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
Given:
$C + 2S \to CS_2 ; \Delta H_f^o = +117.0 \, kJ \, mol^{-1} \dots (1)$
$C + O_2 \to CO_2 ; \Delta H_f^o = -393 \, kJ \, mol^{-1} \dots (2)$
$S + O_2 \to SO_2 ; \Delta H_f^o = -297 \, kJ \, mol^{-1} \dots (3)$
The heat of reaction for $CS_2 + 3O_2 \to CO_2 + 2SO_2$ is:
.....$kJ \, mol^{-1}$
$C + 2S \to CS_2 ; \Delta H_f^o = +117.0 \, kJ \, mol^{-1} \dots (1)$
$C + O_2 \to CO_2 ; \Delta H_f^o = -393 \, kJ \, mol^{-1} \dots (2)$
$S + O_2 \to SO_2 ; \Delta H_f^o = -297 \, kJ \, mol^{-1} \dots (3)$
The heat of reaction for $CS_2 + 3O_2 \to CO_2 + 2SO_2$ is:
.....$kJ \, mol^{-1}$
Medium
View SolutionWhich factor affects the heat of reaction in the Kirchhoff equation?
Easy
View SolutionThe enthalpy change $(\Delta H)$ for the process $N_2H_{4(g)} \to 2N_{(g)} + 4H_{(g)}$ is $1724 \ kJ \ mol^{-1}$. If the bond energy of $N-H$ bond in ammonia is $391 \ kJ \ mol^{-1}$,what is the bond energy of $N-N$ bond in $N_2H_4$ in $kJ \ mol^{-1}$?
Medium
View SolutionFor the reaction $R \rightarrow P$,the following potential energy diagram is given. What will be the enthalpy change $(\Delta H)$ for the given reaction (in $kJ$)?
Calculate the heat change for the reaction $4NH_3(g) + 3O_2(g) \rightarrow 2N_2(g) + 6H_2O(l)$ in $kJ$. The enthalpies of formation of $NH_3(g)$ and $H_2O(l)$ at $298 \ K$ are $-46.0 \ kJ \ mol^{-1}$ and $-286.0 \ kJ \ mol^{-1}$ respectively.
Medium
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