 fanyattehedzg

2022-01-11

The enthalpy of combustion of benzoic acid $\left({C}_{6}{H}_{5}COOH\right)$ is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be -3226.7 kJ/mol. When 1.9862 g of benzoic acid are burned in a calorimeter, the temperature rises from $21.84$ to ${25.67}^{\circ }C$. What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly 2000 g.) ambarakaq8

Step 1
Write a balanced thermochemical equation for the combustion of 1 mol of $\left({C}_{6}{H}_{5}COOH\right)$
$\begin{array}{rlr}{C}_{6}{H}_{5}COOH\left(aq\right)+\frac{1}{2}\phantom{\rule{0.278em}{0ex}}{O}_{2}\left(g\right)& ⇒3{H}_{2}O\left(l\right)+7C{O}_{2}\left(g\right)& \mathrm{\Delta }{H}_{rxn}{}^{\circ }=-3226.7kJ/mol\end{array}$
Step 2
Calculate the molar mass of $\left({C}_{6}{H}_{5}COOH\right)$.
$7\left(12.0107g/mol\right)+6\left(1.0079g/mol\right)+2\left(15.9994g/mol\right)=122.1211g/mo$
Step 3
Calculate the heat of the reaction in joules:
1. Begin with the mass of the $\left({C}_{6}{H}_{5}COOH\right)$.
2. Use the molar mass of $\left({C}_{6}{H}_{5}COOH\right)$ as a conversion factor to convert grams of $\left({C}_{6}{H}_{5}COOH\right)$ to moles of $\left({C}_{6}{H}_{5}COOH\right)$.
3. Use a thermostoichiometric ratio to convert moles of $\left({C}_{6}{H}_{5}COOH\right)$ to kilojoules.
4. Use a conversion factor to convert kilojoules to joules.
$\begin{array}{r}{q}_{rxn}=\begin{array}{cccc}1.9862g{C}_{6}{H}_{5}COOlH& 1mol{C}_{6}{H}_{5}COOH& -3226.7kJ& 1000J\\ & 122.1211\phantom{\rule{0.278em}{0ex}}g{C}_{6}{H}_{5}COOH& 1mol{C}_{6}{H}_{5}COOH& 1kJ\end{array}\\ & =-52480.J\end{array}$
Step 4
Since heat transfers from the reaction to the calorimeter and water, then $\left({q}_{w}+{q}_{cal}\right)$ must be equal in magnitude, but opposite in direction to ${q}_{rxn}$. Expand both sides of the equation to include variables such as mass, temperature change, and specific heat capacity. Then solve for ${C}_{cal}$.
${q}_{rxn}=-\left({q}_{w}+{q}_{cal}\right)$
${q}_{rxn}=-{q}_{w}-{q}_{cal}$
${q}_{rxn}=-{m}_{w}\left(\mathrm{\Delta }T\right)\left({s}_{w}\right)-{C}_{cal}\left(\mathrm{\Delta }T\right)$
${q}_{rxn}+{m}_{w}\left(\mathrm{\Delta }T\right)\left({s}_{w}\right)=-{C}_{cal}\left(\mathrm{\Delta }T\right)$
$⇒$ Don Sumner

Step 1:
To solve the problem, we can use the formula for heat capacity:
$C=\frac{q}{\Delta T}$
where $C$ is the heat capacity of the bomb, $q$ is the heat released by the combustion of benzoic acid, and $\Delta T$ is the change in temperature.
First, we need to calculate the heat released by the combustion of benzoic acid. The enthalpy of combustion of benzoic acid is given as -3226.7 kJ/mol. We can convert the mass of benzoic acid burned to moles using its molar mass.
The molar mass of benzoic acid (${C}_{6}{H}_{5}COOH$) is calculated as follows:
$C=12.01$ g/mol
$H=1.008$ g/mol
$O=16.00$ g/mol
Molar mass of benzoic acid = $\left(6×C\right)+\left(5×H\right)+O+O+H=122.12$ g/mol
Step 2:
Now, we can calculate the number of moles of benzoic acid burned:

