On the basis of the following thermochemical data: $(\Delta_fG^o H^{+}_{(aq)} = 0)$
$H_2O_{(\ell)} \rightarrow H^{+}_{(aq)} + OH^{-}_{(aq)} \,; \, \Delta H = 57.32 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(\ell)} \,; \, \Delta H = -286.20 \, kJ$
The value of enthalpy of formation of $OH^{-}$ ion at $25 \, ^oC$ is : .............. $kJ$
$H_2O_{(\ell)} \rightarrow H^{+}_{(aq)} + OH^{-}_{(aq)} \,; \, \Delta H = 57.32 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(\ell)} \,; \, \Delta H = -286.20 \, kJ$
The value of enthalpy of formation of $OH^{-}$ ion at $25 \, ^oC$ is : .............. $kJ$
- A$-228.88$
- B$+228.88$
- C$-343.52$
- D$-22.88$
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Similar Questions
The heat of formation of water is $260 \ kJ$. How much $H_2O$ is decomposed by $130 \ kJ$ of heat (in $mol$)?
DifficultMHT CET 2010
View SolutionHeat of combustion of two isomers $x$ and $y$ are $17 \ kJ/mol$ and $12 \ kJ/mol$ respectively. From this information,it may be concluded that:
Medium
View SolutionFrom the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:
$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ $\Delta H = 42.09 \ kJ \ mol^{-1}$ $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ $\Delta H = -242 \ kJ \ mol^{-1}$ $H_{2(g)} \rightarrow 2 H_{(g)}$ $\Delta H = 436 \ kJ \ mol^{-1}$ $O_{2(g)} \rightarrow 2 O_{(g)}$ $\Delta H = 496 \ kJ \ mol^{-1}$
| $1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ | $\Delta H = 42.09 \ kJ \ mol^{-1}$ |
| $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ | $\Delta H = -242 \ kJ \ mol^{-1}$ |
| $H_{2(g)} \rightarrow 2 H_{(g)}$ | $\Delta H = 436 \ kJ \ mol^{-1}$ |
| $O_{2(g)} \rightarrow 2 O_{(g)}$ | $\Delta H = 496 \ kJ \ mol^{-1}$ |
Calculate 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) $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -286.2 \ kJ$
DifficultTS EAMCET 2002
View SolutionOn the basis of the following reactions,which one is correct?
$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$
$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$
$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$
$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$
$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$
$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$
Medium
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