The molar heat of reaction is the quantity of energy absorbed or released during the transformation of one mole of a reactant or the formation of one mole of a product.
The mass heat of reaction is the quantity of energy absorbed or released during the transformation of one gram of a reactant or the formation of one gram of a product.
The enthalpy variation of a reaction corresponds to the energy absorbed or released during the reaction. It is generally measured in joules |(\text{J})| or kilojoules |(\text{kJ}).| However, for comparison purposes, it is often useful to express this quantity of energy in relation to the quantity of substance that reacted. The amount of substance can be expressed in grams or moles, which differentiates between the heat capacity and the molar heat of reaction.
When the enthalpy variation |(\Delta H)| is expressed in |\text{kJ/g},| it is referred to as the mass heat of reaction.
The variation in enthalpy can be expressed as a function of the quantity of products or reactants in grams. In this case, it is referred to as the mass heat of reaction. This mass heat expresses the quantity of energy that is absorbed or released during the transformation of one gram of reactant or the formation of one gram of product. The mass heat of reaction is expressed in |\text{kJ/g}.|
If the molar heat of vaporisation of water is |40.8\ \text{kJ/mol},| what is its mass heat of vaporisation?
Knowing that one mole of water is equivalent to |18.02\ \text{g},| we can say that |40.8\ \text{kJ}| is the energy released by the vaporisation of |18.02\ \text{g}.| By a cross product, we can find out how many |\text{kJ}| are released by the vaporisation of |1\ \text{g}| of water.
|\begin{align} \dfrac{40.8\ \text{kJ}}{{18.02}\ \text{g}}&= \dfrac{?}{1\ \text{g}}\\ ?&=\dfrac{40.8\ \text{kJ}\times {1\ \cancel{\text{g}}}}{18.02\ \cancel{\text{g}}}\\ ?&=2.26\ \text{kJ}
\end{align}|
The mass heat of vaporisation of water is therefore |2.26\ \text{kJ/g}.|
When the enthalpy variation |(\Delta H)| is expressed in |\text{kJ/mol},| it is referred to as the molar heat of reaction.
The enthalpy variation can be expressed as a function of the quantity of products or reactants in moles. In this case, it is reffered to as the molar heat of reaction. This molar heat expresses the amount of energy that is absorbed or released during the transformation of one mole of reactant or the formation of one mole of products. The molar heat of reaction is expressed in |\text{kJ/mol}.|
However, even when the molar heat of reaction is expressed in terms of a particular substance, it is important to remember that it is in fact closely related to the reaction being studied.
Methane combustion proceeds according to the following reaction:
|\text{CH}_{4\text{(g)}}+2\text{O}_{2\text{(g)}} \rightarrow \text{CO}_{2\text{(g)}}+2\text{H}_2\text{O}_\text{(l)}|
If the combustion of one mole of |\text{CH}_4| releases |890\ \text{kJ},| we can deduce that :
- the reaction of 2 moles of |\text{O}_2| releases |890\ \text{kJ};|
- the production of 1 mole of |\text{CO}_2| releases |890\ \text{kJ};|
- the production of 2 moles of |\text{H}_2\text{O}| releases |890\ \text{kJ}.|
Thus, the reaction of 1 mole of |\text{O}_2| would release |445\ \text{kJ},| as would the production of 1 mole of |\text{H}_2\text{O}.|
As there are several types of reaction, the molar heat of reaction can be given the specific name of the transformation it describes. Using a calorimeter, it is possible to determine the heat of transformation taking place in an aqueous medium. A distinction is often made between the molar heat of dissolution and the molar heat of neutralisation.
The molar heat of dissolution |(\Delta H_d)| is the amount of energy that is absorbed or released when one mole of solute is dissolved in a solvent.
The molar heat of dissolution of a substance can be calculated from calorimetric experiments in which temperature measurements are recorded and used in heat calculations. Among other things, the molar heat of dissolution can be used to determine the final temperature of a solution after the solute has dissolved.
In a calorimeter containing |150.0\ \text{mL}| of water, we dissolve |6.70\ \text{g}| of lithium hydroxide |(\text{LiOH}_{\text(s)}).| The temperature of the water increases from |25.0°\text{C}| to |37.0°\text{C}.|
What is the molar heat of dissolution of lithium hydroxide |(\text{LiOH})|?
