Standard enthalpy of vapourisation $\Delta_{\text {vap }} H ^{\Theta}$ for water at $100^{\circ} C$ is $40.66\, kJ\, mol ^{-1}$. The internal energy of vaporisation of water at $100^{\circ} C$ (in $kJmol ^{-1}$) is :
(Assume water vapour to behave like an ideal gas)
- 43.76
- 40.66
- 37.56
- -43.76
The Correct Option is C
Solution and Explanation
$\Delta H =\Delta E +\Delta nRT$
$40.66=\Delta E +1 \times \frac{8.314}{1000} \times 373$
$\Delta E =37.5\, kJ$
Top Questions on Enthalpy change
- Given below are two statements:
Statement (I): The radius of isoelectronic species increases in the order: \[ \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} < \text{O}^{2-} \] Statement (II): The magnitude of electron gain enthalpy of halogens decreases in the order: \[ \text{Cl} > \text{F} > \text{Br} > \text{I} \] In light of the above statements, choose the most appropriate answer from the options given below:- JEE Main - 2025
- Chemistry
- Enthalpy change
- The enthalpy of formation for \( \text{H}_2(g) \), \( \text{O}_2(g) \), and \( \text{H}_2O(l) \) are 0, 0, and -285.8 kJ/mol, respectively. What is the enthalpy change for the following reaction: \[ \text{2H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2O(l) \]
- MHT CET - 2025
- Chemistry
- Enthalpy change
- The enthalpy change for the reaction \( \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \) is \( -92.4 \, \text{kJ/mol} \). What is the enthalpy change when 4 moles of nitrogen react?
- MHT CET - 2025
- Chemistry
- Enthalpy change
- The group 14 elements A and B have the first ionisation enthalpy values of 708 and 715 kJ mol$^{-1}$ respectively. The above values are lowest among their group members. The nature of their ions A$^{2+}$ and B$^{4+}$ respectively is:
- JEE Main - 2025
- Chemistry
- Enthalpy change
- The incorrect relationship in the following pairs in relation to ionisation enthalpies is:
- JEE Main - 2025
- Chemistry
- Enthalpy change
Questions Asked in NEET exam
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
- NEET (UG) - 2025
- Current electricity
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
- NEET (UG) - 2025
- Pendulums
- Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is \(q\) and the force of repulsion between them is \(F\). A third identical uncharged conducting sphere C is brought in contact with sphere A first and then with sphere B and finally removed from both. New force of repulsion between spheres A and B (Radius of A and B are negligible compared to the distance of separation so that they can be considered as point charges) is best given as:
- NEET (UG) - 2025
- Coulomb’s Law
- With the help of the given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in the F3 generation.
- NEET (UG) - 2025
- Genetics
- An AC power supply of 220 V at 50 Hz, a resistor of 20\( \Omega \), a capacitor of reactance 25\( \Omega \) and an inductor of reactance 45\( \Omega \) are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively :
- NEET (UG) - 2025
- Alternating Current
Concepts Used:
Enthalpy change
Enthalpy Change refers to the difference between the heat content of the initial and final state of the reaction. Change in enthalpy can prove to be of great importance to find whether the reaction is exothermic or endothermic.
Formula for change in enthalpy is:-
dH = dU + d(PV)
The above equation can be written in the terms of initial and final states of the system which is defined below:
UF – UI = qP –p(VF – VI)
Or qP = (UF + pVF) – (UI + pVI)
Enthalpy (H) can be written as H= U + PV. Putting the value in the above equation, we obtained:
qP = HF – HI = ∆H
Hence, change in enthalpy ∆H = qP, referred to as the heat consumed at a constant pressure by the system. At constant pressure, we can also write,
∆H = ∆U + p∆V
Standard Enthalpy of Reaction
To specify the standard enthalpy of any reaction, it is calculated when all the components participating in the reaction i.e., the reactants and the products are in their standard form. Therefore the standard enthalpy of reaction is the enthalpy change that occurs in a system when a matter is transformed by a chemical reaction under standard conditions.