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 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 is: \[C_6H_5OH + \frac{9}{2}O_2 \rightarrow 6CO_2 + 3H_2O\]. The heat of combustion per gram of phenol is 32.43 kJ/g, and the heat of combustion per mole of phenol is 3055.17 kJ/mol.

Step by step solution

01

Write the balanced chemical equation for the combustion of phenol.

The combustion of phenol (C6H5OH) involves reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this combustion is: \[C_6H_5OH + \frac{9}{2}O_2 \rightarrow 6CO_2 + 3H_2O\]
02

Calculate the heat released.

The total heat capacity of the calorimeter is given as 11.66 kJ/°C. The temperature of the calorimeter plus its contents increased from 21.36°C to 26.37°C. So, the temperature change is: ΔT = 26.37°C - 21.36°C = 5.01°C The heat released (q) by the combustion can be calculated using the formula: q = heat capacity × temperature change q = 11.66 kJ/°C × 5.01°C q = 58.37 kJ
03

Calculate the heat of combustion per gram of phenol.

The mass of the phenol sample is given as 1.800 g. The heat of combustion per gram can be calculated as: Heat of combustion per gram = (Heat released) / (Mass of sample) = 58.37 kJ / 1.800 g = 32.43 kJ/g
04

Calculate the heat of combustion per mole of phenol.

To find the heat of combustion per mole, we first calculate the molar mass of phenol. The molar mass of phenol (C6H5OH) is: (6 × 12.01 g/mol) + (5 × 1.01 g/mol) + 16.00 g/mol + 1.01 g/mol = 94.11 g/mol Now, we can calculate the number of moles of phenol in the sample: number of moles = (mass of sample) / (molar mass) = 1.800 g / 94.11 g/mol = 0.01911 mol Finally, we can calculate the heat of combustion per mole as: Heat of combustion per mole = (Heat released) / (number of moles) = 58.37 kJ / 0.01911 mol = 3055.17 kJ/mol So, the heat of combustion of phenol is 32.43 kJ/g and 3055.17 kJ/mol.

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Most popular questions from this chapter

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{kJ}\) and \(w=-840 \mathrm{J} .\) (b) A system releases 66.1 \(\mathrm{kJ}\) of heat to its surroundings while the surroundings do 44.0 \(\mathrm{kJ}\) of work on the system.

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.

Indicate which of the following is independent of the path by which a change occurs: (a) the change in potential energy when a book is transferred from table to shelf, (b) the heat evolved when a cube of sugar is oxidized to \(\operatorname{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),(\mathbf{c})\) the work accomplished in burning a gallon of gasoline.

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2} .\) From the following enthalpy of reaction data and data in Appendix \(\mathrm{C},\) calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s) :\) $$\begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \Delta H^{\circ}=-127.2 \mathrm{kJ} \end{aligned}$$

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) 1 \(\mathrm{mol} \mathrm{CO}_{2}(s)\) or 1 \(\mathrm{mol} \mathrm{CO}_{2}(g)\) at the same temperature, ( b) 2 \(\mathrm{mol}\) of hydrogen atoms or 1 \(\mathrm{mol}\) of \(\mathrm{H}_{2},(\mathbf{c}) 1 \mathrm{mol} \mathrm{H}_{2}(g)\) and 0.5 \(\mathrm{mol} \mathrm{O}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or 1 \(\mathrm{mol} \mathrm{H}_{2} \mathrm{O}(g)\) at \(25^{\circ} \mathrm{C},(\mathbf{d}) 1 \mathrm{mol} \mathrm{N}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or 1 \(\mathrm{mol} \mathrm{N}_{2}(g)\) at \(300^{\circ} \mathrm{C}\) .

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