The enthalpy of combustion of carbon disulfide $(CS_2)$ is $-110.2 \, kJ \, mol^{-1}$. If the enthalpies of formation of $SO_2$ and $CO_2$ are $-297.4 \, kJ \, mol^{-1}$ and $-394.5 \, kJ \, mol^{-1}$ respectively,the enthalpy of formation of carbon disulfide is:

  • A
    $-823.3 \, kJ \, mol^{-1}$
  • B
    $-825.4 \, kJ \, mol^{-1}$
  • C
    $-840.7 \, kJ \, mol^{-1}$
  • D
    $-879.1 \, kJ \, mol^{-1}$

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The heat of reaction does not depend upon

For the reaction,$C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$,$\Delta U$ is the heat of reaction at constant volume. Then the heat of reaction at constant pressure is:

The heat of neutralization of a strong acid with a strong base is constant and is equal to:

Calculate $\Delta H$ in $kJ$ for the following reaction:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$

For which of the following reactions is the $\Delta H^o$ of the reaction equal to the $\Delta H_f^o$ of the product?

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