At $25^{\circ} C$ and $1 \ atm$ pressure,the enthalpies of combustion are as given below:
Substance $H_{2(g)}$ $C_{\text{(graphite)}}$ $C_{2}H_{6(g)}$ $\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$ $-286.0$ $-394.0$ $-1560.0$
The enthalpy of formation of ethane is ........ .
| Substance | $H_{2(g)}$ | $C_{\text{(graphite)}}$ | $C_{2}H_{6(g)}$ |
| $\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$ | $-286.0$ | $-394.0$ | $-1560.0$ |
The enthalpy of formation of ethane is ........ .
- A$+54.0 \ kJ \ mol^{-1}$
- B$-68.0 \ kJ \ mol^{-1}$
- C$-86.0 \ kJ \ mol^{-1}$
- D$+97.0 \ kJ \ mol^{-1}$
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Similar Questions
For the reaction,$2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_2 O_{(g)}$,$\Delta H^{\circ} = -573.2 \ kJ$. What is the heat of decomposition of water per mol (in $kJ$)?
The enthalpy of combustion of propane,graphite and dihydrogen at $298 \ K$ are: $-2220.0 \ kJ \ mol^{-1}$,$-393.5 \ kJ \ mol^{-1}$ and $-285.8 \ kJ \ mol^{-1}$ respectively. The magnitude of the enthalpy of formation of propane $(C_{3}H_{8})$ is ......... $kJ \ mol^{-1}$. (Nearest integer)
Based on the following thermochemical reactions:
$H_2O_{(g)} + C_{(s)} \rightarrow CO_{(g)} + H_{2(g)} ; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = x \ kJ$
The value of $x$ will be:
$H_2O_{(g)} + C_{(s)} \rightarrow CO_{(g)} + H_{2(g)} ; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = x \ kJ$
The value of $x$ will be:
Difficult
View SolutionIf enthalpies of formation of $C_2H_{4(g)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ at $25 \ ^\circ C$ and $1 \ atm$ pressure are $52$,$-394$ and $-286 \ kJ \ mol^{-1}$ respectively,the enthalpy of combustion of $C_2H_{4(g)}$ will be.....$kJ \ mol^{-1}$.
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