KEY2CHEM

Processes that Change Energy

The energy changes of a chemical system can generally be categorized as temperature changes (heating or cooling a substance without phase change), phase transitions, or chemical reactions. The heat associated with a change in temperature can be determined using $$q = c\times m\times\Delta T$$, where $$q$$ = heat, $$c$$ = specific heat capacity, $$m$$ = mass, and $$\Delta T$$ = change in temperature ($$\Delta T = T_{final} – T_{initial}$$). For energy changes that do not occur with a change in temperature (meaning average kinetic energy is constant), energy changes occur as changes in potential energy and can be determined via $$q = n \times \Delta H$$, where $$\Delta H$$ is the enthalpy of the process (e.g., fusion/vaporization for phase transitions, reaction for a chemical reaction). For chemical processes that result in a change in volume (often an increase in moles of gas), the energy is transferred as both heat and pressure-volume ($$P\Delta V$$) work.

Example 1.

$$50.0 \text{ g}$$ sample of water ($$4.184 \frac{J}{g \cdot ^{\circ}\text{C}}$$) is heated from $$20.0^{\circ}\text{C}$$ to $$50.0^{\circ}\text{C}$$. How much heat is gained by the water?

A. $$6280 \text{ J}$$

B. $$4180 \text{ J}$$

C. $$10,500 \text{ J}$$

Solution

A. $$6280\text{ J}$$

$$q = c\times m \times \Delta T = (4.184 \frac{J}{g \cdot ^{\circ}\text{C}})\times (50.0\text{ g})\times (50.0^{\circ}\text{C} -20.0^{\circ}\text{C} ) = 6276\text{ J}$$

Example 2.

Heating a sample of liquid water from $$33^{\circ}\text{C}$$ to $$65^{\circ}\text{C}$$ ________ the energy of water primarily by changing its ______ energy.

A. increases, kinetic

B. decreases, kinetic

C. increases, potential

Solution

A. increases, kinetic

Increasing the temperature of the water (without undergoing a phase transition) increases its kinetic energy (energy associated with motion).

Example 3.

How much heat is required to vaporize $$1.5 \text{ mol}$$ of water? The $$\Delta {H}_{vap} = 40.7\text{ kJ/mol}$$.

A. $$40.7\text{ kJ}$$

B. $$61.1\text{ kJ}$$

C. $$27.1\text{ kJ}$$

Solution

B. $$61.1\text{ kJ}$$

$$q = n \times \Delta H_{vap} = 1.5\text{ mol} \times 40.7\text{ kJ/mol} = 61.1\text{ kJ}$$