Import Question JSON

$\text{Hint: Hess's Law of Constant Heat Summation}$ $\text{Analysis of Statement I:}$ $\text{As per Hess's Law of Constant Heat Summation, if a reaction takes place in several steps, then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.}$ $\text{This means that enthalpy is a state function, and the total enthalpy change depends only on the initial and final states, not on the pathway taken. Therefore, for a multistep process, the total enthalpy change is indeed the sum of the enthalpy changes of each individual step: } \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \ldots$ $\text{Hence, Statement I is correct.}$ $\text{Analysis of Statement II:}$ $\text{In thermodynamics, there are two sign conventions:}$ $\text{1. In the system-centered convention: Heat absorbed by the system (endothermic) is positive, and heat released by the system (exothermic) is negative.}$ $\text{2. In the surroundings-centered convention: Heat absorbed by the system is negative, and heat released by the system is positive.}$ $\text{The standard convention used in modern thermodynamics is the system-centered approach, where heat absorbed by the system (q) has a positive sign. This convention is used because it aligns with the first law of thermodynamics (} \Delta U = q + w \text{), where an increase in internal energy can result from positive heat transfer to the system.}$ $\text{Therefore, Statement II is incorrect, as heat absorbed by the system has a positive sign, not negative.}$ $\text{Since Statement I is correct and Statement II is incorrect, the answer is option 2.}$