Given:
$(I) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}; \Delta H^o_{298\ K} = -285.9 \ kJ \ mol^{-1}$
$(II) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H^o_{298\ K} = -241.8 \ kJ \ mol^{-1}$
The molar enthalpy of vaporisation of water will be $kJ \ mol^{-1}$.

  • A
    $241.8$
  • B
    $22$
  • C
    $44.1$
  • D
    $527.7$

Explore More

Similar Questions

For the reaction $N_2 + 3 X_2 \longrightarrow 2 NX_3$,where $X = F, Cl$ (the average bond energies are $F-F = 155 \ kJ \ mol^{-1}$,$N-F = 272 \ kJ \ mol^{-1}$,$Cl-Cl = 242 \ kJ \ mol^{-1}$,$N-Cl = 200 \ kJ \ mol^{-1}$ and $N \equiv N = 941 \ kJ \ mol^{-1}$),the heats of formation of $NF_3$ and $NCl_3$ in $kJ \ mol^{-1}$,respectively,are closest to

Hess's law of constant heat summation includes:

The enthalpy of combustion of methane,graphite and dihydrogen at $298 \, K$ are $-890.3 \, kJ \, mol^{-1}$,$-393.5 \, kJ \, mol^{-1}$ and $-285.8 \, kJ \, mol^{-1}$ respectively. The enthalpy of formation of $CH_{4(g)}$ will be:
$(i) -74.8 \, kJ \, mol^{-1}$
$(ii) -52.27 \, kJ \, mol^{-1}$
$(iii) +74.8 \, kJ \, mol^{-1}$
$(iv) +52.26 \, kJ \, mol^{-1}$

Hess's law is based on

The enthalpy change for the reaction $C(s) + O_2(g) \to CO_2(g)$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo