For the allotropic change represented by the equation $C(\text{diamond}) \to C(\text{graphite})$,the enthalpy change is $\Delta H = -1.89 \ kJ$. If $6 \ g$ of diamond and $6 \ g$ of graphite are separately burnt to yield carbon dioxide,the heat liberated in the first case is:

  • A
    Less than in the second case by $1.89 \ kJ$
  • B
    More than in the second case by $1.89 \ kJ$
  • C
    Less than in the second case by $11.34 \ kJ$
  • D
    More than in the second case by $0.945 \ kJ$

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Calculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ if $\Delta_{f} H^{\circ}(CH_{4}) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_{2}) = -390 \ kJ \ mol^{-1}$,and $\Delta_{f} H^{\circ}(H_{2}O) = -286 \ kJ \ mol^{-1}$.

Given the thermochemical reactions:
$C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta H = -110.5 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta H = -283.2 \ kJ$
Calculate the heat of reaction for $C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}$ in $kJ$.

If the enthalpies of atomization of methane and ethane are respectively $320 \ cal$ and $360 \ cal$,then the bond energy of the $C-C$ bond is......$cal$.

Define thermochemical equations and explain the conventions used in them.

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Using the data provided,calculate the bond energy $(kJ \ mol^{-1})$ of a $C \equiv C$ bond in $C_{2}H_{2}$. (Take the bond energy of a $C-H$ bond as $350 \ kJ \ mol^{-1}$)
$2C_{(s)} + H_{2(g)} \longrightarrow C_{2}H_{2(g)} \quad \Delta H = 225 \ kJ \ mol^{-1}$
$2C_{(s)} \longrightarrow 2C_{(g)} \quad \Delta H = 1410 \ kJ \ mol^{-1}$
$H_{2(g)} \longrightarrow 2H_{(g)} \quad \Delta H = 330 \ kJ \ mol^{-1}$

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