Given that:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}, \Delta H = -394 \ kJ$
$2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}, \Delta H = -568 \ kJ$
$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}, \Delta H = -892 \ kJ$
Calculate the heat of formation of $CH_{4(g)}$ in $kJ$.

  • A
    $-70$
  • B
    $71.8$
  • C
    $-76$
  • D
    $+782$

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The bond enthalpies of $H_2$,$X_2$ and $HX$ are in the ratio of $2 : 1 : 2$. If the enthalpy for formation of $HX$ is $-50 \ kJ \ mol^{-1}$,the bond enthalpy of $H_2$ is ..... $kJ \ mol^{-1}$

Use the data from the table to estimate the enthalpy of formation of $CH_3CHO$.
BondBond Enthalpy $(kJ \ mol^{-1})$Enthalpy of formation $(kJ \ mol^{-1})$
$C-H$$400$$C(g): 700$
$C-C$$350$$H(g): 200$
$C=O$$700$$O(g): 250$

$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.

What will be the enthalpy of formation of benzene in $kJ$ for the reaction: $6C_{(s)} + 3H_{2(g)} \rightarrow C_6H_{6(l)}$? Given that the enthalpy of combustion of benzene is $-3268 \ kJ$, the enthalpy of formation of $CO_{2(g)}$ is $-393.5 \ kJ$, and the enthalpy of formation of $H_2O_{(l)}$ is $-285.8 \ kJ$.

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From the given data at $298 \, K$:
$\Delta H_f^o [CO_2, g] = -394 \, kJ/mol$
$\Delta H_f^o [H_2O, l] = -286 \, kJ/mol$
$\Delta H_f^o [propene, g] = 20 \, kJ/mol$
$cyclopropane (g) \to propene (g)$; $\Delta H^o_{isomerisation} = -33 \, kJ/mol$.
Calculate $\Delta H^o_{combustion} [cyclopropane, g]$.
$...... \, kJ/mol$

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