Question

A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 109g of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ?C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

Answer 1

Concepts and reason

Foam coffee cup calorimeter is a constant calorimeter. Heat released by the reaction in the cup is equal to enthalpy change.

The heat released is equal to the product of mass of the solution, specific heat of the solution, and change in temperature of the solution. Assume no heat is lost to the surroundings or to the coffee cup.

Fundamentals

Heat released to the solution by the reaction is $q = m{C_{\rm{p}}}\Delta T$ . Here, $q$ is the amount of heat, $m$ is mass of the solution, ${C_{\rm{p}}}$ is the specific heat of the solution, and $\Delta T$ is the change in temperature of the solution.

Assume the specific heat of the solution is equal to the specific heat of water.

Heat release to the solution at constant pressure is equal to the enthalpy change. That is $q = \Delta H$ . Heat is released in the reaction so the enthalpy of the reaction is negative.

Calculate the heat released to the solution as follows:

$\begin{array}{c}\\q = m{C_{\rm{p}}}\Delta T\\\\ = \left( {109{\rm{ g}}} \right)\left( {4.2{\rm{ J }}{{\rm{g}}^{ - 1}}{{\rm{ }}^{\rm{o}}}{{\rm{C}}^{ - 1}}} \right)\left( {24.70{{\rm{ }}^{\rm{o}}}{\rm{C}} - 21.00{{\rm{ }}^{\rm{o}}}{\rm{C}}} \right)\\\\ = 1693.56{\rm{ J}} \times \frac{{1{\rm{ kJ}}}}{{1000{\rm{ J}}}}\\\\{\rm{ = 1}}{\rm{.69 kJ}}\\\end{array}$

This is the amount of heat released to the solution when 2.00 moles of compound are dissolved in the solution.

Heat is released to the solution so the enthalpy change is negative. Total enthalpy of the reaction is equal to the total heat released to the solution. Therefore,

$\begin{array}{c}\\q = - \Delta H\\\\ = - 1.69{\rm{ kJ}}\\\end{array}$

Divide the total enthalpy with the number of moles of the compound, to get the enthalpy of the reaction.

$\begin{array}{c}\\{\rm{Enthalpy of the reaction = }}\frac{{\Delta H}}{{{\rm{moles of compound}}}}\\\\ = \frac{{ - 1.69{\rm{ kJ}}}}{{2.00{\rm{ mol}}}}\\\\ = - 0.847{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\end{array}$

Ans:

The enthalpy of the reaction is $- 0.847\;{\rm{kJ/mol}}$ .