The enthalpy changes for the following processes are listed below:
$Cl_{2(g)} \to 2Cl_{(g)}$,$\Delta H = 242.3 \ kJ \ mol^{-1}$
$I_{2(g)} \to 2I_{(g)}$,$\Delta H = 151.0 \ kJ \ mol^{-1}$
$ICl_{(g)} \to I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \ kJ \ mol^{-1}$
$I_{2(s)} \to I_{2(g)}$,$\Delta H = 62.76 \ kJ \ mol^{-1}$
Given that the standard states for iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$,the standard enthalpy of formation for $ICl_{(g)}$ is .............. $kJ \ mol^{-1}$

  • A
    $+16.8$
  • B
    $+244.8$
  • C
    $-14.6$
  • D
    $-16.8$

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The heats of formation of $CO_2$ and $H_2O$ are $-97 \, kcal$ and $-68 \, kcal$ respectively. The heat of combustion of benzene is $-783 \, kcal$. What is the heat of formation of benzene in $kcal$?

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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).

The bond dissociation enthalpies of $H_2, Cl_2$ and $HCl$ are $434, 242$ and $431 \, kJ/mol$ respectively. The enthalpy of formation of $HCl$ is ..... $kJ/mol$.

Consider the following cases of standard enthalpy of reaction $\Delta H_{r}^{\circ}$ in $kJ \ mol^{-1}$:
$C_{2}H_{6(g)} + \frac{7}{2} O_{2(g)} \rightarrow 2 CO_{2(g)} + 3 H_{2}O(\ell)$,$\Delta H_{1}^{\circ} = -1550$
$C(\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H_{2}^{\circ} = -393.5$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O(\ell)$,$\Delta H_{3}^{\circ} = -286$
The magnitude of $\Delta H_{f, C_{2}H_{6(g)}}^{\circ}$ is $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.

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 $.... \ kJ$.

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