From the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:
$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ $\Delta H = 42.09 \ kJ \ mol^{-1}$ $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ $\Delta H = -242 \ kJ \ mol^{-1}$ $H_{2(g)} \rightarrow 2 H_{(g)}$ $\Delta H = 436 \ kJ \ mol^{-1}$ $O_{2(g)} \rightarrow 2 O_{(g)}$ $\Delta H = 496 \ kJ \ mol^{-1}$
| $1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ | $\Delta H = 42.09 \ kJ \ mol^{-1}$ |
| $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ | $\Delta H = -242 \ kJ \ mol^{-1}$ |
| $H_{2(g)} \rightarrow 2 H_{(g)}$ | $\Delta H = 436 \ kJ \ mol^{-1}$ |
| $O_{2(g)} \rightarrow 2 O_{(g)}$ | $\Delta H = 496 \ kJ \ mol^{-1}$ |
- A$1174 \ kJ \ mol^{-1}$
- B$742 \ kJ \ mol^{-1}$
- C$926 \ kJ \ mol^{-1}$
- D$690 \ kJ \ mol^{-1}$
Explore More
Similar Questions
Define thermochemical equations and explain the conventions used in them.
Difficult
View SolutionGiven two processes:
$\frac{1}{2} P_{4(s)} + 3 Cl_{2(g)} \to 2 PCl_{3(l)} \;; \Delta H = -635 \ kJ$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)} \;; \Delta H = -137 \ kJ$
The value of the heat of formation of $PCl_{5(s)}$ is ...... $kJ \ mol^{-1}$.
$\frac{1}{2} P_{4(s)} + 3 Cl_{2(g)} \to 2 PCl_{3(l)} \;; \Delta H = -635 \ kJ$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)} \;; \Delta H = -137 \ kJ$
The value of the heat of formation of $PCl_{5(s)}$ is ...... $kJ \ mol^{-1}$.
Difficult
View SolutionThe standard enthalpies of combustion of $C_6H_{6(l)}$,$C(graphite)$ and $H_{2(g)}$ are respectively $-3270 \ kJ \ mol^{-1}$,$-394 \ kJ \ mol^{-1}$ and $-286 \ kJ \ mol^{-1}$. What is the standard enthalpy of formation of $C_6H_{6(l)}$ in $kJ \ mol^{-1}$?
The energy required to break one mole of $Cl-Cl$ bonds in $Cl_2$ is $242 \, kJ \, mol^{-1}.$ The longest wavelength of light capable of breaking a single $Cl-Cl$ bond is $............ \, nm.$
$(c = 3 \times 10^8 \, m \, s^{-1}$ and $N_A = 6.02 \times 10^{23} \, mol^{-1}).$
$(c = 3 \times 10^8 \, m \, s^{-1}$ and $N_A = 6.02 \times 10^{23} \, mol^{-1}).$
MediumAIEEE 2010
View SolutionCalculate the heat of combustion (in $kJ$) of methane from the following data:
$(i)$ $C_{\text{(graphite)}} + 2H_{2(g)} \rightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ$
(ii) $C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
(iii) $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -286.2 \ kJ$
$(i)$ $C_{\text{(graphite)}} + 2H_{2(g)} \rightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ$
(ii) $C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
(iii) $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -286.2 \ kJ$
DifficultAP EAMCET 2002
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo