The enthalpy change for the transition of carbon from diamond to graphite is $\Delta H = -453.5 \ \text{cal}$. What does this indicate?
- AGraphite is chemically different from diamond.
- BGraphite is as stable as diamond.
- CGraphite is more stable than diamond.
- DDiamond is more stable than graphite.
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Similar Questions
The enthalpy of neutralization of $HCN$ by $NaOH$ is $-12.13 \, kJ \, mol^{-1}$. The enthalpy of ionisation of $HCN$ will be $... \, kJ \, mol^{-1}$.
Difficult
View SolutionWhich of the following substances has the highest value of standard molar enthalpy of formation at $298 \ K$?
Bond dissociation enthalpy of $H_2, Cl_2$ and $HCl$ are $434, 242$ and $431 \ kJ \ mol^{-1}$ respectively. Enthalpy of formation of $HCl$ is ............... $kJ \ mol^{-1}$
MediumAIPMT 2008
View SolutionFrom the following bond energies:
$H-H$ bond energy $431.37 \text{ kJ mol}^{-1}$ $C=C$ bond energy $606.10 \text{ kJ mol}^{-1}$ $C-C$ bond energy $336.49 \text{ kJ mol}^{-1}$ $C-H$ bond energy $410.50 \text{ kJ mol}^{-1}$
Enthalpy for the reaction $CH_2=CH_2 + H-H \to CH_3-CH_3$ will be .............. $\text{kJ mol}^{-1}$
| $H-H$ bond energy | $431.37 \text{ kJ mol}^{-1}$ |
| $C=C$ bond energy | $606.10 \text{ kJ mol}^{-1}$ |
| $C-C$ bond energy | $336.49 \text{ kJ mol}^{-1}$ |
| $C-H$ bond energy | $410.50 \text{ kJ mol}^{-1}$ |
Enthalpy for the reaction $CH_2=CH_2 + H-H \to CH_3-CH_3$ will be .............. $\text{kJ mol}^{-1}$
MediumAIPMT 2009
View SolutionAt $25^{\circ} C$ 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 ........ .
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