solution should be an indicator that the previous layers (1–5) in the Structured Learning Approach have not been mastered.
A flask containing two reactants produces enough heat to raise the temperature of 1.25 L of solution from 45 °C to 99 °C. What is the heat of the reaction (in kJ) assuming the density of the solution is exactly 1 g mL–1 and the specific heat of the solution is 4.184 J g–1 °C–1?
- 282
- 0.625
- –0.282
- –282
- –0.548
Solution
Answer: D
Concept: Thermochemistry
Determine the mass of the solution.
\[\begin{align*} m_{\mathrm{soln}} &= dV \\ &= \left ( 1~\mathrm{g~mL^{-1}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \left ( 1.25~\mathrm{L} \right ) \\ &= 1.25\times 10^{3}~\mathrm{g} \end{align*}\]
Next determine the heat of reaction. Note that the reaction flask warms up so the reaction is exothermic (giving off heat). Therefore, the heat of reaction will be negative.
\[\begin{align*} \Delta H_{\mathrm{rxn}} &= -mc\Delta T \\[2ex] &= -\left ( 1.25\times 10^{3}~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C}^{-1} \right ) \left ( 99~^{\circ}\mathrm{C} - 45~^{\circ}\mathrm{C} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\[2ex] &= -282~\mathrm{kJ} \end{align*}\]
What is the change in internal energy (in J) of a system that does 2468.2 J of work on its surroundings and absorbs 25.2 J of heat?
- –2443
- 2443.0
- –2400.0
- 2400.0
- 2.493
Solution
Answer: A
Concept: Thermochemistry
The system aborbs heat; therefore, q is positive. The system does work; therefore, w is negative.
\[\begin{align*} \Delta U &= q + w \\ &= 25.2~\mathrm{J} - 2468.2~\mathrm{J} \\ &= -2443~\mathrm{J} \end{align*}\]
The value of ΔH° for the reaction below is –6,535 kJ mol–1. What change in heat (in kJ) does the combustion reaction of 16.0 g of C6H6(l) experience?
\[2\mathrm{C_6H_6}(l) + \mathrm{15O_2}(g) \longrightarrow \mathrm{12CO_2}(g) + \mathrm{6H_2O}(l)\]- –1.34×103
- 5.23×104
- –669
- 2.68×103
- –6535
Solution
Answer: C
Concept: Thermochemistry
\[\begin{align*} \Delta H_{\mathrm{rxn}} &= \left( \Delta H_{\mathrm{rxn}}^{\circ} \right ) \left ( n_{\mathrm{C_6H_6}} \right ) \\[2ex] &= \left ( \dfrac{-6535~\mathrm{kJ}}{2~\mathrm{mol~C_6H_6}} \right ) \left ( \dfrac{16.0~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{78.12~\mathrm{g}} \right )\\[2ex] &= -669~\mathrm{kJ} \end{align*}\]
The specific heat of liquid bromine is 0.226 J g–1 °C–1. How much heat (in J) is required to raise the temperature of 10.0 mL of bromine from 25.00 °C to 27.30 °C? The density of liquid bromine is 3.12 g mL–1.
- 5.20
- 16.2
- 300
- 32.4
- 10.4
Solution
Answer: B
Concept: Thermochemistry
Find the mass of 10.0 mL of bromine.
\[\begin{align*} m &= dV\\ &= \left ( \dfrac{3.12~\mathrm{g}}{\mathrm{mL}} \right ) \left ( 10.0~\mathrm{mL} \right )\\ &= 31.2~\mathrm{g} \end{align*}\]
Find the heat required.
\[\begin{align*} q &= mc\Delta T \\ &= \left ( 31.2~\mathrm{g} \right ) \left ( 0.226~\mathrm{J~g^{-1}~^{\circ}C} \right ) \left ( 27.30~\mathrm{^{\circ}C} - 25.00~\mathrm{^{\circ}C} \right ) \\ &= 16.2~\mathrm{J} \end{align*}\]
Given the following three reactions
\[\begin{align*} \mathrm{N_2}(g) + \mathrm{O_2}(g) &\longrightarrow \mathrm{2NO}(g) \quad &&\Delta H = -180.5~\mathrm{kJ} \\ \mathrm{N_2}(g) + \mathrm{3H_2}(g) &\longrightarrow \mathrm{2NH_3}(g) \quad &&\Delta H = -91.8~\mathrm{kJ} \\ \mathrm{2H_2}(g) + \mathrm{O_2}(g) &\longrightarrow \mathrm{2H_2O}(g) \quad &&\Delta H = -483.6~\mathrm{kJ} \end{align*}\] calculate the enthalpy of reaction (in kJ) for the following reaction:
\[4\mathrm{NH_3}(g) + \mathrm{5O_2}(g) \longrightarrow \mathrm{4NO}(g) + \mathrm{6H_2O}(g)\]- –1628
- 1628
- –1995
- 1995
- –660
Solution
Answer: A
Concept: Hess’s Law
Transform and combine the given reference reactions to obtain the reaction of interest. Transform ΔH appropriately.
