Thermochemistry Study Guidance – General Chemistry

Q1. Energy that is associated with the position or composition of an object is called:

Background

Topic: Types of Energy

This question tests your understanding of the different forms of energy, specifically potential energy versus kinetic, thermal, and chemical energy.

Key Terms:

• Potential energy: Energy due to position or composition.

• Kinetic energy: Energy due to motion.

• Thermal energy: Energy associated with temperature.

• Chemical energy: Energy stored in chemical bonds.

Step-by-Step Guidance

1. Review the definitions of each energy type listed in the answer choices.

2. Recall that energy associated with position (such as a ball held above the ground) or composition (such as the arrangement of atoms in a molecule) is called potential energy.

3. Compare this to kinetic energy (motion), thermal energy (temperature), and chemical energy (bonding).

Try solving on your own before revealing the answer!

Final Answer: C) Potential energy

Potential energy is the energy associated with the position or composition of an object.

Q2. The water at the top of a waterfall contains ________ energy.

Background

Topic: Types of Energy

This question is about identifying the type of energy present due to the position of an object.

Key Terms:

• Potential energy: Energy due to position.

• Kinetic energy: Energy due to motion.

• Gravitational energy: A form of potential energy due to gravity.

Step-by-Step Guidance

1. Think about the water before it falls – it is not moving, so it does not have kinetic energy yet.

2. Because it is elevated, it has energy due to its position in a gravitational field.

3. Recall that this is a classic example of potential energy.

Try solving on your own before revealing the answer!

Final Answer: C) Potential energy

Water at the top of a waterfall has potential energy due to its position.

Q3. Which of the following signs on q and w represent a system that is doing work on the surroundings as well as losing heat to the surroundings?

Background

Topic: Thermodynamics – Sign Conventions

This question tests your understanding of the sign conventions for heat (q) and work (w) in thermochemistry.

Key Terms:

• q: Heat transferred; negative means heat is lost by the system.

• w: Work done; negative means work is done by the system on the surroundings.

Step-by-Step Guidance

1. Recall that when a system loses heat, q is negative ($q < 0$).

2. When a system does work on the surroundings, w is also negative ($w < 0$).

3. Look for the answer choice where both q and w are negative.

Try solving on your own before revealing the answer!

Final Answer: A) q = -, w = -

Both heat and work are lost by the system, so both signs are negative.

Q6. Calculate the change in internal energy (ΔU) for a system that is giving off 45.0 kJ of heat and is performing 855 J of work on the surroundings.

Background

Topic: First Law of Thermodynamics

This question tests your ability to calculate the change in internal energy using heat and work values.

Key formula:

$\Delta U = q + w$

Where:

• $q$ = heat exchanged (in kJ or J)

• $w$ = work done (in kJ or J)

Step-by-Step Guidance

1. Identify the values: $q = -45.0$ kJ (heat given off, so negative), $w = -855$ J (work done on surroundings, so negative).

2. Convert all units to the same system. Since $q$ is in kJ and $w$ is in J, convert $w$ to kJ: $w = -0.855$ kJ.

3. Plug the values into the formula: $\Delta U = q + w$.

4. Set up the calculation: $\Delta U = -45.0$ kJ $+ (-0.855$ kJ$)$.

Try solving on your own before revealing the answer!

Final Answer: -45.9 kJ

$\Delta U = -45.0$ kJ $+ (-0.855$ kJ$) = -45.9$ kJ

The system loses both heat and work, so the internal energy decreases.

Q17. Calculate the amount of heat (in kJ) required to raise the temperature of a 79.0 g sample of ethanol from 298.0 K to 385.0 K. The specific heat capacity of ethanol is 2.42 J g-1 °C-1.

Background

Topic: Calorimetry – Heat Calculations

This question tests your ability to use the specific heat formula to calculate the heat required for a temperature change.

