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Question:

The reaction between hydrogen and oxygen to yield water vapor has $\Delta{}H^\circ = -484~\mathrm{kJ}$. How much $PV$ work is done, and what is the value of $\Delta{}E$ in kilojoules for the reaction of $0.50\:\mathrm{mol}$ of $\ce{H2}$ with $0.25\:\mathrm{mol}$ of $\ce{O2}$ at atmospheric pressure if the volume change is $-5.6\:\mathrm{L}$?

$$\ce{2H_{2(g)} + O_{2(g)} -> 2H2O_{(g)}}\ \ \ \ \ \Delta{}H^\circ=-484\:\mathrm{kJ}$$

I use the formula $\Delta E=\Delta H-P\Delta V$ to determine $\Delta E$. However, when determing the enthalpy, the solutions manual does this $$\Delta H=\dfrac{-121\:\mathrm{kJ}}{0.50\:\mathrm{mol}~\ce{H2}}.$$

Where does the $-121$ come from? From my understanding, since there are two moles $\ce{H2}$, $\Delta H$ should be $-242\:\mathrm{kJ}$. Or do we take into account all four hydrogen atoms? That would give us $-121\:\mathrm{kJ}$ $\left(\dfrac{-484\:\mathrm{kJ}}{4}\right)=-121\:\mathrm{kJ}$

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  • $\begingroup$ Is ∆E or ∆H change in enthalpy? $\endgroup$ – Dissenter Aug 11 '14 at 16:58
  • $\begingroup$ $\Delta H$ is change in enthalpy $\endgroup$ – Amuna Aug 11 '14 at 17:11
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The confusion is partly caused by careless use of quantities and units. The value $H = -484\ \mathrm{kJ}$ denotes an enthalpy. However, what is actually meant is the enthalpy per one mole; i.e. the molar enthalpy $H_{\mathrm m} = -484\ \mathrm{kJ/mol}$.

In the definition of the molar reaction enthalpy, the ‘per mole’ does not refer to any particular substance in the equation. Instead it refers to the entire reaction as a whole. Therefore, the reaction must be specified for which this quantity applies. In this case, the enthalpy of $484\ \mathrm{kJ}$ is released when $2\ \mathrm{mol}$ of hydrogen gas react with $1\ \mathrm{mol}$ of oxygen gas to form $2\ \mathrm{mol}$ of gaseous water: $$\ce{2H2(g) + O2(g) -> 2H2O(g)}\qquad\Delta H^\circ = -484\ \mathrm{kJ}$$ (By way of comparison, the corresponding value for liquid water is about $-572\ \mathrm{kJ}$.)

Thus, the molar enthalpy relating to the amount of hydrogen is $$\frac{\Delta H^\circ}{n(\ce{H2})}=\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}=\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}=-242\ \mathrm{kJ/mol}$$ i.e. the released enthalpy per $1\ \mathrm{mol}$ of hydrogen gas is $242\ \mathrm{kJ}$.

However, the given question is asking about $n(\ce{H2}) = 0.50\ \mathrm{mol}$ of hydrogen (and accordingly $n(\ce{O2}) = 0.25\ \mathrm{mol}$ of oxygen). The corresponding enthalpy is $$\begin{align} 0.5\ \mathrm{mol}\times\frac{\Delta H^\circ}{n(\ce{H2})}&= 0.5\ \mathrm{mol}\times\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}\\[6pt] &= 0.5\ \mathrm{mol}\times\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}\\[6pt] &= 0.5\ \mathrm{mol}\times -242\ \mathrm{kJ/mol}\\[6pt] &= -121\ \mathrm{kJ} \end{align}$$

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