The molar enthalpies of combustion of $C_2H_{2(g)},$ $C$ (graphite) and $H_{2(g)}$ are $-1300,$ $-394$ and $-286 \ kJ \ mol^{-1},$ respectively. The standard enthalpy of formation of $C_2H_{2(g)}$ is.......$kJ \ mol^{-1}$

  • A
    $-226$
  • B
    $-626$
  • C
    $226$
  • D
    $626$

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If the combustion of $1 \ g$ of graphite produces $20.7 \ kJ$ of heat,what will be the molar enthalpy change? Give the significance of the sign also.

In the reaction $H_2 + Cl_2 \rightarrow 2HCl$,heat is released. The bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $242 \ kJ \ mol^{-1}$ respectively. If the enthalpy of reaction is $-182 \ kJ \ mol^{-1}$,the bond energy of $H-Cl$ is . . . . . . $kJ \ mol^{-1}$.

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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 ...... $kJ$.

Hess's law of constant heat summation includes:

From the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:
$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$$\Delta H = 42.09 \ kJ \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$$\Delta H = -242 \ kJ \ mol^{-1}$
$H_{2(g)} \rightarrow 2 H_{(g)}$$\Delta H = 436 \ kJ \ mol^{-1}$
$O_{2(g)} \rightarrow 2 O_{(g)}$$\Delta H = 496 \ kJ \ mol^{-1}$

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