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The enthalpy and entropy change for a chemical reaction are \(-2.5 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\) and \(7.4 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respec- tively. The reaction at \(298 \mathrm{~K}\). is (a) non-spontaneous (b) spontaneous (c) reversible (d) irreversible

Short Answer

Expert verified
The reaction is spontaneous (b).

Step by step solution

01

Understanding the Gibbs Free Energy Equation

To determine if a reaction is spontaneous, we use the Gibbs free energy equation: \[ \Delta G = \Delta H - T \Delta S \] where \( \Delta G \) is the Gibbs free energy change, \( \Delta H \) is the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the entropy change.
02

Substituting Known Values

We substitute the given values into the Gibbs free energy equation. \( \Delta H = -2.5 \times 10^3 \, \text{J/mol} \), \( \Delta S = 7.4 \, \text{J/K} \, \text{mol} \), and \( T = 298 \, \text{K} \).
03

Calculation of Gibbs Free Energy

Calculate \( \Delta G \) using the equation: \[ \Delta G = -2.5 \times 10^3 - 298 \times 7.4 \] Simplifying it: \[ \Delta G = -2500 - 2205.2 = -4705.2 \, \text{J/mol} \]
04

Analyzing the Results

Since \( \Delta G < 0 \), the Gibbs free energy change is negative, indicating that the reaction is spontaneous at the given conditions.

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Key Concepts

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

Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), refers to the heat absorbed or released in a chemical reaction under constant pressure. It is a critical factor in understanding how energy is transformed during a reaction. When \( \Delta H \) is negative, the reaction is exothermic, meaning it releases heat to the surroundings. Conversely, a positive \( \Delta H \) indicates an endothermic reaction, which absorbs heat.
  • An exothermic reaction often feels warm to the touch because it releases energy.
  • Endothermic reactions may feel cold as they take in energy.

In our example exercise, the enthalpy change is \(-2.5 \times 10^{3}\) J/mol, which is negative, indicating that the reaction releases heat. This release is important in determining the spontaneity of a reaction alongside other factors like entropy and temperature.
Entropy Change
Entropy change, symbolized as \( \Delta S \), measures the disorder or randomness in a system. In the scope of chemical reactions, entropy change refers to the difference in randomness between reactants and products. A positive \( \Delta S \) means increased disorder, while a negative \( \Delta S \) reflects decreased disorder.
  • Reactions that result in a positive entropy change tend to be spontaneous because the universe "favors" increased disorder.
  • Changing states, such as from solid to liquid, often increases entropy.

For the exercise, the entropy change is given as \(7.4 \) J/K mol, which suggests an increase in disorder. Both enthalpy and entropy changes jointly influence the Gibbs Free Energy, conveying whether a reaction will occur spontaneously.
Spontaneous Reaction Conditions
Understanding the conditions under which a reaction becomes spontaneous is crucial in chemistry. A reaction is considered spontaneous if it occurs without the input of additional energy once it has started. This is determined by the Gibbs Free Energy (\( \Delta G \)). According to the equation: \[ \Delta G = \Delta H - T \Delta S \]
  • If \( \Delta G < 0 \), the reaction is spontaneous and will proceed.
  • If \( \Delta G > 0 \), the reaction is non-spontaneous and requires energy to occur.
  • When \( \Delta G = 0 \), the system is at equilibrium, and no net change occurs.

In the provided exercise, \( \Delta G \) is calculated to be \(-4705.2 \) J/mol, clearly showing that the reaction is spontaneous under the conditions of \( 298 \) K. This means the reaction will readily occur, driven by both the enthalpy release and the positive entropy change.

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Most popular questions from this chapter

When \(\mathrm{NH}_{4} \mathrm{Cl}\) is warmed with \(\mathrm{NaOH}\) in a test tube it is a case of (a) Open system (b) Closed system (c) Isolated system (d) Both a, c

If at \(298 \mathrm{~K}\) the bond energies of \(\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}\) and \(\mathrm{H}-\mathrm{H}\) bonds are respectively \(414,347,615\) and \(435 \mathrm{~kJ} \mathrm{~mol}^{-1}\), the value of enthalpy change for the reaction \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}(\mathrm{~g})\) at \(298 \mathrm{~K}\) will be (a) \(+250 \mathrm{~kJ}\) (b) \(-250 \mathrm{~kJ}\) (c) \(+125 \mathrm{~kJ}\) (d) \(-125 \mathrm{~kJ}\)

For the reaction \(\mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) the value of \(\Delta \mathrm{H}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\Delta \mathrm{S}=0.163\) \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\). The free energy change at \(300 \mathrm{~K}\). for the reaction, is (a) \(-289.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(437.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-334.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-291.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Identify the correct statement regarding a spontaneous process. (a) Endothermic processes are never spontaneous (b) Exothermic process are always spontaneous (c) Lowering of energy in the reaction process is the only criterion for spontaneity (d) For a spontaneous process in an isolated system, the change in entropy is positive.

The freezing point of isobutane is \(-160^{\circ} \mathrm{C} \cdot \Delta \mathrm{H}_{\text {(solid } \rightarrow \text { liquid) }}\) is \(+4520 \mathrm{~J} \mathrm{~mol}^{-1}\). For this fusion process, entropy change in \(\mathrm{J} \mathrm{mol}^{-1}\) is \(10 \mathrm{y}\). The value of \(\mathrm{y}\) is

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