Question

An electric current of 0.50A from a 12V supply is passed for 300 sec through a resistance in thermal contact with water and the water is allowed to boil under a pressure of 1.0 atm. The value of enthalpy change during the process, if 0.798 gm of water is vaporised, will be :

• A. $-44 \text{ kJ mol}^{-1}$
• B. $+41 \text{ kJ mol}^{-1}$
• C. $-37 \text{ kJ mol}^{-1}$
• D. $+82 \text{ kJ mol}^{-1}$

Hint:

Pay close attention to thermodynamic signs! Vaporization, melting, and sublimation are always endothermic (positive $\Delta H$). Condensation and freezing are always exothermic (negative $\Delta H$). This immediately eliminates options with negative signs for boiling processes.

Answer 1

The Correct Option is B

Step 1: Concept:
The problem asks for the molar enthalpy of vaporization of water. We are given the electrical energy used to boil a specific mass of water.
We must first calculate the total heat energy supplied by the electrical circuit and then determine the enthalpy change per mole of water vaporized.

Step 2: Key Formula or Approach:
The total electrical energy ($W$ or $Q$) supplied by a circuit is given by Joule's law of heating:
\[ Q = V \times I \times t \]
Where $V$ is voltage, $I$ is current, and $t$ is time in seconds.
The molar enthalpy change ($\Delta H$) is the total heat supplied divided by the number of moles ($n$) vaporized:
\[ \Delta H = \frac{Q}{n} \]

Step 3: Step-by-step Explanation:

• First, calculate the total electrical energy (heat) supplied to the water:
\[ Q = 12 \text{ V} \times 0.50 \text{ A} \times 300 \text{ s} \]
\[ Q = 1800 \text{ Joules} = 1.8 \text{ kJ} \]

• Next, determine the number of moles of water vaporized. The molar mass of water ($H_2O$) is $18 \text{ g mol}^{-1}$.
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.798 \text{ g}}{18 \text{ g mol}^{-1}} \approx 0.04433 \text{ mol} \]

• Now, calculate the enthalpy of vaporization per mole:
\[ \Delta H = \frac{1.8 \text{ kJ}}{0.04433 \text{ mol}} \approx 40.604 \text{ kJ mol}^{-1} \]

• Rounding to the nearest whole number given in the options, we get $41 \text{ kJ mol}^{-1}$.

• Since vaporization requires the absorption of heat (it is an endothermic process), the sign of the enthalpy change must be positive.
\[ \Delta H = +41 \text{ kJ mol}^{-1} \]

Step 4: Final Answer:
The calculated molar enthalpy change is $+41 \text{ kJ mol}^{-1}$, matching option (B).