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. \(61.1\text{ kJ}\)

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