The bond dissociation enthalpies of $H_2$, $Cl_2$, and $HCl$ are $434 \text{ kJ mol}^{-1}$, $242 \text{ kJ mol}^{-1}$, and $431 \text{ kJ mol}^{-1}$ respectively. Calculate the enthalpy of formation of $HCl$.

  • A
    $-93 \text{ kJ mol}^{-1}$
  • B
    $245 \text{ kJ mol}^{-1}$
  • C
    $93 \text{ kJ mol}^{-1}$
  • D
    $-245 \text{ kJ mol}^{-1}$

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Given that $C + O_{2} \longrightarrow CO_{2} ; \Delta H^{\circ} = -x \ kJ$ and $2 CO + O_{2} \longrightarrow 2 CO_{2} ; \Delta H^{\circ} = -y \ kJ$. The heat of formation of carbon monoxide will be

If the standard enthalpies of formation of hydrazine and water are $+50.6 \, kJ/mol$ and $-285.9 \, kJ/mol$ respectively,calculate the $\Delta H$ for the following reaction: $N_2H_{4(g)} + O_{2(g)} \to N_{2(g)} + 2H_2O_{(l)}$ in $kJ \, mol^{-1}$.

Based on the following thermochemical equations:
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = X \ kJ$
The value of $X$ is ...... $kJ$.

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 $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.

Calculate the enthalpy change for the reaction
$H_2 + F_2 \longrightarrow 2HF$
given that
Bond energy of $H-H$ bond $= 434 \ kJ/mol$
Bond energy of $F-F$ bond $= 158 \ kJ/mol$
Bond energy of $H-F$ bond $= 565 \ kJ/mol$
Result in $kJ$.

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