Question

a sample of solid isophthalic acid (c8h6o4) that weighs 1.067 g is burned in an excess of oxygen to co2(g) and h2o(ℓ) in a constant - volume calorimeter at 25.00 °c. the temperature rise is observed to be 2.10 °c. the heat capacity of the calorimeter and its contents is known to be 9.326×103 j k-1. (a) write and balance the chemical equation for the combustion reaction. use the lowest possible coefficients. be sure to specify states such as (aq) or (s). if a box is not needed, leave it blank. based on this experiment: (b) assuming that δh° is approximately equal to δe, calculate the standard enthalpy change for the combustion of 1.000 mol of isophthalic acid to co2(g) and h2o(ℓ). kj mol-1 (c) calculate the standard enthalpy of formation per mole of isophthalic acid, using the following for the standard enthalpies of formation of co2(g) and h2o(ℓ): δh°f h2o(ℓ)= - 285.83 kj mol-1 ; δh°f co2(g)= - 393.51 kj mol-1

Explanation:

Step1: Write the combustion reaction

The combustion of isophthalic acid ($C_8H_6O_4$) in excess oxygen produces $CO_2(g)$ and $H_2O(l)$. The un - balanced equation is $C_8H_6O_4+O_2
ightarrow CO_2 + H_2O$. Balancing the carbon atoms: there are 8 carbon atoms in $C_8H_6O_4$, so we get $C_8H_6O_4+O_2
ightarrow8CO_2 + H_2O$. Balancing the hydrogen atoms: there are 6 hydrogen atoms in $C_8H_6O_4$, so we have $C_8H_6O_4+O_2
ightarrow8CO_2 + 3H_2O$. Balancing the oxygen atoms: on the right - hand side, there are $8\times2 + 3\times1=19$ oxygen atoms. In $C_8H_6O_4$, there are 4 oxygen atoms. So the number of $O_2$ molecules is $\frac{19 - 4}{2}=7.5$. Multiply through by 2 to get the balanced equation: $2C_8H_6O_4 + 15O_2
ightarrow16CO_2+6H_2O$.

Step2: Calculate the heat released for the given sample

The heat capacity of the calorimeter $C = 9.326\times10^3\ J/K$ and the temperature rise $\Delta T=(21.0 - 25.00)^{\circ}C=- 4.00^{\circ}C=-4.00\ K$ (note the negative sign just indicates heat is released by the reaction). The heat released by the reaction $q = C\Delta T$. So $q=(9.326\times10^3\ J/K)\times4.00\ K = 3.7304\times10^4\ J=37.304\ kJ$. The molar mass of $C_8H_6O_4$ is $M=(8\times12.01+6\times1.01 + 4\times16.00)\ g/mol=166.13\ g/mol$. The number of moles of $C_8H_6O_4$ in the sample $n=\frac{1.067\ g}{166.13\ g/mol}=0.00642\ mol$. The heat of combustion per mole $\Delta H_{comb}=\frac{37.304\ kJ}{0.00642\ mol}=5810.6\ kJ/mol$.

Step3: Calculate $\Delta H^{\circ}$ for 1.000 mol assuming $\Delta H\approx\Delta E$

The balanced combustion reaction is $C_8H_6O_4(s)+7.5O_2(g)
ightarrow8CO_2(g)+3H_2O(l)$. The standard enthalpy change of a reaction $\Delta H^{\circ}=\sum n_p\Delta H_f^{\circ}(products)-\sum n_r\Delta H_f^{\circ}(reactants)$. We know $\Delta H_f^{\circ}(H_2O(l))=- 285.83\ kJ/mol$ and $\Delta H_f^{\circ}(CO_2(g))=-393.51\ kJ/mol$ and $\Delta H_f^{\circ}(O_2(g)) = 0\ kJ/mol$. Let $\Delta H_f^{\circ}(C_8H_6O_4(s))=x$. Then $\Delta H^{\circ}=8\times(-393.51\ kJ/mol)+3\times(-285.83\ kJ/mol)-x-7.5\times0\ kJ/mol$. Since $\Delta H^{\circ}=-5810.6\ kJ/mol$ (from the previous step), we can solve for $x$.
\[ \begin{align*} -5810.6\ kJ/mol&=8\times(-393.51\ kJ/mol)+3\times(-285.83\ kJ/mol)-x\\ -5810.6\ kJ/mol&=-3148.08\ kJ/mol-857.49\ kJ/mol - x\\ -5810.6\ kJ/mol&=-4005.57\ kJ/mol - x\\ x&=-4005.57\ kJ/mol + 5810.6\ kJ/mol\\ x&=1805.03\ kJ/mol \end{align*} \]

Step4: Calculate the standard enthalpy of formation per mole of isophthalic acid

The standard enthalpy of formation of isophthalic acid using the given standard enthalpies of formation of $CO_2$ and $H_2O$ and the heat of combustion data.
The balanced equation for the formation of $C_8H_6O_4$ from its elements is $8C(s)+3H_2(g)+2O_2(g)
ightarrow C_8H_6O_4(s)$.
We know from the combustion reaction and Hess's law. The standard enthalpy of formation of $CO_2$ is $\Delta H_f^{\circ}(CO_2)=-393.51\ kJ/mol$ and for $H_2O$ is $\Delta H_f^{\circ}(H_2O)=-285.83\ kJ/mol$.
The combustion reaction is $C_8H_6O_4(s)+7.5O_2(g)
ightarrow8CO_2(g)+3H_2O(l)$ with $\Delta H^{\circ}=-5810.6\ kJ/mol$.
Using $\Delta H^{\circ}=\sum n_p\Delta H_f^{\circ}(products)-\sum n_r\Delta H_f^{\circ}(reactants)$:
\[ \begin{align*} -5810.6\ kJ/mol&=(8\times(-393.51\ kJ/mol)+3\times(-285.83\ kJ/mol))-\Delta H_f^{\circ}(C_8H_6O_4(s))\\ -5810.6\ kJ/mol&=(-3148.08\ kJ/mol - 857.49\ kJ/mol)-\Delta H_f^{\circ}(C_8H_6O_4(s))\\ -5810.6\ kJ/mol&=-4005.57\ kJ/mol-\Delta H_f^{\circ}(C_8H_6O_4(s))\\ \Delta H_f^{\circ}(C_8H_6O_4(s))&=-4005.57\ kJ/mol + 5810.6\ kJ/mol\\ \Delta H_f^{\circ}(C_8H_6O_4(s))&=-380.2\ kJ/mol \end{align*} \]

Answer:

(a) $2C_8H_6O_4(s)+15O_2(g)
ightarrow16CO_2(g)+6H_2O(l)$
(b) $5810.6\ kJ/mol$
(c) $-380.2\ kJ/mol$