Next, we calculate the heat released:
$q=n×\Delta H$
where $\Delta H$ is the enthalpy of combustion.
$q=\left(\frac{1.9862\phantom{\rule{0.167em}{0ex}}\text{g}}{122.12\phantom{\rule{0.167em}{0ex}}\text{g/mol}}\right)×-3226.7\phantom{\rule{0.167em}{0ex}}\text{kJ/mol}$
Step 3:
Next, we need to calculate the change in temperature, $\Delta T$, which is given as $25.67-21.84=3.83$ degrees Celsius.
Finally, we can calculate the heat capacity of the bomb:
$C=\frac{q}{\Delta T}$
$C=\frac{-\left(\frac{1.9862\phantom{\rule{0.167em}{0ex}}\text{g}}{122.12\phantom{\rule{0.167em}{0ex}}\text{g/mol}}\right)×3226.7\phantom{\rule{0.167em}{0ex}}\text{kJ/mol}}{3.83\phantom{\rule{0.167em}{0ex}}\text{K}}$
Simplifying the expression above gives the value of $C$ in kJ/K. nick1337

We know that the heat released by the combustion of benzoic acid is equal to the heat absorbed by the bomb and the water:
${q}_{\text{bomb}}+{q}_{\text{water}}=0$
The heat absorbed by the bomb can be calculated using the formula:
${q}_{\text{bomb}}={C}_{\text{bomb}}·\Delta T$
where ΔT is the change in temperature of the bomb.
The heat absorbed by the water can be calculated using the formula:
${q}_{\text{water}}={m}_{\text{water}}·{C}_{\text{water}}·\Delta T$
where m₋w is the mass of water and C₋w is the specific heat capacity of water.
Since the heat released by the combustion of benzoic acid is equal to the heat absorbed by the bomb and the water, we have:
${C}_{\text{bomb}}·\Delta T+{m}_{\text{water}}·{C}_{\text{water}}·\Delta T=0$
Substituting the given values:
${C}_{\text{bomb}}·\left(25.67-21.84\right)+2000·{C}_{\text{water}}·\left(25.67-21.84\right)=0$
Simplifying the equation:
${C}_{\text{bomb}}·3.83+2000·{C}_{\text{water}}·3.83=0$
Since we know the heat released by the combustion of benzoic acid is -3226.7 kJ/mol, we can calculate the number of moles of benzoic acid burned:

Substituting the given values:
$n=\frac{1.9862\phantom{\rule{0.167em}{0ex}}\text{g}}{122.12\phantom{\rule{0.167em}{0ex}}\text{g/mol}}$
Now we can calculate the heat capacity of the bomb:

Substituting the known values:
${C}_{\text{bomb}}=\frac{-3226.7\phantom{\rule{0.167em}{0ex}}\text{kJ/mol}}{n·\left(25.67-21.84\right)}$
Finally, we can substitute the calculated value of Cᵦ into the first equation to solve for Cw:
${C}_{\text{bomb}}·3.83+2000·{C}_{\text{water}}·3.83=0$
${C}_{\text{water}}=-\frac{{C}_{\text{bomb}}}{2000}$ RizerMix

Solution:
$q=m·C·\Delta T$
where:
- $q$ is the heat transfer
- $m$ is the mass
- $C$ is the heat capacity
- $\Delta T$ is the change in temperature
First, we need to calculate the heat transfer ($q$). We can use the enthalpy of combustion of benzoic acid to determine the heat released during the combustion reaction.
Given that the enthalpy of combustion of benzoic acid is -3226.7 kJ/mol, we can calculate the number of moles of benzoic acid burned:

The molar mass of benzoic acid (${C}_{6}{H}_{5}COOH$) can be calculated as:

The molar masses of carbon, hydrogen, and oxygen are:

Now, we can calculate the moles of benzoic acid burned:

Next, we can calculate the heat transfer ($q$) using the equation:

To calculate the heat capacity ($C$) of the bomb calorimeter, we need to rearrange the equation:
$C=\frac{q}{\Delta T·m}$
Given that the change in temperature ($\Delta T$) is 25.67°C - 21.84°C and the mass ($m$) of the water surrounding the bomb is 2000 g, we can substitute the values into the equation:
$C=\frac{q}{\Delta T·m}$
Finally, we can calculate the heat capacity ($C$) of the bomb calorimeter.

Do you have a similar question?