Problem 102 - Chemistry Matter and Change

Question

Is the following reaction to convert copper(II) sulfide to copper(II) sulfate spontaneous under standard conditions? \(\mathrm{CuS}(\mathrm{s})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CuSO}_{4}(\mathrm{s}) . \Delta H_{\mathrm{rxn}}^{\circ}=-718.3\) \(\mathrm{k} \mathrm{J},\) and \(\Delta S_{\mathrm{rxn}}^{\circ}=-368 \mathrm{J} / \mathrm{K} .\) Explain.

Step-by-Step Solution

Answer

The conversion of copper(II) sulfide to copper(II) sulfate is spontaneous under standard conditions since the calculated Gibbs Free Energy change, \(\Delta G = -608,636\,\text{J}\), is negative.

Step 1: Write down the Gibbs Free Energy equation

The Gibbs Free Energy equation is given by: \[ \Delta G = \Delta H - T\Delta S \] where ∆G is the Gibbs Free Energy change, ∆H is the enthalpy change, T is the temperature in Kelvin, and ∆S is the entropy change.

Step 2: Convert the given values to the same unit of measurement

The enthalpy change is given in kJ, while the entropy change is given in J/K. To keep the units consistent, convert the enthalpy change to J by multiplying it by 1000: \[ \Delta H = -718.3\,\text{kJ} \times 1000 = -718,300\,\text{J} \]

Step 3: Calculate Gibbs Free Energy change at standard conditions

Standard conditions mean that the reaction is taking place at 298K (25°C). Plug the converted value of \(\Delta H\) and the given value of \(\Delta S\) into the Gibbs Free Energy equation, using 298K for T: \[ \Delta G = \Delta H - T\Delta S = -718,300\,\text{J} - (298\,\text{K} \times -368\,\text{J/K}) \] Now, do the calculations: \[ \Delta G = -718,300\,\text{J} + 109,664\,\text{J} = -608,636\,\text{J} \]

Step 4: Determine whether the reaction is spontaneous under standard conditions

A negative value of \(\Delta G\) indicates that the reaction is spontaneous under standard conditions. In this case, \(\Delta G = -608,636\,\text{J}\), which is a negative value. Therefore, the reaction is spontaneous under standard conditions.

Key Concepts

Enthalpy Change

In a chemical reaction, enthalpy change (\( \Delta H \)) is an important factor. It measures the heat change in the system. If \( \Delta H \) is negative, the reaction releases heat, making it exothermic. Comparatively, if \( \Delta H \) is positive, heat is absorbed, making it endothermic.

In the example of converting copper(II) sulfide to copper(II) sulfate, the enthalpy change \( \Delta H \) is -718.3 kJ. Multiplied by 1000, to convert into Joules, you get -718,300 J. This signifies the reaction releases this amount of energy, indicating it as exothermic.

Exothermic reactions tend to be more spontaneous since they increase the surrounding temperature, influencing the system's spontaneous nature. However, to fully understand spontaneity, consider the entropy change as well.

Entropy Change

Entropy change (\( \Delta S \)) describes the change in disorder or randomness in a system during a reaction. A positive entropy indicates more disorder and is generally seen as favorable. Conversely, a negative entropy value reflects an increase in order, considered less favorable for spontaneity.

In the given reaction, \( \Delta S \) is -368 J/K. This negative value indicates the system becomes more ordered from reactants to products.

Although negative entropy can be unfavorable for spontaneity, entropy isn't the sole factor determination. It plays a crucial part when paired with temperature and enthalpy. With the Gibbs Free Energy equation, considering T and \( \Delta S \) helps evaluate the overall reaction spontaneity.

Spontaneous Reaction

A reaction is considered spontaneous if it occurs naturally under given conditions, without requiring energy input beyond the initial. Gibbs Free Energy (\( \Delta G \)) is the definitive factor for assessing spontaneity, utilizing the equation: \[ \Delta G = \Delta H - T\Delta S \] A negative \( \Delta G \) suggests the reaction is spontaneous.

For our reaction, the calculation of \( \Delta G \) gives it as -608,636 J, which is negative. It confirms spontaneity under standard conditions. Though the entropy change is unfavorable, the significantly negative enthalpy compensates, driving spontaneity forward.

In essence, a spontaneous reaction is often driven by combining favorable enthalpy and entropy factors, as shown in the Gibbs Free Energy framework.