If the heat of neutralization of an acid-base reaction is $56 \ kJ \ mol^{-1}$,then the substances could be:

  • A
    $HCl + NH_4OH$
  • B
    $HNO_3 + LiOH$
  • C
    $HCOOH + KOH$
  • D
    $CH_3COOH + NaOH$

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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 ..... $kJ \ mol^{-1}$

If the $\Delta H_{fusion}$ of a substance is $'x'$ and $\Delta H_{vap}$ is $'y'$,then $\Delta H_{sublimation}$ will be:

Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$I. C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -393.5 \ kJ$
$II. H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -286.2 \ kJ$
$III. C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}; \Delta H = -1410.8 \ kJ$ (in $kJ$)

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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