The enthalpies of combustion of $C$(graphite) and $C$(diamond) are $-393.8 \ kJ \ mol^{-1}$ and $-395.3 \ kJ \ mol^{-1}$ respectively. The enthalpy of conversion of $C$(graphite) to $C$(diamond) is

  • A
    $-12.5 \ kJ \ mol^{-1}$
  • B
    $-789.1 \ kJ \ mol^{-1}$
  • C
    $79.0 \ kJ \ mol^{-1}$
  • D
    $1.5 \ kJ \ mol^{-1}$

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Similar Questions

Choose the reaction$(s)$ from the following options,for which the standard enthalpy of reaction is equal to the standard enthalpy of formation.
$(1)$ $\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$
$(2)$ $\frac{1}{8} S_{8(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
$(3)$ $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$
$(4)$ $2 C_{(g)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$

The average $S-F$ bond energy in $kJ \ mol^{-1}$ of $SF_{6}$ is $......$ . (Rounded off to the nearest integer) [Given : The values of standard enthalpy of formation of $SF_{6(g)}$,$S_{(g)}$ and $F_{(g)}$ are $-1100$,$275$ and $80 \ kJ \ mol^{-1}$ respectively.]

On the basis of the thermochemical equations:
$H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
The value of $X$ will be $.... \ kJ$.

Given the bond energies of $N \equiv N$,$H - H$ and $N - H$ bonds are $945$,$436$ and $391 \ kJ \cdot mol^{-1}$ respectively,the enthalpy of the following reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is ...... $kJ$.

If the enthalpy of atomisation for $Br_{2(l)}$ is $x \ kJ/mol$ and the bond enthalpy for $Br_{2(g)}$ is $y \ kJ/mol$,what is the relation between them?

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