Question

At $100{.}^{\circ }\mathrm{C}$ and $1.00\mathrm{atm},\mathrm{\Delta }{H}^{\circ }=40.6\mathrm{kJ}/\mathrm{mol}$ for the vaporization of water. Estimate $\mathrm{\Delta }{G}^{\circ }$ for the vaporization of water at $90{.}^{\circ }\mathrm{C}$ and $110{.}^{\circ }\mathrm{C}$ . Assume $\mathrm{\Delta }{H}^{\circ }$ and $\mathrm{\Delta }{S}^{\circ }$ at $100{.}^{\circ }\mathrm{C}$ and 1.00 $\mathrm{atm}$ do not depend on temperature.

Short answer

The estimated standard Gibbs free energy change for the vaporization of water is approximately $0.94\mathrm{kJ}/\mathrm{mol}$ at ${90}^{\circ }C$ and $-2.69\mathrm{kJ}/\mathrm{mol}$ at ${110}^{\circ }C$.

Step-by-step solution

1. Review the relationship between Gibbs free energy, enthalpy, and entropy

The relationship between the standard Gibbs free energy change (∆G°) standard enthalpy change (∆H°), and standard entropy change (∆S°) is described by the Gibbs-Helmholtz equation: $$\mathrm{\Delta }{G}^{\circ }=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }$$ We are given ∆H° and T at 100°C and need to find ∆G° at different temperatures.

2. Convert the given temperature from Celsius to Kelvin

We first need to convert the given temperature (100°C) to Kelvin: $$T(K)=T{(}^{\circ }C)+273.15$$ For 100°C: $$T({100}^{\circ }C)=100+273.15=373.15\mathrm{K}$$ We know that at 100°C and 1.00 atm, ∆H° = 40.6 kJ/mol

3. Calculate ∆S° at 100°C using the equilibrium condition

At the boiling point, the system is at equilibrium. This means that the Gibbs free energy change (∆G°) is equal to zero: $$\mathrm{\Delta }{G}_{eq}^{\circ }=0$$ We can rearrange the Gibbs-Helmholtz equation to find ∆S°: $$\mathrm{\Delta }{S}^{\circ }=\frac{\mathrm{\Delta }{H}^{\circ }}{T}$$ Using the values of ∆H° and temperature: $$\mathrm{\Delta }{S}^{\circ }=\frac{40.6\mathrm{kJ}/\mathrm{mol}}{373.15\mathrm{K}}=0.1087\mathrm{kJ}/(\mathrm{mol}\cdot \mathrm{K})$$

4. Calculate ∆G° at 90°C and 110°C using the given ∆H° and calculated ∆S°

Now that we have ∆H° and ∆S° at 100°C, we can use the Gibbs-Helmholtz equation to find ∆G° at the other temperatures. First, convert 90°C and 110°C to Kelvin: For 90°C: $$T({90}^{\circ }C)=90+273.15=363.15\mathrm{K}$$ For 110°C: $$T({110}^{\circ }C)=110+273.15=383.15\mathrm{K}$$ Now, use the Gibbs-Helmholtz equation to find ∆G° at each temperature: For 90°C: $$\mathrm{\Delta }{G}^{\circ }({90}^{\circ }C)=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }=40.6\mathrm{kJ}/\mathrm{mol}-(363.15\mathrm{K}\times 0.1087\mathrm{kJ}/(\mathrm{mol}\cdot \mathrm{K}))\approx 0.94\mathrm{kJ}/\mathrm{mol}$$ For 110°C: $$\mathrm{\Delta }{G}^{\circ }({110}^{\circ }C)=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }=40.6\mathrm{kJ}/\mathrm{mol}-(383.15\mathrm{K}\times 0.1087\mathrm{kJ}/(\mathrm{mol}\cdot \mathrm{K}))\approx -2.69\mathrm{kJ}/\mathrm{mol}$$ So, the estimated standard Gibbs free energy change for the vaporization of water is approximately 0.94 kJ/mol at 90°C and -2.69 kJ/mol at 110°C.

Key concepts

Enthalpy

Enthalpy is a measure of the total energy of a thermodynamic system, including both internal energy and the energy required to displace its environment to make room for the system. In chemical reactions, the change in enthalpy ($\mathrm{\Delta }{H}^{\circ }$) can tell us if a reaction is exothermic or endothermic. Vaporation of water at 100°C involves absorbing heat, so it's endothermic, and thus, $\mathrm{\Delta }{H}^{\circ }$ is positive. In this exercise, $\mathrm{\Delta }{H}^{\circ }$ for vaporization is 40.6 kJ/mol, indicating the energy needed to convert water from liquid to vapor at standard conditions. Understanding enthalpy helps us grasp energy flow, which is crucial for determining if reactions are feasible energetically.

Entropy

Entropy ($\mathrm{\Delta }{S}^{\circ }$) is the measure of disorder or randomness in a system. For the vaporization of water, it tells us how the system's randomness increases when water molecules transition from the liquid phase to the gas phase. At equilibrium (such as at the boiling point of water at 100°C), $\mathrm{\Delta }{G}^{\circ }=0$, leading us to calculate entropy using the equation: $\mathrm{\Delta }{S}^{\circ }=\frac{\mathrm{\Delta }{H}^{\circ }}{T}$. Here, we found $\mathrm{\Delta }{S}^{\circ }=0.1087\mathrm{kJ}/(\mathrm{mol}\cdot \mathrm{K})$. Knowing entropy changes supports understanding the spontaneity of processes, which is reasoned by the balance of order and energy.

Temperature Conversion

Temperature conversion, particularly from Celsius to Kelvin, is crucial in thermodynamic calculations. The conversion is simple: just add 273.15 to your Celsius temperature. For example, 100°C becomes 373.15 K. This conversion ensures consistency since Kelvin is the standard unit for thermodynamic temperatures. For temperature-focused calculations like Gibbs free energy, using Kelvin ensures analytical vigor, enabling us to correctly apply formulas like $\mathrm{\Delta }{G}^{\circ }=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }$ without temperature inconsistencies.

Vaporization of Water

The vaporization of water involves transforming water from liquid to gas, requiring energy input (endothermic process). At its boiling point, the energy needed is quantified by the enthalpy of vaporization ($\mathrm{\Delta }{H}^{\circ }$). As water vaporizes, the system's entropy increases because gas molecules are more spread out than those in a liquid. This exercise showed us how to calculate $\mathrm{\Delta }{G}^{\circ }$ at different temperatures (90°C and 110°C), highlighting how the balance of enthalpy and entropy results in positive or negative Gibbs free energy. Understanding water's vaporization provides insight into the behavior of matter under energy changes, with practical applications in everything from weather phenomena to engineering.