The enthalpy change for the transition of carbon from diamond to graphite is $\Delta H = -453.5 \ \text{cal}$. What does this indicate?

  • 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

Given,
$NO_{(g)} + O_{3(g)} \longrightarrow NO_{2(g)} + O_{2(g)}; \Delta H = -198.9 \, kJ/mol$
$O_{3(g)} \longrightarrow 3/2 O_{2(g)}; \Delta H = -142.3 \, kJ/mol$
$O_{2(g)} \longrightarrow 2O_{(g)}; \Delta H = +495.0 \, kJ/mol$
The enthalpy change $(\Delta H)$ for the following reaction is $..... \, kJ/mol$
$NO_{(g)} + O_{(g)} \longrightarrow NO_{2(g)}$

Which of the following is an example of an endothermic reaction?

Given
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta_{r}H^{\theta} = -92.4 \, kJ \, mol^{-1}$
What is the standard enthalpy of formation of $NH_{3}$ gas?

Determine the enthalpy of formation for $H_2O_2(\ell)$,using the listed enthalpies of reaction:
$N_2H_{4(\ell)} + 2H_2O_{2(\ell)} \to N_{2(g)} + 4H_2O_{(\ell)}; \Delta _r H_1^o = -818 \, kJ/mol$
$N_2H_{4(\ell)} + O_{2(g)} \to N_{2(g)} + 2H_2O_{(\ell)}; \Delta _r H_2^o = -622 \, kJ/mol$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(\ell)}; \Delta _r H_3^o = -285 \, kJ/mol$
Calculate the value in $kJ/mol$.

For the reaction,$3 C_2 H_{2(g)} \longrightarrow C_6 H_{6(g)}$,calculate the standard enthalpy change. The values of $\Delta H_f$ for $C_2 H_2$ and $C_6 H_6$ respectively are $250 \ kJ \ mol^{-1}$ and $90 \ kJ \ mol^{-1}$.

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