Exams › JEE Main › Chemistry
Using the following thermochemical equations, determine the enthalpy of vaporisation of water: H₂(g) + 1/2 O₂(g) → H₂O(l), ΔH = -286 kJ H₂(g) + 1/2 O₂(g) → H₂O(g), ΔH = -245.5 kJ
- 6.02 kJ
- 40.5 kJ
- 62.3 kJ
- 1.25 kJ
Correct answer: 62.3 kJ
Solution
The enthalpy of vaporization of water can be calculated by taking the difference between the enthalpy of formation of water in the gas state and the liquid state. This results in 62.3 kJ, which represents the energy required to convert liquid water to vapor.
Related JEE Main Chemistry questions
- Using the following thermochemical equations, determine the enthalpy of vaporisation of water: \[ \mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \to H_2O(l)},\quad \Delta H = -286\ \text{kJ} \] \[ \mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \to H_2O(g)},\quad \Delta H = -245.5\ \text{kJ} \]
- Given that the standard enthalpy of formation of NH$_3$ is $-46.0\ \text{kJ mol}^{-1}$, and the enthalpy of formation of H$_2$ from gaseous atoms is $-436\ \text{kJ mol}^{-1}$ while that of N$_2$ from gaseous atoms is $-712\ \text{kJ mol}^{-1}$, what is the average N–H bond enthalpy in NH$_3$?
- Using the following bond dissociation data, determine the enthalpy change for the conversion $8S(s) + \tfrac32 S_8(g)$.
- Using the bond enthalpies given below, determine the enthalpy change for the hydrogenation reaction shown: H–H bond energy = 431.37 kJ mol$^{-1}$ C–C bond energy = 606.10 kJ mol$^{-1}$ C–H bond energy = 336.49 kJ mol$^{-1}$ C–H bond energy = 410.50 kJ mol$^{-1}$ $C=C + H–H \rightarrow H–C–C–H$ What is the enthalpy change for this reaction?
- For a given reversible reaction, both enthalpy change ($\Delta H$) and entropy change ($\Delta S$) are positive. If $T_1$ denotes the equilibrium temperature, under which condition will the reaction be spontaneous?
- For the spontaneous reaction $2C_2H_6(g) + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g)$, what are the signs of $\Delta H$, $\Delta S$, and $\Delta G$ in that order?
⚔️ Practice JEE Main Chemistry free + battle 1v1 →