Question

For a particular reaction, \(\Delta H=30.0 \mathrm{~kJ}\) and \(\Delta S=90.0 \mathrm{~J} / \mathrm{K}\). Assume that \(\Delta H\) and \(\Delta S\) do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0 ?\) (b) If \(\mathrm{T}\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

Short answer

The reaction has ΔG = 0 at a temperature of approximately 333.33 K. When the temperature is increased from 333.33 K, the reaction becomes spontaneous.

Step-by-step solution

Part (a): Finding the temperature when ΔG = 0

We are given the values of ΔH and ΔS and we need to find the temperature (T) at which ΔG becomes zero. ΔG = ΔH - TΔS Since we need to find T when ΔG = 0, we can rearrange the formula as follows: T = ΔH / ΔS Now, plug in the given values of ΔH and ΔS: T = (30.0 kJ) / (90.0 J/K) Keep in mind that 1 kJ = 1000 J. Therefore, convert ΔH to J: T = (30,000 J) / (90.0 J/K) Now, divide to find the temperature: T = 333.33 K So, the reaction has ΔG = 0 at a temperature of approximately 333.33 K.

Part (b): Spontaneous or non-spontaneous reaction

We have to determine if the reaction becomes spontaneous or non-spontaneous as the temperature increases from 333.33 K (found in part (a)). To do this, we have to consider the effect of increasing temperature on ΔG. ΔG = ΔH - TΔS In this equation, ΔH and ΔS are constants, while the temperature (T) is a variable. As we increase the temperature, the value of TΔS will increase. Since ΔH is positive (given as 30.0 kJ), this means that as T increases, ΔG will become more negative, hence driving the reaction towards spontaneity. So, when T is increased from 333.33 K, the reaction becomes spontaneous.

Key concepts

Enthalpy

Enthalpy, denoted by \( \Delta H \), is a key concept in understanding chemical reactions. It represents the heat content of a system at constant pressure. In the context of chemical reactions, it is the difference between the heat absorbed or released when a reaction occurs.

• Positive \( \Delta H \): Indicates an endothermic reaction, where heat is absorbed.
• Negative \( \Delta H \): Indicates an exothermic reaction, where heat is released.

In the given problem, we have \( \Delta H = 30.0 \) kJ, which is positive, indicating that the reaction absorbs heat and is endothermic. Enthalpy is an important factor in determining the Gibbs Free Energy of a reaction, and it helps us predict whether a reaction will proceed spontaneously under certain conditions.

Entropy

Entropy, symbolized as \( \Delta S \), refers to the degree of disorder or randomness in a system. It is a measure of energy distribution at a particular temperature, and it dictates how energy disperses.Key points to consider about entropy:

• High \( \Delta S \): Implies increased disorder and higher probability of energy distribution.
• Low \( \Delta S \): Implies more order and less likelihood of energy dispersal.

In our exercise, \( \Delta S = 90.0 \) J/K, indicating a fair amount of disorder as the reaction occurs. Entropy plays a crucial role in influencing the spontaneity of a reaction, especially when considered alongside temperature. As temperature changes, the entropy's effect on the Gibbs Free Energy equation helps us determine if a reaction can be spontaneous.

Spontaneity of Reaction

The spontaneity of a chemical reaction is determined by the Gibbs Free Energy, \( \Delta G \). For a reaction to be spontaneous, \( \Delta G \) must be negative. The equation to calculate Gibbs Free Energy is:\[ \Delta G = \Delta H - T\Delta S \]Here's how different factors contribute:

• \( \Delta G < 0 \): Reaction is spontaneous.
• \( \Delta G = 0 \): Reaction is in equilibrium.
• \( \Delta G > 0 \): Reaction is non-spontaneous.

The given exercise asks at what temperature \( \Delta G = 0 \), meaning the reaction is at equilibrium, and calculates this temperature as 333.33 K. When the temperature increases from this value, \( T\Delta S \) becomes significant enough to make \( \Delta G \) negative, driving the reaction towards spontaneity. This explains why increasing temperature can favor the spontaneity of reactions, particularly when \( \Delta H \) is positive and \( \Delta S \) has a significant value.