Key formula:

$q = m \times C \times \Delta T$

Where:

• $q$ = heat absorbed (in J)

• $m$ = mass (in g)

• $C$ = specific heat capacity (in J g-1 °C-1)

• $\Delta T$ = change in temperature (in °C or K)

Step-by-Step Guidance

1. Calculate $\Delta T$: $\Delta T = 385.0$ K $- 298.0$ K $= 87.0$ K (since the change in K and °C is the same).

2. Plug the values into the formula: $q = 79.0$ g $\times 2.42$ J g-1 °C-1 $\times 87.0$ °C.

3. Calculate $q$ in J, then convert to kJ by dividing by 1000.

Try solving on your own before revealing the answer!

Final Answer: 16.6 kJ

$q = 79.0 \times 2.42 \times 87.0 = 16,616.46$ J $\approx 16.6$ kJ

We used the correct formula and converted units properly.

Q22. A balloon is inflated from 0.0100 L to 0.500 L against an external pressure of 10.00 bar. How much work is done in joules? (100 J = 1 L bar)

Background

Topic: Work in Thermodynamics

This question tests your ability to calculate the work done by a system during expansion against a constant external pressure.

Key formula:

$w = -P_{ext} \times \Delta V$

Where:

• $w$ = work (in L·bar or J)

• $P_{ext}$ = external pressure (in bar)

• $\Delta V$ = change in volume (in L)

Step-by-Step Guidance

1. Calculate $\Delta V$: $\Delta V = 0.500$ L $- 0.0100$ L $= 0.490$ L.

2. Plug values into the formula: $w = -10.00$ bar $\times 0.490$ L $= -4.90$ L·bar.

3. Convert L·bar to J using the given conversion: $1$L·bar$= 100$J, so$w = -4.90 \times 100$ J.

Try solving on your own before revealing the answer!

Final Answer: -490 J

Work is negative because the system is doing work on the surroundings during expansion.

Q29. Identify what a bomb calorimeter measures.

Background

Topic: Calorimetry

This question tests your knowledge of the function of a bomb calorimeter in thermochemistry experiments.

Key Terms:

• Bomb calorimeter: Measures the change in internal energy ($\Delta U$) for combustion reactions at constant volume.

• Coffee cup calorimeter: Measures enthalpy change ($\Delta H$) at constant pressure.

Step-by-Step Guidance

1. Recall that a bomb calorimeter operates at constant volume, so it measures $\Delta U$.

2. Bomb calorimeters are typically used for combustion reactions.

3. Compare this to coffee cup calorimeters, which measure $\Delta H$ at constant pressure.

Try solving on your own before revealing the answer!

Final Answer: B) Measures ΔU for combustion reactions

Bomb calorimeters measure the change in internal energy for combustion reactions.

Q31. Calculate the change in internal energy (ΔU) for a system that is giving off 25.0 kJ of heat and is changing from 12.00 L to 6.00 L in volume at 1.50 bar. (Remember that 100 J = 1 L bar)

Background

Topic: First Law of Thermodynamics – Work and Heat

This question tests your ability to calculate internal energy change when both heat and work are involved, including pressure-volume work.

Key formula:

$\Delta U = q + w$

$w = -P_{ext} \times \Delta V$

Where:

• $q$ = heat exchanged (in kJ)

• $w$ = work done (in kJ)

• $P_{ext}$ = external pressure (in bar)

• $\Delta V$ = change in volume (in L)

Step-by-Step Guidance

1. Identify the values: $q = -25.0$ kJ (heat given off, so negative).

2. Calculate $\Delta V$: $\Delta V = 6.00$ L $- 12.00$ L $= -6.00$ L (volume decreases).

3. Calculate work: $w = -P_{ext} \times \Delta V = -1.50$ bar $\times (-6.00$ L$) = +9.00$ L·bar.

4. Convert work to kJ: $9.00$ L·bar $\times 100$ J/L·bar $= 900$ J $= 0.900$ kJ.

5. Plug values into $\Delta U = q + w$ and set up the calculation.

Try solving on your own before revealing the answer!

Final Answer: -24.1 kJ

$\Delta U = -25.0$ kJ $+ 0.900$ kJ $= -24.1$ kJ

The system loses heat but gains energy from work done on it.