Question

A sample of solid benzoic acid (C₆H₃OOOH)that weighs 0.800gis burned in an excess of oxygen to CO₂(g)and H₂O(l) in a constant-volume calorimeter at 25⁰C. The temperature increase is observed to be 2.15⁰C. The heat capacity of the calorimeter and its contents is known to be 9382JK^-1.

(a) Write and balance the equation for the combustion of benzoic acid.

(b) Calculate the standard change in internal energy ($\Delta U^{\circ })$for the combustion of 1.000molbenzoic acid to CO₂ (g)and H₂O (l)at 25⁰C.

(c) Calculate the standard enthalpy change $\left(\Delta H^{\circ }\right)$for the same reaction as in part (b).

(d) Calculate the standard enthalpy of formation per mole of benzoic acid, using data for the standard enthalpies of formation of CO₂ (g)and H₂O (l)from Appendix D.

Short answer

(a) The answer is $C_6{H}_{5}COOH\left(s\right)+15/2O_2\left(g\right)\to 7C{O}_2\left(g\right)+3H_{2}O.$

(b) The answer is $\Delta U=-3083.53kJ/mol$.

(c) The answer is $\Delta H=-3084.77kJ.$

(d) The answer is localid="1663685883018" $\Delta H_f(C_6{H}_{5}COOH,s)=-527.3kJ/mol.$

Step-by-step solution

Given data

The amount of benzoic acid $m\left(C_6{H}_{5}COOH\right)=0.800g$

$T={25}^{°}C$

$$\begin{gathered} \Delta T(Changeintemperature)is2.{15}^{°}C. \\ {C}_p(Volume-constantcalorimeter)=9832J{K}^{-1} \end{gathered}$$

The data obtained from Appendix D:

$$\begin{gathered} \Delta H_f\left(C{O}_2\right)=-393.51kJ/mol \\ \Delta H_f\left(H_{2}O\right)=-285.83"kJ/mol \end{gathered}$$

Concept of enthalpy

Enthalpy change is amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as the difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in the chemical reaction.

It is represented by $\Delta H$.

Balanced equation for the combustion of benzoic acid

(a)

$$\begin{gathered} Reactants:C_6{H}_{5}COOH\left(s\right)and{O}_2\left(g\right) \\ Products:C{O}_2\left(g\right)and{H}_{2}O\left(l\right) \end{gathered}$$

$C_6{H}_{5}COOH\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(l\right)$

Balance the carbon atoms:

As we can see we have 7 carbon atoms on the left side and only 1 on the right side.

$C_6{H}_{5}COOH\left(s\right)+O_2\left(g\right)\to 7C{O}_2\left(g\right)+H_{2}O\left(l\right)$

Balance the hydrogen atoms:

On the left side we have 6 hydrogen atoms and on the right we have only 2.

$C_6{H}_{5}COOH\left(s\right)+O_2\left(g\right)\to 7C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

Balance the oxygen atoms:

Now to balance out the left side of the reaction.

17 oxygen atoms on the right side, On the left side 2 oxygen atoms in the benzoic acid$=9832J{K}^{-1}$, these will not be change.

The final equation is,$C_6{H}_{5}COOH\left(s\right)+15/2O_2\left(g\right)\to 7C{O}_2\left(g\right)+3H_{2}O\left(l\right)$.

Calculate the number of moles

(b)

Calculate the number of moles in order to calculate heat.

Molar mass of benzoic acid is $M\left(C_6{H}_{5}COOH\right)=122.12g/mol$.

Amount of benzoic acid is $m\left(C_6{H}_{5}COOH\right)=0.800g$.

$$\begin{gathered} n\left(C_6{H}_{5}COOH\right)=\frac{m\left(C_6{H}_{5}COOH\right)}{M\left(C_6{H}_{5}COOH\right)} \\ =\frac{0.800}{122.12} \\ =6.55\times {10}^{-3}mol \end{gathered}$$

Calculate the heat

The formula for the heat is$,q=n{C}_p\Delta T$.

$$\begin{gathered} q=6.55\times {10}^{-3}\times 9382\times 2.15 \\ q=3083.53kJ/mol \end{gathered}$$

Since this is the heat absorbed by the calorimeter, the heat that is released from q=-3083.53kJ/mol the system is the same in value, but different in sign.

Calculate the internal energy change

The formula for an internal energy change is, $\Delta U=q+w$.

Since the process takes place in a volume-constant calorimeter, the work done will be equal to 0 (no change in volume=no work).

This means that the change in internal energy will be equal to the heat transferred, $\Delta U=q$.

Change in internal energy is, $\Delta U=q$.

So, $\Delta U=-3083.53kJ/mol$.

Calculate the value of Δng

(c)

The formula that includes the change in internal energy, $\Delta H=\Delta U+\Delta \left(pV\right).$

$\Delta \left(pV\right)$Can be calculated from the ideal gas equation, pV = nRT.

$\Delta \left(pV\right)=\Delta n_{g}RT$

Where $\Delta n_g$is the change in the number of gaseous reactants of products.

Determine $\Delta n_g$:

The only gaseous participants are O₂ and CO₂.

$$\begin{gathered} \Delta n_g=n_g\left(products\right)-n_g\left(reac\tan ts\right) \\ \Delta n_g=7-\frac{15}{2} \\ =-0.5mol \end{gathered}$$

Calculate the enthalpy

Calculate the enthalpy of the benzoic acid by the given data as:

$$\begin{gathered} \Delta U=-3083.53kJ/mol \\ \Delta n_g=-0.5mol \\ T=298K \\ \Delta H=\Delta U+RT\Delta n_g \end{gathered}$$

Calculate further as shown below.

$\begin{array}{l}\Delta H=-3083.53\times {10}^3+8.314x298\times \left(-0.5\right)\\ \Delta H=-3084.77kJ\end{array}$

Calculate the overall change in enthalpy

(d)

The data obtained from Appendix D.

$$\begin{gathered} \Delta H_f\left(C{O}_2\right)=-393.51kJ/mol \\ \Delta H_f\left(H_{2}O\right)=-285.83kJ/mol \end{gathered}$$

The overall change in enthalpy is calculated by, $\Delta H=\Delta H_f\left(products\right)-\Delta H_f\left(reactants\right).$

Given reaction: $C_6{H}_{5}COOH\left(s\right)+15/2O_2\left(g\right)\to 7C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

The overall enthalpy change equation would be, $\Delta H=7\times \Delta H_f(C{O}_2,g)+3\times \Delta H_f(H_{2}O,l)-\Delta H_f(C_6{H}_{5}COOH,s)-15/2\Delta H_f(O_2,g).$

The standard enthalpy of formation of oxygen (=0). The standard enthalpy of formation of benzoic acid would then be:

role="math" localid="1663685867619" $$\begin{gathered} \Delta H_f\left(C_6{H}_{5}COOH\right)=7\times \Delta H_f(C{O}_2,g)+3\times \Delta H_f(H_{2}O,l)-\Delta H \\ =7\times (-393.51)+3\times (-285.83)-(-3084.77) \\ =-527.3kJ/mol \end{gathered}$$