Questions: Assume you produce 2 moles of water in the neutralization reaction of HCl and NaOH with a temperature increase of 5 degrees. What is your calorimeter constant? Add your answer Integer, decimal, or Enotation allowed

Solution Steps

The neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is:

\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

This reaction produces water and releases energy in the form of heat. The heat released can be calculated using the formula:

\[ q = n \cdot \Delta H_{\text{neutralization}} \]

where \( n \) is the number of moles of water produced, and \(\Delta H_{\text{neutralization}}\) is the enthalpy change of the reaction, typically around \(-57.32 \, \text{kJ/mol}\) for strong acid-strong base reactions.

Given that 2 moles of water are produced, the heat released is:

\[ q = 2 \, \text{mol} \times (-57.32 \, \text{kJ/mol}) = -114.64 \, \text{kJ} \]

Since the calorimeter constant is typically expressed in joules per degree Celsius, convert the heat from kilojoules to joules:

\[ q = -114.64 \, \text{kJ} \times 1000 \, \text{J/kJ} = -114640 \, \text{J} \]

The calorimeter constant \( C \) is calculated using the formula:

\[ C = \frac{q}{\Delta T} \]

where \(\Delta T\) is the temperature change, which is 5 degrees Celsius in this case.

\[ C = \frac{-114640 \, \text{J}}{5 \, \text{°C}} = -22928 \, \text{J/°C} \]

Since the calorimeter constant is typically expressed as a positive value, we take the absolute value:

\[ C = 22928 \, \text{J/°C} \]

Final Answer