Enthalpy of formation is a special case of enthalpy of reaction. Which of the following reactions does $NOT$ represent the enthalpy of formation of the product?

  • A
    $CaO_{(s)} + CO_{2_{(g)}} \to CaCO_{3_{(s)}}$
  • B
    $Ca_{(s)} + \frac{3}{2}O_{2_{(g)}} + C_{(graphite)} \to CaCO_{3_{(s)}}$
  • C
    $H_{2_{(g)}} + S_{(rhombic)} + 2O_{2_{(g)}} \to H_2SO_{4_{(l)}}$
  • D
    $6C_{(graphite)} + 6H_{2_{(g)}} + 3O_{2_{(g)}} \to C_6H_{12}O_{6_{(s)}}$

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$Fe_2O_{3(s)} + \frac{3}{2} C_{(s)} \to \frac{3}{2} CO_{2(g)} + 2Fe_{(s)}$
$\Delta H^o = +234.1 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
$\Delta H^o = -393.5 \ kJ$
Use these equations and $\Delta H^o$ values to calculate $\Delta H^o$ for this reaction:
$4Fe_{(s)} + 3O_{2(g)} \to 2Fe_2O_{3(s)}$
..... $kJ$

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The enthalpy of atomization of $PH_3(g)$ is $228 \, kcal \, mol^{-1}$ and that of $P_2H_4(g)$ is $355 \, kcal \, mol^{-1}$. The $P-P$ bond energy (in $kcal \, mol^{-1}$) is:

Calculate the heat of formation of $Ca(OH)_{2(s)}$ at $1.8\,^{\circ}C$ from the following data:
$CaO_{(s)} + H_2O_{(l)} \to Ca(OH)_{2(s)}$; $\Delta H_{1.8\,^{\circ}C} = -15.26\,K\,cal$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H_{1.8\,^{\circ}C} = -68.37\,K\,cal$
$Ca_{(s)} + \frac{1}{2}O_{2(g)} \to CaO_{(s)}$; $\Delta H_{1.8\,^{\circ}C} = -151.80\,K\,cal$

The heat of neutralisation of $NH_4OH$ and $HCl$ is:

The heat of formation of $H_2O$ is $-286 \, kJ/mol$ and $H_2O_2$ is $-188 \, kJ/mol$. The enthalpy change for the reaction $2H_2O_2 \to 2H_2O + O_2$ is......$kJ$.

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