First, we need to calculate the energy released by the calorimeter.
|\begin{align}Q&=mc\Delta T\\Q&=150.0\ \cancel{\text{g}} \times 4.19\ \text{J/}\cancel{\text{g}}\cdot{\cancel{°\text{C}}}\times (37.0\cancel{°\text{C}}-25.0\cancel{°\text{C}})\\
Q&=7\;542\ \text{J} \approx 7.54 \times 10^3\ \text{J}\\
Q&= 7.54\ \text{kJ}
\end{align}|
As energy is released, the enthalpy variation |(\Delta H)| is negative, which gives |-7.54\ \text{kJ}.|
Next, we need to find out how many moles correspond to |6{,}70\ \text{g},| knowing that one mole of lithium hydroxide |(\text{LiOH})| corresponds to |23.95\ \text{g}.|
|\begin{align}\dfrac{?\ \text{mol}}{6.70\ \text{g}}&=\dfrac{1\ \text{mol}}{23.95\ \text{g}}\\
?\ \text{mol}&= \dfrac{1\ \text{mol} \times 6.70\ \cancel{\text{g}}}{23.95\ \cancel{\text{g}}}\\
?\ \text{mol} &= 0.280\ \text{mol}
\end{align}|
Finally, we just need to find out how much energy would be released for 1 mole of lithium hydroxide |(\text{LiOH})| .
|\begin{align}\dfrac{-7.54\ \text{kJ}}{0.280\ \text{mol}}&=\dfrac{?\ \text{kJ}}{1\ \text{mol}}\\?\ \text{kJ} &=\dfrac{-7.54\ \text{kJ}\times1\ \cancel{\text{mol}}}{0.280\ \cancel{\text{mol}}}\\
?\ \text{kJ}&=-26.9\ \text{kJ}
\end{align}|
Thus, the molar heat of dissolution of lithium hydroxide |(\text{LiOH})| is |-26.9\ \text{kJ}.|
The molar heat of neutralisation |(\Delta H_n)| is the amount of energy that is absorbed or released during the neutralisation of one mole of acid or one mole of base.
A neutralisation reaction is accompanied by a transfer of heat resulting from the interaction between the reacting ions. Neutralisation reactions can be studied in the laboratory using a calorimeter. Since acidic and basic solutions are diluted, their density and specific heat capacity are assumed to be equivalent to that of water.
When the formula |Q=mc\Delta T| is used for neutralisation, the quantities of acidic and basic solutions must be added together to find the value of the mass to be included in the equation.
In a calorimeter, |100\ \text{mL}| of an aqueous solution of |\text{NaOH}| at |0.5\ \text{mol/L}| is completely neutralised by adding |100\ \text{mL}| of |\text{HCl}| at |0.5\ \text{mol/L}.| The initial temperature of the solutions before neutralisation is |22.5°\text{C}.| The highest temperature obtained during neutralisation (after mixing) is |25.9°\text{C}.| What is the molar heat of neutralisation of |\text{NaOH}|?
It is assumed that |100\ \text{mL}| of solution weighs |100\ \text{g}.|
First, we need to calculate the energy involved in neutralisation.
|\begin{align} Q&=mc \Delta T\\
Q&=(100\ \cancel{\text{g}} + 100\ \cancel{\text{g}})\times 4.19\ \text{J/}\cancel{\text{g}}\cdot \cancel{°\text{C}} \times (25.9\cancel{°\text{C}}-22.5\cancel{°\text{C}})\\
Q&=2\;849.2\ \text{J}
\end{align}|
Since the temperature of the water has risen, this means that neutralisation has released energy. So, |\Delta H = -2\; 849.2\ \text{J}.|
Next, we need to find the number of moles of |\text{NaOH}| neutralised using the concentration of the solution, which is |0.5\ \text{mol/L}.|
|\begin{align}C &= \dfrac{n}{V}\\
0.5\ \text{mol/}\cancel{\text{L}}&= \dfrac{n}{0.1\ \cancel{\text{L}}}\\
n&= 0.05\ \text{mol}
\end{align}|
Finally, we need to calculate the molar heat of neutralisation.
|\begin{align}\dfrac{-2\; 849.2\ \text{J}}{0.05\ \cancel{\text{mol}}}&= \dfrac{?\ \text{J}}{1\ \cancel{\text{mol}}}\\
?\ \text{J}&= \dfrac{-2\;849.2\ \text{J}\times 1\ \cancel{\text{mol}}}{0.05\ \cancel{\text{mol}}}\\
?\ \text{J}&\approx -56\; 984\ \text{J}\\
?\ \text{J}&\approx -57.0\ \text{kJ}
\end{align}|
Therefore, the molar heat of neutralization of |\text{NaOH}| is approximately |-57\ \text{kJ/mol}.|