In the reaction for the transition of carbon in the diamond form to carbon in the graphite form,$\Delta H = -453.5 \ cal$. This points out that

  • A
    Graphite is chemically different from diamond
  • B
    Graphite is as stable as diamond
  • C
    Graphite is more stable than diamond
  • D
    Diamond is more stable than graphite

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Similar Questions

The standard enthalpies of formation of $Al_{2}O_{3}$ and $CaO$ are $-1675 \ kJ \ mol^{-1}$ and $-635 \ kJ \ mol^{-1}$ respectively.
For the reaction $3 CaO + 2 Al \rightarrow 3 Ca + Al_{2}O_{3}$,the standard reaction enthalpy $\Delta_{r}H^{\circ}$ is .......... $kJ$.
(Round off to the Nearest Integer).

Fill in the blanks:
$(i)$ The enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states is called the ...... of that compound.
$(ii)$ The total heat change in a chemical reaction is equal to the algebraic sum of the heat changes of the individual steps of the reaction. This law was given by ............. .
$(iii)$ $A$ process during which there is no exchange of heat between the system and the surroundings is called an ................ process.
$(iv)$ The standard enthalpy value of any element in its standard state is considered to be ...... .

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For $NaCl_{(s)}$,the enthalpy of solution is $4 \ kJ \ mol^{-1}$ and the lattice enthalpy is $790 \ kJ \ mol^{-1}$. What is the hydration enthalpy of $NaCl$?

$XeF_{2(g)} + H_{2(g)} \to 2HF_{(g)} + Xe_{(g)}$,$\Delta H^o = -430 \ kJ$
Bond energy:
$H-H = 435 \ kJ/mol$
$H-F = 565 \ kJ/mol$
Calculate the average bond energy of the $Xe-F$ bond in $kJ/mol$.

Based on the following thermochemical equations:
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = X \ kJ$
The value of $X$ is ...... $kJ$.

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