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NEET CHEMISTRYThermodynamicsMedium

Question

Consider the following reactions: (i) H+(aq)+OH(aq)H2O(l);ΔH=x1 kJ mol1H^+(aq) + OH^-(aq) \rightarrow H_2O(l); \Delta H = -x_1 \text{ kJ mol}^{-1} (ii) H2(g)+12O2(g)H2O(l);ΔH=x2 kJ mol1H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l); \Delta H = -x_2 \text{ kJ mol}^{-1} (iii) CO2(g)+H2(g)CO(g)+H2O(l);ΔH=x3 kJ mol1CO_2(g) + H_2(g) \rightarrow CO(g) + H_2O(l); \Delta H = -x_3 \text{ kJ mol}^{-1} (iv) C2H5(g)+52O2(g)2CO2(g)+H2O(l);ΔH=x4 kJ mol1C_2H_5(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(l); \Delta H = -x_4 \text{ kJ mol}^{-1} Enthalpy of formation of H2O(l)H_2O(l) is:

A

x2 kJ mol1-x_2 \text{ kJ mol}^{-1}

B

+x3 kJ mol1+x_3 \text{ kJ mol}^{-1}

C

x4 kJ mol1-x_4 \text{ kJ mol}^{-1}

D

x1 kJ mol1-x_1 \text{ kJ mol}^{-1}

Step-by-Step Solution

The standard enthalpy of formation (ΔfH\Delta_f H^\circ) is defined as the enthalpy change when exactly one mole of a compound is formed from its constituent elements in their most stable states of aggregation (reference states) . For liquid water (H2OH_2O), the constituent elements in their standard states are hydrogen gas (H2H_2) and oxygen gas (O2O_2). The reaction representing the formation of one mole of H2O(l)H_2O(l) is: H2(g)+12O2(g)H2O(l)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) This matches the second reaction provided in the question, where the enthalpy change is given as x2 kJ mol1-x_2 \text{ kJ mol}^{-1}. Note: Reaction (i) represents the enthalpy of neutralization, not formation.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsconsiderfollowingreactionsrightarrowfracog

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