Steps:
- Multiply the first reaction by 2 to get 4NO(g) on the right.
- Reverse and multiply the second reaction by 2 to get 4NH3(g) on the left.
- Multiply the third reaction by 3 to get 3O2(g) on the left for a total of 5O2(g) overall.
\[\begin{align*} \mathrm{2N_2}(g) + \mathrm{2O_2}(g) &\longrightarrow \mathrm{4NO}(g) \quad &&\Delta H = -361~\mathrm{kJ} \\ \mathrm{4NH_3}(g) &\longrightarrow \mathrm{2N_2}(g) + \mathrm{6H_2}(g) \quad &&\Delta H = 183.6~\mathrm{kJ} \\ \mathrm{6H_2}(g) + \mathrm{3O_2}(g) &\longrightarrow \mathrm{6H_2O}(g) \quad &&\Delta H = -1450.8~\mathrm{kJ}\\ \end{align*}\]
Adding the reactions together gives
\[\begin{align*} 4\mathrm{NH_3}(g) + \mathrm{5O_2}(g) \longrightarrow \mathrm{4NO}(g) + \mathrm{6H_2O}(g) \quad \Delta H = -1628~\mathrm{kJ} \end{align*}\]
You are hoping to use specific heat to determine the identity of a metal. You weigh the piece of metal and find that it is 0.015 kg. You find that it requires 60.3 J to raise the temperature of this piece of metal by 10 °C. Of the options in the table, which is the most likely identity of the piece of metal?
Metal Specific Heat (J g–1 °C–1) Gold
0.125604
Copper
0.376812
Iron
0.460548
Brass (Yellow)
0.4019328
Indium
0.2386476
- Brass
- Gold
- Copper
- Iron
- Indium
Solution
Answer: A
Concept: Specific heat
Determine the mass of the metal in g.
\[\begin{align*} m &= \left ( \dfrac{0.015~\mathrm{kg}}{} \right ) \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) = 15~\mathrm{g} \end{align*}\]
Find the specific heat of the metal.
\[\begin{align*} q &= mc\Delta T \longrightarrow \\[2ex] c &= \dfrac{q}{c \Delta T}\\[2ex] &= \dfrac{60.3~\mathrm{J}}{\left ( 15~\mathrm{g} \right ) \left ( 10~^{\circ}\mathrm{C} \right )} \\[2ex] &= 0.402~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \end{align*}\]
The metal is brass.
Which of the following statements best describes a state function?
- The final and initial states are the only ones that matter.
- The path, or the steps taken to get from point A to point B, matter.
- Only the initial state is taken into consideration.
- Only the final state is taken into consideration.
- The magnitude, but not the sign, of the values matters.
Solution
Answer: A
Concept: State function
Using the data given, calculate the standard reaction enthalpy (in kJ mol–1) for a precipitation reaction when Pb(NO3)2 and NaCl solutions are mixed.
Compound ΔHf° (kJ mol–1) Pb(NO3)2
-454.9
NaCl
-411.2
PbCl2
-359.4
NaNO3
-467
- –16.1 kJ mol–1
- 16.1 kJ mol–1
- 39.7 kJ mol–1
- –39.7 kJ mol–1
- Not enough information given
Solution
Answer: A
Concept: Enthalpy of reaction
Write out a balanced chemical equation.
\[\mathrm{Pb(NO_3)_2}(aq) + \mathrm{2NaCl}(aq) \longrightarrow \mathrm{PbCl_2}(s) + \mathrm{2NaNO_3}\]
Determine the enthalpy of reaction.
\[\begin{align*} \Delta H_{\mathrm{rxn}} &= &&\Sigma \left (\Delta H_{\mathrm{products}} \right ) - \Sigma \left (\Delta H_{\mathrm{reactants}} \right )\\[2ex] &= &&\left [ \left ( \Delta H_f^{\circ}(\mathrm{PbCl_2}) \right ) + \left ( 2 \times \Delta H_f^{\circ}(\mathrm{NaNO_3}) \right ) \right ] - \\ &\phantom{=} &&\left [ \left ( \Delta H_f^{\circ}(\mathrm{Pb(NO_3)_2}) \right ) + \left ( 2 \times \Delta H_f^{\circ}(\mathrm{NaCl}) \right ) \right ] \\[2ex] &= &&\left [ \left ( -359.4 \right ) + \left ( 2 \times -467.0 \right ) \right ] - \\ &\phantom{=} &&\left [ \left ( -454.9 \right ) + \left ( 2 \times -411.2 \right ) \right ]~\mathrm{kJ~mol^{-1}} \\[2ex] &= &&-16.1~\mathrm{kJ~mol^{-1}} \end{align*}\]
A combustion reaction is conducted in a bomb calorimeter with a heat capacity of 1522 J °C–1. If the calorimeter temperature rises from 20.00 °C to 21.23 °C, how much thermal energy (in J) is released from the combustion reaction?
