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The value of \(\Delta G^{\circ}\) for the reaction $$ 2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(l) $$ is \(-5490 . \mathrm{kJ} .\) Use this value and data from Appendix 4 to calculate the standard free energy of formation for \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\).

Short Answer

Expert verified
The standard free energy of formation for butane gas (C4H10(g)) is -481.4 kJ/mol.

Step by step solution

01

Write down the reaction and known information

Write down the reaction: $$ 2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(l) $$ We are given: $$ \Delta G^{\circ}_{reaction}=-5490 \mathrm{kJ} $$ We need to find the \(\Delta G^{\circ}_{C_{4}H_{10}(g)}\).
02

Write down the standard free energy change formula and the known values

Write down the standard free energy change formula: $$\Delta G^{\circ}_{reaction} = \sum n \Delta G^{\circ}_{products} - \sum n \Delta G^{\circ}_{reactants}$$ Using the values from Appendix 4, we have: $$ \Delta G^{\circ}_{O_2(g)} = 0 \space \mathrm{kJ \space mol^{-1}} $$ $$ \Delta G^{\circ}_{CO_2(g)} = -394.4 \space \mathrm{kJ \space mol^{-1}} $$ $$ \Delta G^{\circ}_{H_2O(l)} = -237.2 \space \mathrm{kJ \space mol^{-1}} $$
03

Substitute the values into the formula and solve for \(\Delta G^{\circ}_{C_{4}H_{10}(g)}\)

Substitute the known values into the standard free energy change formula: $$ -5490 = \left[ 8 \times (-394.4) + 10 \times (-237.2) \right] - \left[ 2 \times \Delta G^{\circ}_{C_{4}H_{10}(g)} + 0 \right] $$ Solve for \(\Delta G^{\circ}_{C_{4}H_{10}(g)}\): $$ \Delta G^{\circ}_{C_{4}H_{10}(g)} = \frac{-5490 - 8 \times (-394.4) - 10 \times (-237.2)}{2} $$ $$ \Delta G^{\circ}_{C_{4}H_{10}(g)} = \frac{-5490 + 3155.2 + 2372}{2} $$ $$ \Delta G^{\circ}_{C_{4}H_{10}(g)} = -481.4 \mathrm{kJ \space mol^{-1}} $$
04

State the answer

The standard free energy of formation for butane gas (C4H10(g)) is -481.4 kJ/mol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamics: Fundamental Concepts and Laws
Thermodynamics is a branch of physics that deals with the relationships between heat, energy, and work. This field is hugely important in understanding
  • how energy is exchanged in chemical reactions,
  • how systems behave under different thermal conditions, and
  • how engines and similar systems operate efficiently.
One vital part of thermodynamics is understanding how energy transitions relate to temperatures and physical or chemical changes. Thermodynamics centers around four key laws:
- The Zeroth Law, which touches on thermal equilibrium
- The First Law, known as the law of energy conservation
- The Second Law, which introduces the concept of entropy
- The Third Law, which deals with the absolute zero of temperature
For students, realizing these laws' implications means appreciating why certain chemical reactions happen spontaneously and others do not. All reactions and physical processes abide by these principles, guiding how energy changes occur in everyday and industrial processes.
Understanding Gibbs Free Energy
Gibbs free energy, denoted as \(\Delta G\), is a thermodynamic quantity that helps predict whether a chemical reaction will occur spontaneously. It is used extensively in chemistry to determine reaction feasibility.
The equation for calculating Gibbs free energy is:
  • \( \Delta G = \Delta H - T\Delta S \)
where \( \Delta H \) is the change in enthalpy (heat absorbed or released), \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy (measure of disorder). A negative \( \Delta G \) means the process is spontaneous under constant pressure and temperature.
In the context of calculating reaction energies, \( \Delta G^{\circ}_{reaction} \) is used when conditions are standard—meaning
  • 1 atm of pressure,
  • pure substances, and
  • the temperature is typically 298.15 K (25°C).
Gibbs free energy links the concepts of enthalpy (total energy) and entropy (degree of randomness), helping interpret physical and chemical processes' driving forces.
Chemical Reactions Energy Calculations
Energy calculations for chemical reactions help chemists predict how energy will change during the process. One common approach in these calculations is using the standard free energy change formula:

\( \Delta G^{\circ}_{reaction} = \sum n \Delta G^{\circ}_{products} - \sum n \Delta G^{\circ}_{reactants} \)

This formula allows for estimating the energy efficiency and feasibility of reactions. When solving problems like calculating the standard free energy of formation for a compound like butane, the formula takes into account the
  • number of moles (n) of each reactant and product,
  • their respective standard free energy of formation values.
For reactants like \( \mathrm{O}_2 \), their \( \Delta G^{\circ} \) is 0 because they are elemental and naturally stable. This simplifies some calculations.
In practice, understanding how to apply this formula requires good algebra skills and knowledge of chemical thermodynamics. Being attentive to units and ensuring all values are within the same system is crucial to achieving accurate results.

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Most popular questions from this chapter

For ammonia \(\left(\mathrm{NH}_{3}\right)\), the enthalpy of fusion is \(5.65 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of fusion is \(28.9 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). a. Will \(\mathrm{NH}_{3}(s)\) spontaneously melt at \(200 . \mathrm{K}\) ? b. What is the approximate melting point of ammonia?

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ}\), for a reaction at \(25^{\circ} \mathrm{C}\). How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C} ?\) What assumptions are made?

The equilibrium constant for a certain reaction decreases from \(8.84\) to \(3.25 \times 10^{-2}\) when the temperature increases from \(25^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C}\). Estimate the temperature where \(K=1.00\) for this reaction. Estimate the value of \(\Delta S^{\circ}\) for this reaction. (Hint: Manipulate the equation in Exercise \(79 .\) )

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {surr }}\) ? \(\Delta S\) ? \(\Delta S_{\text {univ }} ?\) Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

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