Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

For vaporization of water at $1$ atmospheric pressure, the values of $\Delta H$ and $\Delta S$ are $40.63\text{ kJ mol}^{-1}$ and $108.8\text{ J K}^{-1}\text{ mol}^{-1}$ , respectively. The temperature when Gibbs energy change ( $\Delta G$ ) for this transformation will be zero, is:

$393.4\text{ K}$

$373.4\text{ K}$

$293.4\text{ K}$

$273.4\text{ K}$

Step-by-Step Solution

According to the Gibbs-Helmholtz equation, $\Delta G = \Delta H - T\Delta S$ . When the Gibbs energy change is zero ( $\Delta G = 0$ ), the system is at equilibrium. At this point: $\Delta H = T\Delta S$ $\Rightarrow T = \frac{\Delta H}{\Delta S}$ Given: $\Delta H = 40.63\text{ kJ mol}^{-1} = 40630\text{ J mol}^{-1}$ (Note: Convert kJ to J to match the unit of $\Delta S$ ) $\Delta S = 108.8\text{ J K}^{-1}\text{ mol}^{-1}$ Substituting the values: $T = \frac{40630\text{ J mol}^{-1}}{108.8\text{ J K}^{-1}\text{ mol}^{-1}} = 373.4375\text{ K} \approx 373.4\text{ K}$ .

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