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$690 \ kJ \ mol^{-1}$
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Similar Questions
Calculate the amount of methane formed by the liberation of $149.6 \ kJ$ of heat using the following equation:
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ/mol$ (in $g$)
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ/mol$ (in $g$)
Enthalpy of formation is a special case of enthalpy of reaction. Which of the following reactions does $NOT$ represent the enthalpy of formation of the product?
Easy
View SolutionEnthalpy is an extensive property. In general,if enthalpy of an overall reaction $A \to B$ along one route is $\Delta_r H$ and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$ represent enthalpies of intermediate reactions leading to product $B$. What will be the relation between $\Delta_r H$ for overall reaction and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$ etc. for intermediate reactions?
Difficult
View SolutionCalculate the enthalpy change in $kJ$ for the reaction: $2C_{(graphite)} + 2H_{2(g)} \to C_2H_{4(g)}$
$C_{(graphite)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
$C_2H_{4(g)} + 3O_{2(g)} \to 2CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H = -1410.9 \ kJ$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(l)} \quad \Delta H = -285.8 \ kJ$
$C_{(graphite)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
$C_2H_{4(g)} + 3O_{2(g)} \to 2CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H = -1410.9 \ kJ$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(l)} \quad \Delta H = -285.8 \ kJ$
Medium
View SolutionIf $S + O_2 \to SO_2; (\Delta H = -298.2 \ kJ)$,$SO_2 + \frac{1}{2} O_2 \to SO_3; (\Delta H = -98.2 \ kJ)$,$SO_3 + H_2O \to H_2SO_4; (\Delta H = -130.2 \ kJ)$,$H_2 + \frac{1}{2} O_2 \to H_2O; (\Delta H = -287.3 \ kJ)$,then the enthalpy of formation of $H_2SO_4$ at $298 \ K$ will be......$kJ$.
Medium
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