Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g),\) is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) How much heat is produced in burning 1 \(\mathrm{mol}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) under standard conditions if both reactants and products are brought to 298 \(\mathrm{K?}\) (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?

Short Answer

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The balanced chemical equation for the combustion of acetylene gas (\(\mathrm{C}_{2} \mathrm{H}_{2}(g)\)) is: \[ \mathrm{C}_{2} \mathrm{H}_{2}(g) + \frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \] The heat produced when burning 1 mol of acetylene gas under standard conditions is -1290.1 kJ/mol. The maximum amount of useful work that can be accomplished under standard conditions by the combustion of acetylene gas is -1235.0 kJ/mol.

Step by step solution

01

Write the balanced chemical equation for the combustion of acetylene gas

We need to create a balanced chemical equation for the combustion of acetylene gas. Acetylene gas is given as \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\). Upon combustion, it produces carbon dioxide (\(\mathrm{CO}_{2}(g)\)) and water (\(\mathrm{H}_{2} \mathrm{O}(l)\)). The balanced chemical equation will be: \[ \mathrm{C}_{2} \mathrm{H}_{2}(g) + \frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \]
02

Calculate the heat produced in burning 1 mol of acetylene gas

Under standard conditions, we can use the enthalpy change of the reaction (\(\Delta H^\circ\)) to calculate the heat produced when 1 mol of acetylene gas is burned. First, we need to determine the enthalpy changes of formation (\(\Delta H_f^\circ\)) of the reactants and products. The standard enthalpy change of formation for acetylene gas \(\mathrm{C}_{2} \mathrm{H}_{2}\) is +226.7 kJ/mol, for oxygen gas \(\mathrm{O}_{2}\) is 0 kJ/mol, for carbon dioxide gas \(\mathrm{CO}_{2}\) is -393.5 kJ/mol, and for liquid water \(\mathrm{H}_{2} \mathrm{O}\) is -285.8 kJ/mol. To calculate the enthalpy change of the reaction (\(\Delta H^\circ\)), we can use the formula: \[ \Delta H^\circ = [\Sigma n \Delta H_f^\circ(\mathrm{products})] - [\Sigma n \Delta H_f^\circ(\mathrm{reactants})] \] Where \(n\) represents the stoichiometric coefficients and \(\Delta H_f^\circ\) represents the standard enthalpy changes of formation. Using the stoichiometric coefficients and the given enthalpy changes of formation, we get: \[ \Delta H^\circ = [(2(-393.5 \mathrm{kJ/mol}) + 1(-285.8 \mathrm{kJ/mol})] - [(1(+226.7 \mathrm{kJ/mol}) + \frac{5}{2}(0 \mathrm{kJ/mol})] \] \[ \Delta H^\circ = -1290.1 \mathrm{kJ/mol\, of\, C_{2}H_{2}} \] So, the amount of heat produced in burning 1 mol of acetylene gas under standard conditions is -1290.1 kJ/mol.
03

Calculate the maximum amount of useful work under standard conditions

Under standard conditions, the maximum amount of useful work that can be accomplished by a reaction is given by the change in Gibbs free energy (\(\Delta G^\circ\)). We can find the change in Gibbs free energy using the equation: \[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \] Where \(T\) is the temperature in Kelvin and \(\Delta S^\circ\) is the entropy change of the reaction. In this problem, both reactants and products are at 298 K. The standard entropy values of \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), \(\mathrm{O}_{2}(g)\), \(\mathrm{CO}_{2}(g)\), and \(\mathrm{H}_{2} \mathrm{O}(l)\) are 200.9, 205.0, 213.6, and 69.9 J/mol*K, respectively. To calculate the entropy change of the reaction, we can use a similar formula as in step 2: \[ \Delta S^\circ = [\Sigma n S^\circ(\mathrm{products})] - [\Sigma n S^\circ(\mathrm{reactants})] \] Using the entropy values and stoichiometric coefficients, we get: \[ \Delta S^\circ = [(2(213.6) + 69.9) - (200.9 + \frac{5}{2}(205.0))] \, \mathrm{J/mol\,*\, K} \] \[ \Delta S^\circ = -188.1 \, \mathrm{J/mol\, *\, K} \] Now we can calculate the change in Gibbs free energy: \[ \Delta G^\circ = -1290.1\, \mathrm{kJ/mol} - 298\, \mathrm{K} \times (-188.1 \,\mathrm{J/mol\, *\, K}) \] \[ \Delta G^\circ = -1290.1\, \mathrm{kJ/mol} - (-55.1\, \mathrm{kJ/mol}) \] \[ \Delta G^\circ = -1235.0\, \mathrm{kJ/mol\, of\, C_{2}H_{2}} \] So, the maximum amount of useful work that can be accomplished under standard conditions by the combustion of acetylene gas is -1235.0 kJ/mol.

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Most popular questions from this chapter

The oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in body tissue produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2} .\) (a) Using data given in Appendix \(\mathrm{C},\) compare the equilibrium constants for the following reactions: $$ \begin{array}{c}{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)} \\ {\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g)}\end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

Predict which member of each of the following pairs has the greater standard entropy at \(25^{\circ} \mathrm{C} :(\mathbf{a}) \operatorname{Sc}(s)\) or \(\operatorname{Sc}(g)\) (b) \(\mathrm{NH}_{3}(g)\) or \(\mathrm{NH}_{3}(a q),(\mathbf{c}) \mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g),(\mathbf{d}) \mathrm{C}(\mathrm{graphite})\) or \(\mathrm{C}(\) diamond). Use Appendix \(\mathrm{C}\) to find the standard entropy of each substance.

As shown here, one type of computer keyboard cleaner contains liquefied \(1,1\) -difluoroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{F}_{2}\right),\) which is a gas at atmospheric pressure. When the nozzle is squeezed, the \(1,1\) -difluoroethane vaporizes out of the nozzle at high pressure, blowing dust out of objects. (a) Based on your experience, is the vaporization a spontaneous process at room temperature? (b) Defining the \(1,1\) -difluoroethane as the system, do you expect \(q_{s y s}\) for the process to be positive or negative? (c) Predict whether \DeltaS is positive or negative for this process. (d) Given your answers to (a), (b), and (c), do you think the operation of this product depends more on enthalpy or entropy? [Sections 19.1 and 19.2\(]\)

(a) What sign for \(\Delta S\) do you expect when the pressure on 0.600 mol of an ideal gas at 350 \(\mathrm{K}\) is increased isothermally from an initial pressure of 0.750 atm? (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

Indicate whether each of the following statements is true or false. If it is false, correct it. (a) The feasibility of manufacturing \(\mathrm{NH}_{3}\) from \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) depends entirely on the value of \(\Delta H\) for the process \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) The reaction of \(\mathrm{Na}(s)\) with \(\mathrm{Cl}_{2}(g)\) to form \(\mathrm{NaCl}(s)\) is a spontaneous process.(c) A spontaneous process can in principle be conducted reversibly. (d) Spontaneous processes in general require that work be done to force them to proceed. (e) Spontaneous processes are those that are exothermic and that lead to a higher degree of order in the system.

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