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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Heat of transition is the heat evolved or absorbed when a substance is converted from

Given: $S + \frac{3}{2} O_2 \to SO_3 + 2x \ \text{kcal}$,$\Delta H = -2x \ \text{kcal}$ and $SO_2 + \frac{1}{2} O_2 \to SO_3 + y \ \text{kcal}$,$\Delta H = -y \ \text{kcal}$. Find the heat of formation of $SO_2$.

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Which of the following reactions represents the standard heat of formation of the product?

Given $:$
$\Delta H^{\ominus}_{sub}[C(graphite)] = 710 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C-H} = 414 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{H-H} = 436 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C=C} = 611 \ kJ \ mol^{-1}$
The $\Delta H^{\ominus}_{f}$ for $CH_2=CH_2$ is $............ \ kJ \ mol^{-1}$ $(nearest \ integer \ value)$

In the reaction $S + \frac{3}{2} O_{2} \rightarrow SO_{3} + 2x \ kJ$ and $SO_{2} + \frac{1}{2} O_{2} \rightarrow SO_{3} + y \ kJ$,the heat of formation of $SO_{2}$ is

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