One mole of $C_2H_5OH_{(l)}$ was completely burnt in oxygen to form $CO_{2(g)}$ and $H_2O_{(l)}$. The standard enthalpy of formation $\Delta_fH^{\ominus}$ of $C_2H_5OH_{(l)}, CO_{2(g)}$ and $H_2O_{(l)}$ is $x, y, z \ kJ \ mol^{-1}$ respectively. What is $\Delta_rH^{\ominus}$ (in $kJ \ mol^{-1}$) for this reaction?

  • A
    $2y + 3z - x$
  • B
    $2y - 3z + x$
  • C
    $x - 2y - 3z$
  • D
    $2y + 3z + x$

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Identify the law for the statement "Overall the enthalpy change for a reaction is equal to the sum of enthalpy changes of individual steps in the reaction".

When $1 \, \text{mol}$ of anhydrous salt $AB$ is dissolved in water,$21.0 \, J \, \text{mol}^{-1}$ of heat is released. The enthalpy of hydration of $AB$ is $-29.4 \, J \, \text{mol}^{-1}$. What is the enthalpy of solution of the hydrated salt $AB \cdot 2H_2O_{(s)}$ in $J \, \text{mol}^{-1}$ (in $.4$)?

If the bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio of $1 : 1 : 0.5$ and $\Delta_f H$ for the formation of $XY$ is $-200 \ kJ \ mol^{-1}$,the bond dissociation energy of $X_2$ will be in $kJ \ mol^{-1}$:

The heat of solution of $Na_2SO_{4(s)}$ and $Na_2SO_4 \cdot 10H_2O_{(s)}$ are $-2.34 \ kJ \ mol^{-1}$ and $78.87 \ kJ \ mol^{-1}$ respectively. The heat of hydration of $Na_2SO_{4(s)}$ is:

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The enthalpy change at $298 \ K$ for the decomposition of water is given in the following two steps:
Step $I$: $H_2O_{(g)} \to H_{(g)} + OH_{(g)}$; $\Delta H = 498 \ kJ \ mol^{-1}$
Step $II$: $OH_{(g)} \to H_{(g)} + O_{(g)}$; $\Delta H = 428 \ kJ \ mol^{-1}$
The average bond enthalpy of the $O-H$ bond is $.... \ kJ \ mol^{-1}$

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