Electrical Energy to Water Vaporization Enthalpy Calculation

This section calculates the enthalpy change for water vaporization using the provided electrical energy parameters. The goal is to find the energy required per mole to vaporize water.

Calculate Electrical Energy Supplied

Electrical energy is converted into heat. Calculate the total electrical energy supplied using the formula:

E = V × I × t

Given values:

• Voltage (V) = 12 V
• Current (I) = 0.50 A
• Time (t) = 300 s

Calculation:

E = $12 \text{ V} \times 0.50 \text{ A} \times 300 \text{ s}$ = $1800 \text{ J}$

This total energy is absorbed as heat (Q) by the water for vaporization: Q = $1800 \text{ J}$

Calculate Moles of Water Vaporized

To find the enthalpy change per mole, first determine the number of moles of water vaporized.

Number of moles (n) = $\frac{\text{Mass}}{\text{Molar Mass}}$

Given values:

• Mass of water vaporized = 0.798 gm
• Molar Mass of water ($H_2O$) $\approx 18.015$ g/mol

Calculation:

n = $\frac{0.798 \text{ gm}}{18.015 \text{ g mol}^{-1}} \approx 0.0443 \text{ mol}$

Determine Enthalpy Change per Mole

Calculate the enthalpy change ($\Delta H$) by dividing the total heat absorbed by the number of moles vaporized.

Enthalpy Change ($\Delta H$) = $\frac{Q}{n}$

Calculation:

$\Delta H = \frac{1800 \text{ J}}{0.0443 \text{ mol}} \approx 40609 \text{ J mol}^{-1}$

Convert the result from Joules per mole to Kilojoules per mole:

$\Delta H \approx \frac{40609}{1000} \text{ kJ mol}^{-1} \approx 40.6 \text{ kJ mol}^{-1}$

Since vaporization is an endothermic process (requires energy input), the enthalpy change is positive. The calculated value of approximately $40.6 \text{ kJ mol}^{-1}$ is closest to $+41 \text{ kJ mol}^{-1}$.