- 1,872
- 1,906
- 2,211
- 30,440
- 32,310
Solution
Answer: A
Concept: Thermochemistry
\[\begin{align*} q &= C\Delta T \\ &= \left ( 1522~\mathrm{J~^{\circ}C} \right ) \left ( 21.23~\mathrm{^{\circ}C} - 20.00~\mathrm{^{\circ}C} \right ) \\ &= 1872~\mathrm{J} \end{align*}\]
Identify the true statement about the following reaction:
\[\mathrm{H_2} + \mathrm{O_2} \longrightarrow \mathrm{2H_2O} \quad \Delta H = -286~\mathrm{kJ~mol^{-1}}\]- This reaction is an endothermic reaction.
- This reaction is an exothermic reaction.
- This reaction is not balanced.
- This reaction requires 2 moles of O2 to make 2 moles of H2O.
- There is no true statement.
Solution
Answer: B
Concept: Thermochemistry
A 60.5 g piece of copper (cCu = 0.377 J g–1 °C–1) was heated to 85.6 °C and submerged in 22.4 °C water of unknown mass. The metal and water reached an equilibrated temperature of 29.9 °C. Find the mass (in g) of the water if cwater = 4.184 J g–1 °C–1.
- 40.5
- 60.5
- 33.6
- 15.2
- 0.56
Solution
Answer: A
Concept: Thermochemistry
\[\begin{align*} -q_{\mathrm{Cu}} &= q_{\mathrm{water}} \\[2ex] -(mc\Delta T)_{\mathrm{Cu}} &= (mc\Delta T)_{\mathrm{water}}\\[2ex] m_{\mathrm{water}} &= \dfrac{-(mc\Delta T)_{\mathrm{Cu}}}{(c\Delta T)_{\mathrm{water}}}\\[2ex] &= \dfrac{-(60.5~\mathrm{g})(0.377~\mathrm{J~g^{-1}~^{\circ}C^{-1}}) (29.9~\mathrm{~^{\circ}C} - 85.6~\mathrm{~^{\circ}C})} {(4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}}) (29.9~\mathrm{~^{\circ}C} - 22.4~\mathrm{~^{\circ}C})}\\[2ex] &= 40.5~\mathrm{g} \end{align*}\]
Use the given bond energies (in kJ mol–1) to calculate the approximate enthalpy change (ΔH in kJ) for the given reaction.
CO(g) + 2H2(g) → CH3OH(g)
C≡O: 1080
H–H: 436
C–H: 415
C–O: 350
O–H: 464- 107
- 723
- –243
- 243
- –107
Solution
Answer: A
Concept: Heat of reaction from bond enthalpies
Determine the number and types of bonds in each reactant and product.
CO- 1 C≡O
- 2 × 1 H–H
- 3 C–H
- 1 C–O
- 1 O–H
Determine the reaction enthalpy from the bond energies.
\[\begin{align*} \Delta H_{\mathrm{rxn}} &= \Sigma (\mathrm{product~bond~enthalpies}) - \\ &\phantom{=} ~~ \Sigma (\mathrm{reactant~bond~enthalpies}) \\[2ex] &= \left [ \left ( 3\times 415 \right ) + \left ( 350 \right ) + \left ( 464 \right ) \right ] \\ &\phantom{=} ~~ \left [ \left ( 1080 \right ) + \left ( 2\times 436 \right ) \right ] \\[2ex] &= 107~\mathrm{kJ} \end{align*}\]
Which ionic compound in each of the following pairs is expected to have a lower melting temperature (i.e. lower lattice energy)?
i. NaF and CsBr
ii. MgO and RbS
iii. CaS and Al2O3
iv. CaO and CaS- i: NaF ii: MgO iii: Al2O3 iv: CaO
- i: CsBr ii: MgO iii: Al2O3 iv: CaS
- i: CsBr ii: MgO iii: Al2O3 iv: CaO
- i: NaF ii: MgO iii: CaS iv: CaS
- i: CsBr ii: RbS iii: CaS iv: CaS
Solution
Answer: E
Concept: Lattice energy