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Chapter 5: Problem 58

A 1.800 -g sample of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) was burned in a bomb calorimeter whose total heat capacity is \(11.66 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter plus contents increased from \(21.36^{\circ} \mathrm{C}\) to \(26.37{ }^{\circ} \mathrm{C}\). (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol? Per mole of phenol?

Short Answer

Expert verified
The balanced chemical equation for the combustion of phenol (C6H5OH) is: \(C_6H_5OH_{(l)} + 7.5O_2_{(g)} \rightarrow 6CO_2_{(g)} + 3H_2O_{(l)}\) The total heat released during the combustion process is \(58.29\, \mathrm{kJ}\). The heat of combustion per gram of phenol is \(32.39\, \mathrm{kJ\, g^{-1}}\). The heat of combustion per mole of phenol is \(3,050.26\, \mathrm{kJ\, mol^{-1}}\).

Step by step solution

01

(1. Write the balanced chemical equation)

The balanced chemical equation for the combustion of phenol (C6H5OH) is: \(C_6H_5OH_{(l)} + 7.5O_2_{(g)} \rightarrow 6CO_2_{(g)} + 3H_2O_{(l)}\)
02

(2. Calculate the total heat released)

To find the total heat released (q) during the combustion process, we'll use the formula: \(q = C_p \Delta T\), where \(C_p\) is the total heat capacity of the calorimeter, and \(\Delta T\) is the change in temperature. Given that \(C_p = 11.66 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), and the change in temperature is \(\Delta T = 26.37^{\circ} \mathrm{C} - 21.36^{\circ} \mathrm{C} = 5.01^{\circ} \mathrm{C}\). Now we can calculate q: \(q = (11.66\, \mathrm{kJ\,} { }^{\circ} \mathrm{C^{-1}})(5.01^{\circ} \mathrm{C}) \) \(q = 58.29\, \mathrm{kJ}\)
03

(3. Calculate the heat of combustion per gram of phenol)

We are given the mass of phenol as \(1.800\, \mathrm{g}\). To find the heat of combustion per gram, we'll divide the total heat released (q) by the mass of the phenol sample: Heat of combustion per gram = \(\frac{58.29\, \mathrm{kJ}}{1.800\, \mathrm{g}}\) Heat of combustion per gram = \(32.39\, \mathrm{kJ\, g^{-1}}\)
04

(4. Calculate the heat of combustion per mole of phenol)

To find the heat of combustion per mole of phenol, we'll first calculate the molar mass of phenol. The molar mass of phenol (C6H5OH): \((6 \times 12.01) + (6 \times 1.01) + (1 \times 16.00) = 94.11\, \mathrm{g}\, \mathrm{mol^{-1}}\) Now, we can calculate the heat of combustion per mole by dividing the total heat released (q) by the number of moles of phenol burned: Number of moles of phenol = \(\frac{1.800\, \mathrm{g}}{94.11\, \mathrm{g\, mol^{-1}}}\) Number of moles of phenol = \(0.0191\, \mathrm{mol}\) Heat of combustion per mole of phenol = \(\frac{58.29\, \mathrm{kJ}}{0.0191\, \mathrm{mol}}\) Heat of combustion per mole of phenol = \(3,050.26\, \mathrm{kJ\, mol^{-1}}\) So, the heat of combustion of phenol is \(32.39\, \mathrm{kJ\, g^{-1}}\) and \(3,050.26\, \mathrm{kJ\, mol^{-1}}\).

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Most popular questions from this chapter

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is currently blended with gasoline as an automobile fuel. (a) Write a balanced equation for the combustion of liquid ethanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced per liter of ethanol by combustion of ethanol under constant pressure. Ethanol has a density of \(0.789 \mathrm{~g} / \mathrm{mL}\). (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per \(\mathrm{kJ}\) of heat emitted.

Consider a system consisting of the following apparatus, in which gas is confined in one flask and there is a vacuum in the other flask. The flasks are separated by a valve. Assume that the flasks are perfectly insulated and will not allow the flow of heat into or out of the flasks to the surroundings. When the valve is opened, gas flows from the filled flask to the evacuated one. (a) Is work performed during the expansion of the gas? (b) Why or why not? (c) Can you determine the value of \(\Delta E\) for the process?

Consider the following unbalanced oxidation-reduction reactions in aqueous solution: $$ \begin{aligned} \mathrm{Ag}^{+}(a q)+\mathrm{Li}(s) & \longrightarrow \mathrm{Ag}(s)+\mathrm{Li}^{+}(a q) \\ \mathrm{Fe}(s)+\mathrm{Na}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Na}(s) \\ \mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{KOH}(a q)+\mathrm{H}_{2}(g) \end{aligned} $$ (a) Balance each of the reactions. (b) By using data in Appen\(\operatorname{dix} C,\) calculate \(\Delta H^{\circ}\) for each of the reactions. (c) Based on the values you obtain for \(\Delta H^{\circ}\), which of the reactions would you expect to be thermodynamically favored? (d) Use the activity series to predict which of these reactions should occur. (Section 4.4) Are these results in accord with your conclusion in part (c) of this problem?

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ}\) For this reaction, calculate \(\Delta H\) for the formation of (a) 1.36 mol of \(\mathrm{O}_{2}\) and (b) \(10.4 \mathrm{~g}\) of \(\mathrm{KCl}\) (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2},\) is likely to be feasible under ordinary conditions? Explain your answer.

Consider the decomposition of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) to gaseous acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g):\) $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \quad \Delta H=+630 \mathrm{~kJ} $$ (a) What is the enthalpy change for the reverse reaction? (b) What is \(\Delta H\) for the formation of 1 mol of acetylene? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) were consumed instead of \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.
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