The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O (l) 
are produced and 3267.0 kJ of heat is liberated.
The standard enthalpy of formation, 
fH of benzene is:
(Standard enthalpies of formation of CO
2(g) and H2O (l) are –393.5 kJ mol–1 and – 285.83 kJ mol–1 respectively.)

1. 54. 24 kJ mol–1
2. 48. 51 kJ mol–1
3. 66. 11 kJ mol–1
4. 15. 21 kJ mol–1

Subtopic:  Hess's Law |
 67%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

Given

\(\begin{aligned} &\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \\ &\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &H_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ &\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \\ &\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=+890.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\)

Based on the above thermochemical equations, the value of ΔrH° at 298 K for the reaction

C(graphite)+2H2(g)CH4(g)will be:

1. –74.8 kJ mol–1

2. –144.0 kJ mol–1

3. +74.8 kJ mol–1

4. +144.0 kJ mol–1

Subtopic:  Hess's Law |
 76%
From NCERT
Please attempt this question first.
Hints
Please attempt this question first.